Running head: INTERNATIONAL BUSINESS LAW 1
INTERNATIONAL BUSINESS LAW 6
International business law case
Name: Abdullah Alshaya
Voest-Alpine Trading USA Corp v. Bank of China
Introduction
This paper addresses the case of Voest-Alpine Trading USA Corp v. Bank of China, case number 167F.Supp.2d 940(2000). The case deals with issues in international business transactions as it is between Voest-Alpine Trading USA Corp, an American-based company and Jiangyin Foreign Trade Corporation, a Chinese-based company. The relationship between the plaintiff (Voest-Alpine Trading USA Corp) and Jiangyin Foreign Trade Corporation emerges from a contractual agreement dated June 23, 1995. Voest-Alpine was to supply or sell 1000 metric tons of styrene monomer at total of price of $ 1.2 million. The defendant Bank of China entered the picture when JFTC applied for a letter of credit to it in order to fund the transaction with the plaintiff upon delivery to Zhangjiagang. After a series of delays and miscommunication, the Bank of China did not honor the letter of credit presented by the plaintiff. However, the court ruled the case in favor of the plaintiff Voest-Alpine after assessing the list of discrepancies presented by the Bank of China.
The companies
The companies involved in the case include Voest-Alpine Trading USA Corp, the Jiangyin Foreign Trade Corporation, the Bank of China and the Texas Commerce Bank. The Voest Alpine Trade Corporation deals in technology and capital goods. The company supplies steel and other metals to companies in the aerospace, automotive and consumer goods market. The Jiangyin Foreign Trade Corporation deals in imports and exports of products within the light industry and thus sourcing of the 1000 metric tons of styrene monomer was for the purpose of production. The Bank of China which is the defendant in the case is a large commercial bank in China owned by the government and deals in financial services. The bank’s headquarters is in Beijing, China. It was tasked with allocating payment to Voest-Alpine Trading USA Corp upon end of transaction by honoring the letter of credit applied by Jiangyin Foreign Trade Corporation. The case took root, when JFTC asked for a price concession after the significant drop in the market price of styrene monomer; however, Voest Alpine Trade USA Corp declined and instead shipped the styrene monomer on July 18 1995. The Texas Commerce bank had the role of assessing the letter of credit and documents presented by Voest –Alpine but identified some discrepancies. Even so, Voest-Alpine ordered the bank to forward the documents and letter of credit to the Bank of China. However, the Bank of China declined to pay the letter of credit and highlighted to the Texas Bank, “informing them of seven alleged discrepancies between the lette.
Fleet v. Bank of America case from California Court of AppealLegalDocsPro
This Fleet v. Bank of America case was recently decided by a California Court of Appeal. This case was decided by Division Three of the Fourth District Court of Appeal on August 25, 2014, on September 23, 2014 the Court granted the request of several parties for publication. The case involved allegations by the Fleets of fraud on the part of Bank of America during the loan modification process.
The Court of Appeal reversed the Judgment entered in the case and reversed the order sustaining the demurrer to the cause of action for fraud as to Bank of America and several other individual defendants, as well as reversing the order sustaining demurrers to the breach of contract and promissory estoppel causes of action against Bank of America athough the Court did affirm the order sustaining the demurrers without leave to amend against several other defendants including Recon Trust. The Court also affirmed the order sustaining the demurrer to the cause of action for accounting without leave to amend. This case is very good news in my opinion as this case may represent a turning point as it is the only published case from California that I am aware of in which an appeals Court appears to be at least considering the possibility that the big banks may be engaging in a pattern of fraud and deceit.
Motion to Dismiss 12 B 5 FILING Stamped-1 July 2021.pdfFrankEkejija1
Frank Ekejija and NVC Fund evidence supporting the Court filings exposed and debunked the SEC's wrongful actions and false assumptions. The facts are clear and on record.
Libor Lawsuit - In Re _ LIBOR Antitrust Litigation vs. Bank of America, JPMor...Umesh Heendeniya
LIBOR Lawsuit - In Re: LIBOR Antitrust Litigation vs. Bank of America, JPMorgan Chase, Royal Bank of Scotland, UBS AG, Barclays, Citigroup, Credit Suisse, Deutsche Bank, HSBC Holdings, WestLB AG, Royal Bank of Canada.
Asbestos Trusts and Asbestos Bankruptcies HB Asbestos Litigation Conferenc...Kirk Hartley
Neutral, fact oriented presentation identifying a wide range of information sources for persons concerned about ise of trusts to resolve tort claims, especially mass tort claiming and asbestos trusts.
Case Analysis · Post a brief case analysis of a listed problem f.docxwendolynhalbert
Case Analysis
· Post a brief case analysis of a listed problem for the week in the corresponding weeks assignment dropbox. The case assignments will be posted by professor in the Announcements each week. In a large class some students may have duplicate cases assigned to other students.
· The assignment should consist of a presentable and entertaining presentation (Power Point or other medium) and will be delivered in some form of participative medium (webex/on-site/or alternative as determined by professor) . It should include a summary of the relevant facts, the law, judicial opinion and answer the case questions. All that is necessary for an understanding of the case is important and required.
· The report must go beyond the discussion of the problem posed in the textbook, to achieve a superior grade. Do research outside the textbook- this must include research outside the case citation such as the Lexus-Nexis in the DeVry Library or FindLaw.com, do research on the parties and circumstances of the case itself and incorporate some audio-visual modality as a part of the case analysis.something about one of the parties, as well as some background contained in the legal opinion. Doing significant research outside the textbook is essential.
· Utilize the case format below.
· Your grade comes from the content contained on the actual submission.
Case Analysis Format
1. Read and understand the case or question assigned. Show your Analysis and Reasoning and make it clear you understand the material. Be sure to incorporate the concepts of the chapter we are studying to show your reasoning. Dedicate at least one heading to each following outline topic:
Parties [Identify the plaintiff and the defendant]
Facts [Summarize only those facts critical to the outcome of the case]
Procedure [Who brought the appeal? What was the outcome in the lower court(s)?]
Issue [Note the central question or questions on which the case turns]
Explain the applicable law(s). Use the textbook here. The law should come from the same chapter as the case. Be sure to use citations from the textbook including page numbers.
Holding [How did the court resolve the issue(s)? Who won?]
Reasoning [Explain the logic that supported the court's decision]
2. Do significant research outside of the book and demonstrate that you have in a very obvious way. This refers to research beyond the legal research. This involves something about the parties or other interesting related area. Show something you have discovered about the case, parties or other important element from your own research. Be sure this is obvious and adds value beyond the legal reasoning of the case.
3. Dedicate 1 slide to each of the case question(s) immediately following the case, if there are any. Be sure to state and fully answer the questions in the presentation.
4. Quality in terms of substance, form, grammar and context. Be entertaining! Use excellent audio-visual material and backgrounds!
5. Wrap up with a Conclusi ...
Ethanolv.DrizinUnited States District Court, N.D. Iowa, Eastern .docxelbanglis
Ethanolv.Drizin
United States District Court, N.D. Iowa, Eastern DivisionFeb 7, 2006
No. C03-2021 (N.D. Iowa Feb. 7, 2006) Copy Citation
No. C03-2021.
February 7, 2006
Be a better lawyer. Casetext is legal research for lawyers who want do their best work.
ORDER
JOHN JARVEY, Magistrate Judge
This matter comes before the court pursuant to trial on the merits which commenced on January 23, 2006. The above-described parties have consented to jurisdiction before a United States Magistrate Judge pursuant to 28 U.S.C. § 636(c). The court finds in favor of the plaintiff and awards compensatory damages in the amount of $3,800,000 and punitive damages in the amount of $7,600,000.
NATURE OF THE CASE
In this case, the plaintiff brings numerous theories of recovery against defendant Jerry Drizin arising out of the misappropriation of escrow funds that were to serve as security for financing for the construction of an ethanol plant in Manchester, Iowa. The plaintiff contends that defendant Drizin, in concert with others, knowingly converted funds from an escrow account that were not to have been spent on anything without the plaintiff's prior written permission. Defendant Drizin contends that his only client and only duty of loyalty was to a Nigerian citizen living in Munich who caused the funds to be sent to bank accounts controlled by Defendant Drizin. The court makes the following findings of fact and conclusions of law.
FINDINGS OF FACT
In 2000 in Manchester, Iowa, farmer and President of the local Co-op, Douglas Bishop, began meeting with representatives of the United States Department of Agriculture to explore the feasibility of building an ethanol plant in the Manchester area. The idea was to assist farmers in the area in getting more value for their crops. An ethanol plant produces ethanol and feed grain which can be sold at a profit exceeding that associated with the mere sale of grain.
A series of 40 local meetings culminated in a membership drive. The Plaintiff, Northeast Iowa Ethanol, L.L.C., was later formed in order to sell 2500 shares of stock in the L.L.C. to raise funds for the financing of the plant. The construction of the plant was expected to cost $21 Million. It would have a capacity for producing 15 million gallons of ethanol per year. Through the meetings, Mr. Bishop and others raised $2,365,000. The average investor purchased two shares.
The membership drive ended in September 2001. The original plan was to begin construction in the fall of 2001 and have the plant operating by the fall of 2002. However, the issue of financing for the plant was more problematic than plaintiff had anticipated. Traditional lenders (banks) demanded that the plaintiff raise forty percent of the construction costs. It was clear that the plaintiff could not raise $8 Million. Plaintiff's proposed marketing partner, Williams Ethanol Services, agreed to invest $1 Million in the project. The contractor anticipated to build the facility, North Centr ...
Fleet v. Bank of America case from California Court of AppealLegalDocsPro
This Fleet v. Bank of America case was recently decided by a California Court of Appeal. This case was decided by Division Three of the Fourth District Court of Appeal on August 25, 2014, on September 23, 2014 the Court granted the request of several parties for publication. The case involved allegations by the Fleets of fraud on the part of Bank of America during the loan modification process.
The Court of Appeal reversed the Judgment entered in the case and reversed the order sustaining the demurrer to the cause of action for fraud as to Bank of America and several other individual defendants, as well as reversing the order sustaining demurrers to the breach of contract and promissory estoppel causes of action against Bank of America athough the Court did affirm the order sustaining the demurrers without leave to amend against several other defendants including Recon Trust. The Court also affirmed the order sustaining the demurrer to the cause of action for accounting without leave to amend. This case is very good news in my opinion as this case may represent a turning point as it is the only published case from California that I am aware of in which an appeals Court appears to be at least considering the possibility that the big banks may be engaging in a pattern of fraud and deceit.
Motion to Dismiss 12 B 5 FILING Stamped-1 July 2021.pdfFrankEkejija1
Frank Ekejija and NVC Fund evidence supporting the Court filings exposed and debunked the SEC's wrongful actions and false assumptions. The facts are clear and on record.
Libor Lawsuit - In Re _ LIBOR Antitrust Litigation vs. Bank of America, JPMor...Umesh Heendeniya
LIBOR Lawsuit - In Re: LIBOR Antitrust Litigation vs. Bank of America, JPMorgan Chase, Royal Bank of Scotland, UBS AG, Barclays, Citigroup, Credit Suisse, Deutsche Bank, HSBC Holdings, WestLB AG, Royal Bank of Canada.
Asbestos Trusts and Asbestos Bankruptcies HB Asbestos Litigation Conferenc...Kirk Hartley
Neutral, fact oriented presentation identifying a wide range of information sources for persons concerned about ise of trusts to resolve tort claims, especially mass tort claiming and asbestos trusts.
Case Analysis · Post a brief case analysis of a listed problem f.docxwendolynhalbert
Case Analysis
· Post a brief case analysis of a listed problem for the week in the corresponding weeks assignment dropbox. The case assignments will be posted by professor in the Announcements each week. In a large class some students may have duplicate cases assigned to other students.
· The assignment should consist of a presentable and entertaining presentation (Power Point or other medium) and will be delivered in some form of participative medium (webex/on-site/or alternative as determined by professor) . It should include a summary of the relevant facts, the law, judicial opinion and answer the case questions. All that is necessary for an understanding of the case is important and required.
· The report must go beyond the discussion of the problem posed in the textbook, to achieve a superior grade. Do research outside the textbook- this must include research outside the case citation such as the Lexus-Nexis in the DeVry Library or FindLaw.com, do research on the parties and circumstances of the case itself and incorporate some audio-visual modality as a part of the case analysis.something about one of the parties, as well as some background contained in the legal opinion. Doing significant research outside the textbook is essential.
· Utilize the case format below.
· Your grade comes from the content contained on the actual submission.
Case Analysis Format
1. Read and understand the case or question assigned. Show your Analysis and Reasoning and make it clear you understand the material. Be sure to incorporate the concepts of the chapter we are studying to show your reasoning. Dedicate at least one heading to each following outline topic:
Parties [Identify the plaintiff and the defendant]
Facts [Summarize only those facts critical to the outcome of the case]
Procedure [Who brought the appeal? What was the outcome in the lower court(s)?]
Issue [Note the central question or questions on which the case turns]
Explain the applicable law(s). Use the textbook here. The law should come from the same chapter as the case. Be sure to use citations from the textbook including page numbers.
Holding [How did the court resolve the issue(s)? Who won?]
Reasoning [Explain the logic that supported the court's decision]
2. Do significant research outside of the book and demonstrate that you have in a very obvious way. This refers to research beyond the legal research. This involves something about the parties or other interesting related area. Show something you have discovered about the case, parties or other important element from your own research. Be sure this is obvious and adds value beyond the legal reasoning of the case.
3. Dedicate 1 slide to each of the case question(s) immediately following the case, if there are any. Be sure to state and fully answer the questions in the presentation.
4. Quality in terms of substance, form, grammar and context. Be entertaining! Use excellent audio-visual material and backgrounds!
5. Wrap up with a Conclusi ...
Ethanolv.DrizinUnited States District Court, N.D. Iowa, Eastern .docxelbanglis
Ethanolv.Drizin
United States District Court, N.D. Iowa, Eastern DivisionFeb 7, 2006
No. C03-2021 (N.D. Iowa Feb. 7, 2006) Copy Citation
No. C03-2021.
February 7, 2006
Be a better lawyer. Casetext is legal research for lawyers who want do their best work.
ORDER
JOHN JARVEY, Magistrate Judge
This matter comes before the court pursuant to trial on the merits which commenced on January 23, 2006. The above-described parties have consented to jurisdiction before a United States Magistrate Judge pursuant to 28 U.S.C. § 636(c). The court finds in favor of the plaintiff and awards compensatory damages in the amount of $3,800,000 and punitive damages in the amount of $7,600,000.
NATURE OF THE CASE
In this case, the plaintiff brings numerous theories of recovery against defendant Jerry Drizin arising out of the misappropriation of escrow funds that were to serve as security for financing for the construction of an ethanol plant in Manchester, Iowa. The plaintiff contends that defendant Drizin, in concert with others, knowingly converted funds from an escrow account that were not to have been spent on anything without the plaintiff's prior written permission. Defendant Drizin contends that his only client and only duty of loyalty was to a Nigerian citizen living in Munich who caused the funds to be sent to bank accounts controlled by Defendant Drizin. The court makes the following findings of fact and conclusions of law.
FINDINGS OF FACT
In 2000 in Manchester, Iowa, farmer and President of the local Co-op, Douglas Bishop, began meeting with representatives of the United States Department of Agriculture to explore the feasibility of building an ethanol plant in the Manchester area. The idea was to assist farmers in the area in getting more value for their crops. An ethanol plant produces ethanol and feed grain which can be sold at a profit exceeding that associated with the mere sale of grain.
A series of 40 local meetings culminated in a membership drive. The Plaintiff, Northeast Iowa Ethanol, L.L.C., was later formed in order to sell 2500 shares of stock in the L.L.C. to raise funds for the financing of the plant. The construction of the plant was expected to cost $21 Million. It would have a capacity for producing 15 million gallons of ethanol per year. Through the meetings, Mr. Bishop and others raised $2,365,000. The average investor purchased two shares.
The membership drive ended in September 2001. The original plan was to begin construction in the fall of 2001 and have the plant operating by the fall of 2002. However, the issue of financing for the plant was more problematic than plaintiff had anticipated. Traditional lenders (banks) demanded that the plaintiff raise forty percent of the construction costs. It was clear that the plaintiff could not raise $8 Million. Plaintiff's proposed marketing partner, Williams Ethanol Services, agreed to invest $1 Million in the project. The contractor anticipated to build the facility, North Centr ...
Job_Profit_LossFancy Yachts: Job Profit / (Loss)LC:Hull No.70-07Broker Info:Customer Name:Third Party BuyerAddressCity, State, ZipPhoneCommission:Contract Date:11/25/11Sales PriceBase Boat:$ 1,450,000.00DateAmountOptions:Deposits:11/28/11$ 145,000.00Duty Fees:4/22/12$ 36,250.00Discounts:$ - 05/22/12$ 1,238,750.00Adjustments:5/22/12$ 30,000.00Total Sales Price:$ 1,450,000.00Less: Factory Invoice$ 1,198,140.00Gross Profit:$ 251,860.00Total Deposit:$ 1,450,000.00Less: Vend/Jost Cost$ 483,262.84Total Sales Price$ 1,450,000.00Less: Freight$ 70,000.00Less Payments:$ (1,450,000.00)Net Profit/ (Loss)$ (301,402.84)A/R$ - 0Total Cost$ 1,751,402.84Shipped Date:Arrival Date:Sales / Title Transfer12/1/12
CashReceipt_Vistas, LLCDateLine DescriptionAmount8/5/08Initial deposit163,000.005/7/09 progress deposit100,000.006/1/09 progress deposit63,000.0011/5/09 progress deposit25,000.0011/6/09 progress deposit25,000.0011/12/09 progress deposit15,000.0011/13/09 progress deposit30,000.0011/25/09 progress deposit20,000.0011/27/09 progress deposit10,000.003/16/10reimbursement for duty/customs for importation of Davits314.66451,314.66Deposits from 1st buyer
Fancy Yachts
Cash Receipts Report
Default Buyer
CashReceipts_NewBuyerDateLine DescriptionAmount11/28/111st deposit 70-07145,000.004/22/12Payment direct to Broker commission36,250.005/22/12Deposit made directly to pay down LC1,238,750.005/22/12Prepaid Repair Account for warranty 3 yrs.30,000.001,450,000.00Deposits from 2nd buyer
Fancy Yachts
Cash Receipts
Sergi
Fancy Yachts - Case Study
On August 1, 2008, Fancy Yachts entered into a contract with Vistas, LLC to deliver a 70 foot yacht in exchange for consideration of $1,600,000. This boat was given the hull number 70-07. Fancy Yachts received a deposit of $163,000 with the signed contract. The yacht was manufactured and shipped from the Taiwanese factory on October 6, 2009 and arrived in the United States (Port of Everglades, Miami) on October 29, 2009. The buyer had 72 hours to inspect the boat and make arrangements for the final payment (less a small hold-out to be paid upon title transfer) due on October 31, 2009.
On the required date of final payment, the customer indicated that he was having difficulty with financing and asked for an extension of time. The owners of Fancy Yachts agreed. It was not uncommon for a customer to need additional time; a week or so to liquidate investments or transfer funds. They also had no other choice given the large sum of money involved!
Keep in mind, that this contract was entered into a month before the September, 2008 financial collapse. Even a year later, when the final balance was due, the economy had not yet recovered. The once available credit markets had all dried up. According to the principal of Vistas, LLC, his net worth decreased from $8 million to $2 million dollars. It was not a good time to liquidate investments and banks were not lending. Especially for luxury item ...
Ethanolv.DrizinUnited States District Court, N.D. Iowa, Eastern .docxdebishakespeare
Ethanolv.Drizin
United States District Court, N.D. Iowa, Eastern DivisionFeb 7, 2006
No. C03-2021 (N.D. Iowa Feb. 7, 2006) Copy Citation
No. C03-2021.
February 7, 2006
Be a better lawyer. Casetext is legal research for lawyers who want do their best work.
ORDER
JOHN JARVEY, Magistrate Judge
This matter comes before the court pursuant to trial on the merits which commenced on January 23, 2006. The above-described parties have consented to jurisdiction before a United States Magistrate Judge pursuant to 28 U.S.C. § 636(c). The court finds in favor of the plaintiff and awards compensatory damages in the amount of $3,800,000 and punitive damages in the amount of $7,600,000.
In this case, the plaintiff brings numerous theories of recovery against defendant Jerry Drizin arising out of the misappropriation of escrow funds that were to serve as security for financing for the construction of an ethanol plant in Manchester, Iowa. The plaintiff contends that defendant Drizin, in concert with others, knowingly converted funds from an escrow account that were not to have been spent on anything without the plaintiff's prior written permission. Defendant Drizin contends that his only client and only duty of loyalty was to a Nigerian citizen living in Munich who caused the funds to be sent to bank accounts controlled by Defendant Drizin. The court makes the following findings of fact and conclusions of law.
In 2000 in Manchester, Iowa, farmer and President of the local Co-op, Douglas Bishop, began meeting with representatives of the United States Department of Agriculture to explore the feasibility of building an ethanol plant in the Manchester area. The idea was to assist farmers in the area in getting more value for their crops. An ethanol plant produces ethanol and feed grain which can be sold at a profit exceeding that associated with the mere sale of grain.
A series of 40 local meetings culminated in a membership drive. The Plaintiff, Northeast Iowa Ethanol, L.L.C., was later formed in order to sell 2500 shares of stock in the L.L.C. to raise funds for the financing of the plant. The construction of the plant was expected to cost $21 Million. It would have a capacity for producing 15 million gallons of ethanol per year. Through the meetings, Mr. Bishop and others raised $2,365,000. The average investor purchased two shares.
The membership drive ended in September 2001. The original plan was to begin construction in the fall of 2001 and have the plant operating by the fall of 2002. However, the issue of financing for the plant was more problematic than plaintiff had anticipated. Traditional lenders (banks) demanded that the plaintiff raise forty percent of the construction costs. It was clear that the plaintiff could not raise $8 Million. Plaintiff's proposed marketing partner, Williams Ethanol Services, agreed to invest $1 Million in the project. The contractor anticipated to build the facility, North Central Construction from North Dakota,.
CASE BRIEF 7.2 Tiffany and Company v. Andrew 2012 W.docxdewhirstichabod
CASE BRIEF 7.2
Tiffany and Company v. Andrew
2012 WL 5451259 (S.D.N.Y.)
FACTS: Tiffany (plaintiffs) allege that Andrew and others (defendants) sold counterfeit Tiffany
products through several websites hosted in the United States. Andrew accepted payment in U.S.
dollars, used PayPal, Inc. to process customers' credit card transactions, then transferred the sales
proceeds to accounts held by the Bank of China (“BOC”), Industrial and Commercial Bank of
China (“ICBC”), and China Merchants Bank (“CMB”) (“Banks”).
Andrew defaulted on the suit, and Tiffany sought discovery from the Banks by serving subpoenas
seeking the identities of the holders of the accounts into which the proceeds of the counterfeit sales
were transferred and the subsequent disposition of those proceeds. The Banks involved all
maintained branch offices in the Southern District of New York, and the subpoenas were served
on those branch offices.
The Banks responded to the subpoenas by explaining that the information sought was all
maintained in China and that the New York branches of the Banks lacked the ability to access the
requested information. China's internal laws prohibited the disclosure of the information except
under certain conditions. The Banks proposed that the plaintiffs pursue the requested discovery
pursuant to the Hague Convention.
The court concluded that Tiffany should pursue discovery through the Hague Convention. Tiffany
submitted its Hague Convention application to China's Central Authority in November 2010, and
on August 7, 2011, the Ministry of Justice of the People's Republic of China (“MOJ”) responded
by producing some of the documents requested. For each of the Banks, the MOJ produced account
opening documents (including the government identification card of the account holder), written
confirmation of certain transfers into the accounts and a list of transfers out of the accounts. With
respect to CMB, the records indicate that all funds in the account were withdrawn through cash
transactions at either an ATM or through a teller. BOC and CMB each produced documents
concerning a single account; ICBC produced documents for three accounts.
In its cover letter, the MOJ noted that it was not producing all documents requested. Specifically,
the letter stated, “Concerning your request for taking of evidence for the Tiffany case, the Chinese
competent authority holds that some evidence required lacks direct and close connections with the
litigation. As the Chinese government has declared at its accession to the Hague Evidence
Convention that for the request issued for the purpose of the pre-trial discovery of documents only
the request for obtaining discovery of the documents clearly enumerated in the Letters of Request
and of direct and close connection with the subject matter of the litigation will be executed, the
Chinese competent authority has partly executed the requests which it d.
353 F.3d 668 STANDARD CONCRETE PRODUCTS INC., Plaintiff-Ap.docxtamicawaysmith
353 F.3d 668
STANDARD CONCRETE PRODUCTS INC., Plaintiff-Appellee,
v.
GENERAL TRUCK DRIVERS, OFFICE, FOOD AND WAREHOUSE UNION, LOCAL 952,
Defendant-Appellant.
Standard Concrete Products Inc., Plaintiff-Appellant,
v.
General Truck Drivers, Office, Food and Warehouse Union, Local 952, Defendant-Appellee.
No. 01-57256.
No. 01-57257.
United States Court of Appeals, Ninth Circuit.
Argued and Submitted April 8, 2003 — Pasadena, California.
Filed December 18, 2003.
Howard C. Hay, (argued and brief) and Glenn L. Briggs (brief), Paul, Hastings, Janofsky &
Walker, LLP, Costa Mesa, California, CA, for the plaintiff/appellee/appellant.
Appeals from the United States District Court for the Central District of California; Alicemarie
H. Stotler, District Judge, Presiding. D.C. No. CV-00-00016-AHS.
Before Harry PREGERSON, A. Wallace TASHIMA, and Richard R. CLIFTON, Circuit Judges.
OPINION
PREGERSON, Circuit Judge:
1
Plaintiff Standard Concrete Products ("Standard Concrete") delivers concrete throughout
Southern California. Relevant to this appeal, Standard Concrete has facilities in Riverside
County and in Orange County. The International Brotherhood of the Teamsters, General Truck
Drivers, Office, Food & Warehouse Union, Local 952 represents Standard Concrete's employees
at its Corona facility in Riverside County ("Corona bargaining unit"). Local 952 also represents
Standard Concrete's employees at its three Orange County facilities ("Orange County bargaining
unit") under a separate collective bargaining agreement.
2
In January 2000, the Corona bargaining unit went on strike against Standard Concrete. The strike
was called because the Corona bargaining unit believed that Standard Concrete was negotiating
with Local 952 in bad faith. Members of the Corona bargaining unit established picket lines at
the Corona facility. On the second day of the strike, the Corona bargaining unit extended its
picket lines to Standard Concrete's three facilities in Orange County. Members of the Orange
County bargaining unit honored the Corona bargaining unit's picket lines.
3
At issue in this case is whether the Orange County bargaining unit violated its Collective
Bargaining Agreement ("CBA") with Standard Concrete when members of the Orange County
bargaining unit honored the Corona bargaining unit's picket lines at the Standard Concrete
facilities in Orange County. On January 6, 2000, Standard Concrete filed a complaint against
Local 952 in the United States District Court for the Central District of California. The complaint
alleged that Local 952 breached the no-strike clause in the Orange County CBA by participating
in and encouraging the Orange County bargaining unit members to honor the Corona bargaining
unit's picket lines. On April 27, 2000, Local 952 filed a motion to dismiss the complaint, arguing
that Standard Concrete violated the Orange County CBA by failing to submit the d ...
353 F.3d 668 STANDARD CONCRETE PRODUCTS INC., Plaintiff-Ap.docxgilbertkpeters11344
353 F.3d 668
STANDARD CONCRETE PRODUCTS INC., Plaintiff-Appellee,
v.
GENERAL TRUCK DRIVERS, OFFICE, FOOD AND WAREHOUSE UNION, LOCAL 952,
Defendant-Appellant.
Standard Concrete Products Inc., Plaintiff-Appellant,
v.
General Truck Drivers, Office, Food and Warehouse Union, Local 952, Defendant-Appellee.
No. 01-57256.
No. 01-57257.
United States Court of Appeals, Ninth Circuit.
Argued and Submitted April 8, 2003 — Pasadena, California.
Filed December 18, 2003.
Howard C. Hay, (argued and brief) and Glenn L. Briggs (brief), Paul, Hastings, Janofsky &
Walker, LLP, Costa Mesa, California, CA, for the plaintiff/appellee/appellant.
Appeals from the United States District Court for the Central District of California; Alicemarie
H. Stotler, District Judge, Presiding. D.C. No. CV-00-00016-AHS.
Before Harry PREGERSON, A. Wallace TASHIMA, and Richard R. CLIFTON, Circuit Judges.
OPINION
PREGERSON, Circuit Judge:
1
Plaintiff Standard Concrete Products ("Standard Concrete") delivers concrete throughout
Southern California. Relevant to this appeal, Standard Concrete has facilities in Riverside
County and in Orange County. The International Brotherhood of the Teamsters, General Truck
Drivers, Office, Food & Warehouse Union, Local 952 represents Standard Concrete's employees
at its Corona facility in Riverside County ("Corona bargaining unit"). Local 952 also represents
Standard Concrete's employees at its three Orange County facilities ("Orange County bargaining
unit") under a separate collective bargaining agreement.
2
In January 2000, the Corona bargaining unit went on strike against Standard Concrete. The strike
was called because the Corona bargaining unit believed that Standard Concrete was negotiating
with Local 952 in bad faith. Members of the Corona bargaining unit established picket lines at
the Corona facility. On the second day of the strike, the Corona bargaining unit extended its
picket lines to Standard Concrete's three facilities in Orange County. Members of the Orange
County bargaining unit honored the Corona bargaining unit's picket lines.
3
At issue in this case is whether the Orange County bargaining unit violated its Collective
Bargaining Agreement ("CBA") with Standard Concrete when members of the Orange County
bargaining unit honored the Corona bargaining unit's picket lines at the Standard Concrete
facilities in Orange County. On January 6, 2000, Standard Concrete filed a complaint against
Local 952 in the United States District Court for the Central District of California. The complaint
alleged that Local 952 breached the no-strike clause in the Orange County CBA by participating
in and encouraging the Orange County bargaining unit members to honor the Corona bargaining
unit's picket lines. On April 27, 2000, Local 952 filed a motion to dismiss the complaint, arguing
that Standard Concrete violated the Orange County CBA by failing to submit the d.
All product and company names mentioned herein are for identification and educational purposes only and are the property of, and may be trademarks of, their respective owners.
Running head MARKETING ANALYSIS ASSIGNMENTS .docxwlynn1
Running head: MARKETING ANALYSIS ASSIGNMENTS 1
MARKETING ANALYSIS ASSIGNMENTS 6
Researching Marketing Questions
MKT/571
Melissa Simmons
Roberto Ancis
Part 1: Memorandum
TO: Senior Vice President (Marketing)
FROM: Jacob Glenns
DATE: August 19, 2018
SUBJECT: Marketing Analysis
Summary Analysis
This analysis of the market report that was presented the market analyst provides detailed insights from the data that may help in formulating an effective marketing strategy. The key information include: revenue performance for the first half between 2015 and 2016 and revenue trends over the same period. This information help in deciding whether to the organization should continue with its growth strategy or to reverse the decline.
Revenue Analysis
Analysis of the company’s semiannual performance- between January and June- indicates that there was an increase of 10.18 percent in the generated revenues per day from 96,000 dollars to 105,768 dollars in 2015 and 2016 respectively. The revenues per day, domestic market, were 93,683 dollars and 85,181 dollars in 2016 and 2015 respectively, over the same period. Overall, the semiannual revenue for the year 2016 was 13,644,073 dollars with the United States market contributing 12,085,137 dollars, which is approximately 88.6 percent of the semiannual revenue. The international market contributed 1,558,936 dollars, which is 11.4 percent of the total revenue. The average gross profit per day was 8.3 percent for the six months between January and June, 2016. For the three months of April, May and June, 2016 the total revenue was 7,024,096 dollars with the domestic market contributing 6,145,978 dollars and the international market contributing 878,119 dollars. The gross profit was 6.5 percent.
Revenue Trends
With regards to customer class, commercial customers contributed 7,195,592 dollars in the six months of January to June, 2016. The revenue per day was 55,780 dollars, an increase of 5,008 dollars compared to 50,772 dollars realized over the same period in 2015. At the second place was the municipal segment with 1,634,643 dollars. The revenue per day for the first six months was 12,672 in 2016 compared to 12,034 in 2015. The international market segment contributed 1,535,905 dollars and the revenue per day was 11,906 dollars and 11,700 dollars in 2016 and 2015 respectively. The other important segments- resellers, industrial labs, government, resell, education and others- also registered increments in the revenue per day for the first 6 months between 2015 and 2016. The revenue trend for the second quarter (between April and May) illustrate that commercial market contributed 1,130,973 dollars which is 50 percent of the total revenue from the customer class segment. The international market contributed 323,990 follo.
Running head MANAGING A DIVERSE WORKFORCE1MANAGING A DIVERSE.docxwlynn1
Running head: MANAGING A DIVERSE WORKFORCE 1
MANAGING A DIVERSE WORKFORCE 6
Managing a diverse workforce
Name
Institutional affiliation
What does it mean to be an effective manager in a diverse workforce?
According to Chip Conley, the workforce diversity is characterized of gender, ethnicity and age; which needs a much keener attention. He points out that an effective manager should realize that age diversity makes a company stronger and that different generations within a workplace should focus on mentoring one another at work. He emphasizes on the need to allow openness with one another so that wisdom; knowledge, experience and skills from the young to the old and vice versa. According to Chip Conley, the current 60s is the new 40s and that the current 30s is the new 50s; a key note to take on how effective relationship in a workplace could enrichen a company with greater shared wisdom and skills. Every manager need to relate such knowledge in ensuring effective making of modern elders from the millennials.
According to Chip, an effective manager should establish a learning environment for the boomers and the millennials. Each generation should see the other as assets from which they can derive wisdom. Moreover, Chip calls for both the millennials and the boomers to fix their ego, perhaps so that they can enhance their relationship and get to learn from one another. He calls for the need of the managers to enhance a growth mindset in a workplace and the need for the employees to be curious of getting to know what the other generation can offer, and trying to oneself. Chip states that “Curiosity is the elixir for life”
Working on the psychological empowerment of specifics groups and ensuring mental flexibility is very important for various generations to work coherently effectively. Additionally, a manager in charge of a diverse workforce should ensure that the differences existing between the BB and X generations, and the Y and Z generations should be harmonized so that they do not tamper with the achievement of the organizations set goals and objectives (Toro, Labrador-Fernández & De Nicolas, 2019).
Maintaining a positive working environment helps in enhancing the performance of a diverse workforce. Looking at the small business managers, workforce diversity can be well managed if the owner’s manager supports the existing generational interconnections and the variations as a result of the general difference defining these groups by valuing their differences and the similarities. An effective manager is therefore required to cause a diversity openness among the workforce. Such ensure the performance at all levels, i.e. both the organizational and individual. A manager should, therefore, have the ability to effectively enforce the eradication of the internal communication barriers existing as a result generational, racial, gender, ethnic, age, personality tenure, cognitive style, education among other dissimilarities .
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Job_Profit_LossFancy Yachts: Job Profit / (Loss)LC:Hull No.70-07Broker Info:Customer Name:Third Party BuyerAddressCity, State, ZipPhoneCommission:Contract Date:11/25/11Sales PriceBase Boat:$ 1,450,000.00DateAmountOptions:Deposits:11/28/11$ 145,000.00Duty Fees:4/22/12$ 36,250.00Discounts:$ - 05/22/12$ 1,238,750.00Adjustments:5/22/12$ 30,000.00Total Sales Price:$ 1,450,000.00Less: Factory Invoice$ 1,198,140.00Gross Profit:$ 251,860.00Total Deposit:$ 1,450,000.00Less: Vend/Jost Cost$ 483,262.84Total Sales Price$ 1,450,000.00Less: Freight$ 70,000.00Less Payments:$ (1,450,000.00)Net Profit/ (Loss)$ (301,402.84)A/R$ - 0Total Cost$ 1,751,402.84Shipped Date:Arrival Date:Sales / Title Transfer12/1/12
CashReceipt_Vistas, LLCDateLine DescriptionAmount8/5/08Initial deposit163,000.005/7/09 progress deposit100,000.006/1/09 progress deposit63,000.0011/5/09 progress deposit25,000.0011/6/09 progress deposit25,000.0011/12/09 progress deposit15,000.0011/13/09 progress deposit30,000.0011/25/09 progress deposit20,000.0011/27/09 progress deposit10,000.003/16/10reimbursement for duty/customs for importation of Davits314.66451,314.66Deposits from 1st buyer
Fancy Yachts
Cash Receipts Report
Default Buyer
CashReceipts_NewBuyerDateLine DescriptionAmount11/28/111st deposit 70-07145,000.004/22/12Payment direct to Broker commission36,250.005/22/12Deposit made directly to pay down LC1,238,750.005/22/12Prepaid Repair Account for warranty 3 yrs.30,000.001,450,000.00Deposits from 2nd buyer
Fancy Yachts
Cash Receipts
Sergi
Fancy Yachts - Case Study
On August 1, 2008, Fancy Yachts entered into a contract with Vistas, LLC to deliver a 70 foot yacht in exchange for consideration of $1,600,000. This boat was given the hull number 70-07. Fancy Yachts received a deposit of $163,000 with the signed contract. The yacht was manufactured and shipped from the Taiwanese factory on October 6, 2009 and arrived in the United States (Port of Everglades, Miami) on October 29, 2009. The buyer had 72 hours to inspect the boat and make arrangements for the final payment (less a small hold-out to be paid upon title transfer) due on October 31, 2009.
On the required date of final payment, the customer indicated that he was having difficulty with financing and asked for an extension of time. The owners of Fancy Yachts agreed. It was not uncommon for a customer to need additional time; a week or so to liquidate investments or transfer funds. They also had no other choice given the large sum of money involved!
Keep in mind, that this contract was entered into a month before the September, 2008 financial collapse. Even a year later, when the final balance was due, the economy had not yet recovered. The once available credit markets had all dried up. According to the principal of Vistas, LLC, his net worth decreased from $8 million to $2 million dollars. It was not a good time to liquidate investments and banks were not lending. Especially for luxury item ...
Ethanolv.DrizinUnited States District Court, N.D. Iowa, Eastern .docxdebishakespeare
Ethanolv.Drizin
United States District Court, N.D. Iowa, Eastern DivisionFeb 7, 2006
No. C03-2021 (N.D. Iowa Feb. 7, 2006) Copy Citation
No. C03-2021.
February 7, 2006
Be a better lawyer. Casetext is legal research for lawyers who want do their best work.
ORDER
JOHN JARVEY, Magistrate Judge
This matter comes before the court pursuant to trial on the merits which commenced on January 23, 2006. The above-described parties have consented to jurisdiction before a United States Magistrate Judge pursuant to 28 U.S.C. § 636(c). The court finds in favor of the plaintiff and awards compensatory damages in the amount of $3,800,000 and punitive damages in the amount of $7,600,000.
In this case, the plaintiff brings numerous theories of recovery against defendant Jerry Drizin arising out of the misappropriation of escrow funds that were to serve as security for financing for the construction of an ethanol plant in Manchester, Iowa. The plaintiff contends that defendant Drizin, in concert with others, knowingly converted funds from an escrow account that were not to have been spent on anything without the plaintiff's prior written permission. Defendant Drizin contends that his only client and only duty of loyalty was to a Nigerian citizen living in Munich who caused the funds to be sent to bank accounts controlled by Defendant Drizin. The court makes the following findings of fact and conclusions of law.
In 2000 in Manchester, Iowa, farmer and President of the local Co-op, Douglas Bishop, began meeting with representatives of the United States Department of Agriculture to explore the feasibility of building an ethanol plant in the Manchester area. The idea was to assist farmers in the area in getting more value for their crops. An ethanol plant produces ethanol and feed grain which can be sold at a profit exceeding that associated with the mere sale of grain.
A series of 40 local meetings culminated in a membership drive. The Plaintiff, Northeast Iowa Ethanol, L.L.C., was later formed in order to sell 2500 shares of stock in the L.L.C. to raise funds for the financing of the plant. The construction of the plant was expected to cost $21 Million. It would have a capacity for producing 15 million gallons of ethanol per year. Through the meetings, Mr. Bishop and others raised $2,365,000. The average investor purchased two shares.
The membership drive ended in September 2001. The original plan was to begin construction in the fall of 2001 and have the plant operating by the fall of 2002. However, the issue of financing for the plant was more problematic than plaintiff had anticipated. Traditional lenders (banks) demanded that the plaintiff raise forty percent of the construction costs. It was clear that the plaintiff could not raise $8 Million. Plaintiff's proposed marketing partner, Williams Ethanol Services, agreed to invest $1 Million in the project. The contractor anticipated to build the facility, North Central Construction from North Dakota,.
CASE BRIEF 7.2 Tiffany and Company v. Andrew 2012 W.docxdewhirstichabod
CASE BRIEF 7.2
Tiffany and Company v. Andrew
2012 WL 5451259 (S.D.N.Y.)
FACTS: Tiffany (plaintiffs) allege that Andrew and others (defendants) sold counterfeit Tiffany
products through several websites hosted in the United States. Andrew accepted payment in U.S.
dollars, used PayPal, Inc. to process customers' credit card transactions, then transferred the sales
proceeds to accounts held by the Bank of China (“BOC”), Industrial and Commercial Bank of
China (“ICBC”), and China Merchants Bank (“CMB”) (“Banks”).
Andrew defaulted on the suit, and Tiffany sought discovery from the Banks by serving subpoenas
seeking the identities of the holders of the accounts into which the proceeds of the counterfeit sales
were transferred and the subsequent disposition of those proceeds. The Banks involved all
maintained branch offices in the Southern District of New York, and the subpoenas were served
on those branch offices.
The Banks responded to the subpoenas by explaining that the information sought was all
maintained in China and that the New York branches of the Banks lacked the ability to access the
requested information. China's internal laws prohibited the disclosure of the information except
under certain conditions. The Banks proposed that the plaintiffs pursue the requested discovery
pursuant to the Hague Convention.
The court concluded that Tiffany should pursue discovery through the Hague Convention. Tiffany
submitted its Hague Convention application to China's Central Authority in November 2010, and
on August 7, 2011, the Ministry of Justice of the People's Republic of China (“MOJ”) responded
by producing some of the documents requested. For each of the Banks, the MOJ produced account
opening documents (including the government identification card of the account holder), written
confirmation of certain transfers into the accounts and a list of transfers out of the accounts. With
respect to CMB, the records indicate that all funds in the account were withdrawn through cash
transactions at either an ATM or through a teller. BOC and CMB each produced documents
concerning a single account; ICBC produced documents for three accounts.
In its cover letter, the MOJ noted that it was not producing all documents requested. Specifically,
the letter stated, “Concerning your request for taking of evidence for the Tiffany case, the Chinese
competent authority holds that some evidence required lacks direct and close connections with the
litigation. As the Chinese government has declared at its accession to the Hague Evidence
Convention that for the request issued for the purpose of the pre-trial discovery of documents only
the request for obtaining discovery of the documents clearly enumerated in the Letters of Request
and of direct and close connection with the subject matter of the litigation will be executed, the
Chinese competent authority has partly executed the requests which it d.
353 F.3d 668 STANDARD CONCRETE PRODUCTS INC., Plaintiff-Ap.docxtamicawaysmith
353 F.3d 668
STANDARD CONCRETE PRODUCTS INC., Plaintiff-Appellee,
v.
GENERAL TRUCK DRIVERS, OFFICE, FOOD AND WAREHOUSE UNION, LOCAL 952,
Defendant-Appellant.
Standard Concrete Products Inc., Plaintiff-Appellant,
v.
General Truck Drivers, Office, Food and Warehouse Union, Local 952, Defendant-Appellee.
No. 01-57256.
No. 01-57257.
United States Court of Appeals, Ninth Circuit.
Argued and Submitted April 8, 2003 — Pasadena, California.
Filed December 18, 2003.
Howard C. Hay, (argued and brief) and Glenn L. Briggs (brief), Paul, Hastings, Janofsky &
Walker, LLP, Costa Mesa, California, CA, for the plaintiff/appellee/appellant.
Appeals from the United States District Court for the Central District of California; Alicemarie
H. Stotler, District Judge, Presiding. D.C. No. CV-00-00016-AHS.
Before Harry PREGERSON, A. Wallace TASHIMA, and Richard R. CLIFTON, Circuit Judges.
OPINION
PREGERSON, Circuit Judge:
1
Plaintiff Standard Concrete Products ("Standard Concrete") delivers concrete throughout
Southern California. Relevant to this appeal, Standard Concrete has facilities in Riverside
County and in Orange County. The International Brotherhood of the Teamsters, General Truck
Drivers, Office, Food & Warehouse Union, Local 952 represents Standard Concrete's employees
at its Corona facility in Riverside County ("Corona bargaining unit"). Local 952 also represents
Standard Concrete's employees at its three Orange County facilities ("Orange County bargaining
unit") under a separate collective bargaining agreement.
2
In January 2000, the Corona bargaining unit went on strike against Standard Concrete. The strike
was called because the Corona bargaining unit believed that Standard Concrete was negotiating
with Local 952 in bad faith. Members of the Corona bargaining unit established picket lines at
the Corona facility. On the second day of the strike, the Corona bargaining unit extended its
picket lines to Standard Concrete's three facilities in Orange County. Members of the Orange
County bargaining unit honored the Corona bargaining unit's picket lines.
3
At issue in this case is whether the Orange County bargaining unit violated its Collective
Bargaining Agreement ("CBA") with Standard Concrete when members of the Orange County
bargaining unit honored the Corona bargaining unit's picket lines at the Standard Concrete
facilities in Orange County. On January 6, 2000, Standard Concrete filed a complaint against
Local 952 in the United States District Court for the Central District of California. The complaint
alleged that Local 952 breached the no-strike clause in the Orange County CBA by participating
in and encouraging the Orange County bargaining unit members to honor the Corona bargaining
unit's picket lines. On April 27, 2000, Local 952 filed a motion to dismiss the complaint, arguing
that Standard Concrete violated the Orange County CBA by failing to submit the d ...
353 F.3d 668 STANDARD CONCRETE PRODUCTS INC., Plaintiff-Ap.docxgilbertkpeters11344
353 F.3d 668
STANDARD CONCRETE PRODUCTS INC., Plaintiff-Appellee,
v.
GENERAL TRUCK DRIVERS, OFFICE, FOOD AND WAREHOUSE UNION, LOCAL 952,
Defendant-Appellant.
Standard Concrete Products Inc., Plaintiff-Appellant,
v.
General Truck Drivers, Office, Food and Warehouse Union, Local 952, Defendant-Appellee.
No. 01-57256.
No. 01-57257.
United States Court of Appeals, Ninth Circuit.
Argued and Submitted April 8, 2003 — Pasadena, California.
Filed December 18, 2003.
Howard C. Hay, (argued and brief) and Glenn L. Briggs (brief), Paul, Hastings, Janofsky &
Walker, LLP, Costa Mesa, California, CA, for the plaintiff/appellee/appellant.
Appeals from the United States District Court for the Central District of California; Alicemarie
H. Stotler, District Judge, Presiding. D.C. No. CV-00-00016-AHS.
Before Harry PREGERSON, A. Wallace TASHIMA, and Richard R. CLIFTON, Circuit Judges.
OPINION
PREGERSON, Circuit Judge:
1
Plaintiff Standard Concrete Products ("Standard Concrete") delivers concrete throughout
Southern California. Relevant to this appeal, Standard Concrete has facilities in Riverside
County and in Orange County. The International Brotherhood of the Teamsters, General Truck
Drivers, Office, Food & Warehouse Union, Local 952 represents Standard Concrete's employees
at its Corona facility in Riverside County ("Corona bargaining unit"). Local 952 also represents
Standard Concrete's employees at its three Orange County facilities ("Orange County bargaining
unit") under a separate collective bargaining agreement.
2
In January 2000, the Corona bargaining unit went on strike against Standard Concrete. The strike
was called because the Corona bargaining unit believed that Standard Concrete was negotiating
with Local 952 in bad faith. Members of the Corona bargaining unit established picket lines at
the Corona facility. On the second day of the strike, the Corona bargaining unit extended its
picket lines to Standard Concrete's three facilities in Orange County. Members of the Orange
County bargaining unit honored the Corona bargaining unit's picket lines.
3
At issue in this case is whether the Orange County bargaining unit violated its Collective
Bargaining Agreement ("CBA") with Standard Concrete when members of the Orange County
bargaining unit honored the Corona bargaining unit's picket lines at the Standard Concrete
facilities in Orange County. On January 6, 2000, Standard Concrete filed a complaint against
Local 952 in the United States District Court for the Central District of California. The complaint
alleged that Local 952 breached the no-strike clause in the Orange County CBA by participating
in and encouraging the Orange County bargaining unit members to honor the Corona bargaining
unit's picket lines. On April 27, 2000, Local 952 filed a motion to dismiss the complaint, arguing
that Standard Concrete violated the Orange County CBA by failing to submit the d.
All product and company names mentioned herein are for identification and educational purposes only and are the property of, and may be trademarks of, their respective owners.
Running head MARKETING ANALYSIS ASSIGNMENTS .docxwlynn1
Running head: MARKETING ANALYSIS ASSIGNMENTS 1
MARKETING ANALYSIS ASSIGNMENTS 6
Researching Marketing Questions
MKT/571
Melissa Simmons
Roberto Ancis
Part 1: Memorandum
TO: Senior Vice President (Marketing)
FROM: Jacob Glenns
DATE: August 19, 2018
SUBJECT: Marketing Analysis
Summary Analysis
This analysis of the market report that was presented the market analyst provides detailed insights from the data that may help in formulating an effective marketing strategy. The key information include: revenue performance for the first half between 2015 and 2016 and revenue trends over the same period. This information help in deciding whether to the organization should continue with its growth strategy or to reverse the decline.
Revenue Analysis
Analysis of the company’s semiannual performance- between January and June- indicates that there was an increase of 10.18 percent in the generated revenues per day from 96,000 dollars to 105,768 dollars in 2015 and 2016 respectively. The revenues per day, domestic market, were 93,683 dollars and 85,181 dollars in 2016 and 2015 respectively, over the same period. Overall, the semiannual revenue for the year 2016 was 13,644,073 dollars with the United States market contributing 12,085,137 dollars, which is approximately 88.6 percent of the semiannual revenue. The international market contributed 1,558,936 dollars, which is 11.4 percent of the total revenue. The average gross profit per day was 8.3 percent for the six months between January and June, 2016. For the three months of April, May and June, 2016 the total revenue was 7,024,096 dollars with the domestic market contributing 6,145,978 dollars and the international market contributing 878,119 dollars. The gross profit was 6.5 percent.
Revenue Trends
With regards to customer class, commercial customers contributed 7,195,592 dollars in the six months of January to June, 2016. The revenue per day was 55,780 dollars, an increase of 5,008 dollars compared to 50,772 dollars realized over the same period in 2015. At the second place was the municipal segment with 1,634,643 dollars. The revenue per day for the first six months was 12,672 in 2016 compared to 12,034 in 2015. The international market segment contributed 1,535,905 dollars and the revenue per day was 11,906 dollars and 11,700 dollars in 2016 and 2015 respectively. The other important segments- resellers, industrial labs, government, resell, education and others- also registered increments in the revenue per day for the first 6 months between 2015 and 2016. The revenue trend for the second quarter (between April and May) illustrate that commercial market contributed 1,130,973 dollars which is 50 percent of the total revenue from the customer class segment. The international market contributed 323,990 follo.
Running head MANAGING A DIVERSE WORKFORCE1MANAGING A DIVERSE.docxwlynn1
Running head: MANAGING A DIVERSE WORKFORCE 1
MANAGING A DIVERSE WORKFORCE 6
Managing a diverse workforce
Name
Institutional affiliation
What does it mean to be an effective manager in a diverse workforce?
According to Chip Conley, the workforce diversity is characterized of gender, ethnicity and age; which needs a much keener attention. He points out that an effective manager should realize that age diversity makes a company stronger and that different generations within a workplace should focus on mentoring one another at work. He emphasizes on the need to allow openness with one another so that wisdom; knowledge, experience and skills from the young to the old and vice versa. According to Chip Conley, the current 60s is the new 40s and that the current 30s is the new 50s; a key note to take on how effective relationship in a workplace could enrichen a company with greater shared wisdom and skills. Every manager need to relate such knowledge in ensuring effective making of modern elders from the millennials.
According to Chip, an effective manager should establish a learning environment for the boomers and the millennials. Each generation should see the other as assets from which they can derive wisdom. Moreover, Chip calls for both the millennials and the boomers to fix their ego, perhaps so that they can enhance their relationship and get to learn from one another. He calls for the need of the managers to enhance a growth mindset in a workplace and the need for the employees to be curious of getting to know what the other generation can offer, and trying to oneself. Chip states that “Curiosity is the elixir for life”
Working on the psychological empowerment of specifics groups and ensuring mental flexibility is very important for various generations to work coherently effectively. Additionally, a manager in charge of a diverse workforce should ensure that the differences existing between the BB and X generations, and the Y and Z generations should be harmonized so that they do not tamper with the achievement of the organizations set goals and objectives (Toro, Labrador-Fernández & De Nicolas, 2019).
Maintaining a positive working environment helps in enhancing the performance of a diverse workforce. Looking at the small business managers, workforce diversity can be well managed if the owner’s manager supports the existing generational interconnections and the variations as a result of the general difference defining these groups by valuing their differences and the similarities. An effective manager is therefore required to cause a diversity openness among the workforce. Such ensure the performance at all levels, i.e. both the organizational and individual. A manager should, therefore, have the ability to effectively enforce the eradication of the internal communication barriers existing as a result generational, racial, gender, ethnic, age, personality tenure, cognitive style, education among other dissimilarities .
Running head MANAGING TECHNOLOGICAL INNOVATION IN DIGITAL BUS.docxwlynn1
Running head: MANAGING TECHNOLOGICAL INNOVATION IN DIGITAL BUSINESS
ENVIRONMENTS 1
Managing Technological Innovation in Digital Business Environments
Yolanda McNeil
ENGL 602 Field Project: Final Product
Liberty University
MANAGING TECHNOLOGICAL INNOVATION IN DIGITAL BUSINESS
ENVIRONMENTS 2
Introduction
Background of the Research
Innovation plays a critical role in assisting businesses to sustain and grow their market
shares. It takes place in dissimilar functions and parts of the business and it is significant to
understand the best way to create and manage it effectively. Digital technologies have been
regularly used in business and this has led to digitized workplaces that demand the need to invent
to remain at the top in the market (Kay & Willman, 2018). Digitizing places of work has played
a key role in changing the way business is usually managed and this has similarly affected how
innovation must be managed and embraced in such a novel business atmosphere. Therefore, the
best way to understand technological innovation in the digital business atmosphere is the need to
understand how technology has been shaping the business world.
The reason for choosing technological innovation in digital business environments is that
business owners play a critical role in the identification and application of new technologies. By
investing in initiatives that permit them to deliver efficient and effective services and products,
they discover innovative solutions to complex challenges (Camisón & Villar-López, 2014).
Successful technological innovation needs collaboration, expert project management, planning,
and execution. Worldwide competition and rigorous demand to bring commodities to market
very fast affect decisions.
Research Purpose
1
2
Tess Stockslager @ 2020-03-06T10:07:25-08:00
This wording seems a bit circular: "the best way to understand...is the need to understand." Is there a clearer way you could state this?
Tess Stockslager @ 2020-03-06T10:09:33-08:00
Even without the word "I," you're indirectly referring to yourself here, which isn't necessary in this paper. You don't need to explain why you chose your topic; instead, you should explain why the topic is important in the field (which is exactly what you did in this sentence--you just need to frame it differently).
MANAGING TECHNOLOGICAL INNOVATION IN DIGITAL BUSINESS
ENVIRONMENTS 3
The purpose of this research is to explore the role and importance of managing
technological innovation in the digital business environment. Technological innovation strategies
that a firm pursues can either break or make the company. The current business landscape is
increasingly multifaceted. For an organization to succeed in the modern business environment, it
is critical that it adopts digital innovation which can assist to attain its goals and remain at the top
in the competition (Camisón & Villar-Lóp.
Running head MANAGERIAL REPORT FOR SUPERVISING MANAGER 1MAN.docxwlynn1
Running head: MANAGERIAL REPORT FOR SUPERVISING MANAGER
1
MANAGERIAL REPORT FOR SUPERVISING MANAGER
7
Managerial Report
HMGT 300 6380 Introduction to the U.S Health Care Sector 2205
Taneshia Davis
UMGC
Professor: Todd Price
May 31, 2020
Manager's Name and Role:
Name: The patient experience-supervising manager is Mr. Aleo Brandford
Roles:
The supervising manager ensures that all patients are fully engaged in inpatient experience activities under the supervision of highly experienced healthcare professionals. The manager also ensures that all healthcare professionals are compliant with policies, rules, and regulations that govern patients, healthcare practice, healthcare organizations, government, and the corporate world. Moreover, the supervisor conducts monitoring and evaluation of the healthcare providers to ensure they are delivering high-quality services within the set time. The manager also monitors and evaluates the healthcare systems in the organization to ensure that they are affirmative to rules, policies, and standards set for healthcare service facilities and providers as a to deliver satisfactory high-quality services. The manager, together with respective departments and personnel, initiates, improves, and implements patient experience programs that equip personnel with relevant patient experience skills, knowledge, and competencies necessary for satisfactory healthcare service provision. One other key role of the manager is the contact point for all inquiries, explanations, experiences, and feedbacks associated with patients and the healthcare facility.
Healthcare Setting:
The Minnesota Healthcare Facility is a county facility that offers preventive and curative healthcare services for in- and out-patients. It serves the entire region with all healthcare needs. It has both children and adults wings with fully functional departments and equipment. It is the only healthcare facility in rural with a population capacity of 200 per day. It is well equipped with childbirth and immunization facilities and serves the general public healthcare needs.
Managerial Issue:
Determining MeaslesSpread Rate
The manager needs to task-relevant departments to collect patient and exposed children information from children's care centers, schools, attendance lists, and health facilities. The information will help determine the rate of immunization, the number of patients, and approximate exposed children and other adults. The number of children vaccinated against measles, 21 days before its eruption should be identified from the Immunization Information System of Minnesota, and facility children's care center information System. The challenge will be on the follow up of the exposed children and administering necessary interventions. This is necessary for checking further spread of the disease in the community (Hall et al., 2017).
Impact & Details: Restrict Public Gathering
To restrict the mingling of children in healthcare faciliti.
Running head MANAGING DYNAMIC ENVIRONMENTS FINAL .docxwlynn1
Running head: MANAGING DYNAMIC ENVIRONMENTS FINAL
1
MANAGING DYNAMIC ENVIRONMENTS FINAL
2
Managing Dynamic Environments Final
Managing Dynamic Environments Final
Introduction
The for-profit organization which will be analyzed in this report is a famous casual dining restaurant and bar called Buffalo Wild Wings Restaurant and Sports Bar. This is an international organization which has various outlets in different parts of the world such as in the United States, Mexico, Canada, Panama, India, and the Philippines among other countries. The reason why Buffalo Wild Wings is the target organization for this report is that it recently received a new president, Lyle Tick, who set an objective to improve the brand image of the restaurant so that it can attract more customers (Romeo, 2018). Due to this, the organization is undertaking some changes in its marketing which is an important component of the internal operations of the business. The change of focus is implementing a social media marketing campaign to increase the number of new customers for the restaurant. This report will evaluate different factors, positive and negative issues, and challenges, which can affect the change process as well as analyze different concepts which can be used to improve change management and change process so as to result to the desired outcomes.
Identify the role of strategic renewal in propelling change.
Strategic renewal is important in creating change interventions which will impact the team members and the organization positively. This is an important process which helps change managers to evaluate the existing progress of the change process and focus on how to improve the change process so that the desired outcome may be achieved. One of the roles of strategic renewal in propelling change is by revisiting and improving the change strategies. Strategic renewal ensures that the organization is able to develop a strategic game plan which will be used to promote different growth objectives during change management. This enhances change since the organization is able to focus on having a competitive advantage against other competitors and satisfying the customers’ needs to the best of its abilities. In the case of Buffalo Wild Wing Restaurant, it focused on adopting new growth objective which aimed at attracting more millennial customers to ensure it increases the size of the target market for the restaurant.
Strategic renewal helps in concentrating all the efforts in brainstorming and identification of solutions to challenges which may impact the change action plan. The organization and its employees are able to focus on finding different approaches which can be used to improve the experience resulting from the change process. This pushes change since the organization is able to avoid certain pitfalls which the organizations would have experienced. This aspect has been achieved by Buffalo Wild Wings Restaurant whereby the organization.
Running head MANAGING DONUT FRANCHISES1MANAGING DONUT FRANCHIS.docxwlynn1
Running head: MANAGING DONUT FRANCHISES 1
MANAGING DONUT FRANCHISES 2
Managing Donuts
Joyce Crow
Ashford University
MGT 330 Management for Organization
Jill Heaney
May 10, 2020
District Manager of Five Dunkin’ Donut Franchises
Introduction
As the new District Manager, I intend to build and structure the foundation of workers for all the five Dunkin' Donuts establishments. My goal is to increase the fiscal profits for every unit to establish extra legacies to the company's brand. The paper analyzes the following categories of Dunkin' Donuts: job design including job analysis, job description and job specification, and organizational design. Workers job designs will be assessed with the use of a divisional structure for Bakers, Crewmembers, and managers. Inside of Dunkin' Donuts will be analyzed to decide the needs for recruiting and selecting applicants. Also, the essay discusses the training and performance appraisals for the value of significance to the franchise.
Job Design
Job design refers to the process of organizing duties and roles into a productive unit of work. The job design will include job analysis, job description and job specification. Job design occurs when managers decide the duties to be completed, the people who will do them and the selection approach to be adopted in choosing workers (Reilly, Minnick, & Baack, 2011). Below, I have used job analysis, job description, and job specification to discuss the job design of the five new establishments.
Job Analysis
The process of assigning tasks will be undertaken by the HR department and the departmental managers. I will be adapting the extermination model of job analysis. Every branch will have 5 to 8 workers per shift, with one being a manager, one may be a shift leader and the rest will include crewmembers and bakers. They will be in charge of food handling, housekeeping and sales. Each worker's qualification will include preparing donuts, coffee, frozen meals, and working on the cash register.
Job Description
For job descriptions, the current Dunkin' Donuts models will be appropriate for the Crewmembers, Bakers, and Management (https://www.peopleanswers.com/pa/testSplashPageEntry.do?splashURL=portalDunkinDonuts1&src=825452). Most roles at the organization are entry-level positions, which need filling customer orders through preparing drinks and baked food. Applicants will need to show their readiness to take directions and interact with the clients regularly.
Job Specification
Bakers, Crewmembers, and Shift Leaders – These are the entry-level spots that will need minimal requirements. Basic requirements include at least a High School Diploma (GED or equivalent), inclination to take direction and intermingle with clients, and interpersonal working capabilities. These roles are trainable on the job. The position of shift leader will be achievable by an existing baker or crewmember .
Running head MANAGEMENT DILEMMAS1MANAGEMENT DILEMMAS6.docxwlynn1
Running head: MANAGEMENT DILEMMAS 1
MANAGEMENT DILEMMAS 6
Management Dilemmas
Name
Institutional Affiliation
Management Dilemmas
Part I: Research Questions
1. Should student athletes receive a stipend by the universities as reimbursement for participating in sports? Are there policies under the ISSF that guide on how best students should be compensated for their participation in different sports?
2. What challenges do coaches face in managing their respective teams? Is there an approved ISSF standard management structure that would allow coaches to participate and interact more with their players such that they are not only constrained to their managerial duties?
Part II: Research Topic
Problem Statement
Professional athletes earn large sums of money, though considered unethical; due to the fact that most of the times these athletes are students who are “exploited”. The estimated value rose through college athletics is considered to be roughly more than a billion dollars yearly, with this revenue being generated from an estimated 25 football schools and 64 basketball schools respectively (Brown & Williams, 2019). The concern raised is that the students do not get to see the money earned; but instead are offered athletic scholarships, allowing them to get free college education. The concerning factor is that most students use this opportunity as a chance to qualify for professional leagues, without considering the beneficial factors that their education offers. They are continuously to sacrifice their class and study hours such that they can practice and travel for their sports (Brown & Williams, 2019). Even though a scholarship seems like a good deal for some of these college athletes, what criteria is used to reward those athletes who are often viewed as celebrities and exploited for their affiliation with different institution to earn money for them?
Quite often, managers are faced with the dilemma of relating with their athletes mainly because they are absorbed in managerial duties that limit their interactions with their players. As a result, the element of teamwork is ignored and disregarded, leading to lack of communication, lack of trust, and continued conflict, which may affect the effectiveness of the team (Rollnick, Fader, Breckon, & Moyers, 2019). Sometimes the coaches aspect of caring is viewed as interference because there is no connection between the players and their coach, with coaches feeling left out of most decisions made by the players. This in mind, the study focuses on finding new strategies that can be applied by all coaches in every sport, such that the aspect of unity and communication is achieved, with coaches participating more in their respective projects.
Importance of the Study
Given the dynamic scope of this industry, it is important to do more research to understand the depth of the dilemma within the industry, with the use of previous and current research to provide insight on different pers.
Running head MANAGERIAL ACCOUNTING 1MANAGERIAL ACCOUNTING.docxwlynn1
Running head: MANAGERIAL ACCOUNTING
1
MANAGERIAL ACCOUNTING
2
Managerial Accounting
Accounting can be defined as the procedure of keeping monetary financial records. Accounting can be group as financial and managerial accounting. For businesses to be successful, they need to be having both managerial and financial accounting experts. Impeccable managerial and financial bookkeeping are important to the progress and constant survival of any corporate. Structurally, economically, and lawfully, bookkeeping is an essential section in any institute, and the necessity for an extremely skilled accounting squad is unconditionally crucial. Despite the similarities between financial and managerial accounting, there are also differences between them.
The managerial accounting works through measuring, analyzing and reporting monetary and non-monetary information that aids directors to make judgements to accomplish the objectives of an organization. Managerial accounting emphasizes on the internal broadcasting and is not regulated by generally accepted accounting principles (GAAP). Management accounting is known for its much efforts to focus on the future rather than paying much attention to what happened in the past (Kinicki & Fugate, 2016). This type of accounting is so influential to the performance of directors and other workers as opposed to principally reporting financial events. There are no principles which guide the operations of management accounting.
Management accounting permits executives to charge attention on owners’ principal to aid judge a division’s presentation, although this may not be allowed by generally accepted accounting principles. Managerial accounting comprises assets or liabilities which may not be recognized by generally accepted accounting principles and it makes use of asset or liability quantifying rules like present values or resale prices which is not acceptable under GAAP.
Financial accounting on the other hand emphasizes on commentary to exterior events like shareholders, government interventions, and banks. It evaluates and registers business dealings and provides fiscal reports that are grounded on generally accepted accounting principles (GAAP). Financial bookkeeping is controlled by commonly accepted accounting principles (Weygandt, Kimmel & Kieso, 2015). Financial accounting comprises of sending monetary reports like income reports or balance sheets, to outside bodies like creditors, tax specialists, shareholders, and the Interior Revenue Service.
The managerial accounting positions out profit and loss accounts, job costing accounts, and operating resources, financial accounting conveys facts only for those on the external who want to decide the company's marketplace assessment. Managerial accounting emphases on issues and answers within an institute while financial accounting is worried with productivity from without. Managerial accountants make internal working reports, while financial accountants generat.
Running head: LOGISTIC REGRESSION 1
LOGISTIC REGRESSION 2
Logistic Regression
Student Name
Institution
Course
Instructor
Date
Question (a)
Categorical variables are useful in classifying data that usually takes only one form. An example where categorical variables can be used is when classifying the ages of different individual based on the gender of the participants. The use of n-1 variable in categorical variables makes the classification easier since variables take either of the quantitative provided. In these situations, the variables are limited to take either one or zero as the quantitative value to ease the classification process (Bühlmann & Dezeure, 2016). Classification based on n-1 variable tends to be faster and also saves time and does not have many problems. When a particular variable takes 1 is assumed to be quantitative but when it takes zero the assumption made is that the variable is absent. Categorical variables involving n variables, the n-1 variables are the only important variables since they classify the data given accordingly to the required quantitative values which I either 1 or 0.
Classification of information based on categorical valuables, the n variables tend to have problems. The n value can sometimes lead to problems that may end up prolonging the classification process and also make it difficult. The n variable has problem in resulting to multi-co linearity in classifying (Guo & Berkhahn, 2016). The problem results when there is similar interconnections between the variables this create a problem in interpreting the information. The interconnection of the n variables can result in the prediction of the other variable from the other. Another problem resulting in from categorical variables is that n variable is intuitively meaning that variables can be classified based on the interests or feelings of the research. Lastly, the n variables are redundant that is do not have updated information.
Question (b)
In statistics, logistic regressions are used in classification of variable that tend to have different forms either positive or negative values. Logistic regressions classify data consisting of dependent variables with and more than two or more independent variables. The classifications are based on pacing several variables at their different level of existence (van Smeden et al., 2016). Logistic regression predict the relationship of variables that can either take 1or 0 in the classification. Logistic regressions is concerned in giving descriptions to the data and give detailed information relationship between one independent variable and more nominal independent variables. For instance, logistic regression can be used in financial institutions to clarify financial defaulters. In classification of the data, logistic re.
Running head MANAGEMENT OF CONGESTIVE HEART FAILURE THROUGH MO .docxwlynn1
Running head: MANAGEMENT OF CONGESTIVE HEART FAILURE THROUGH MO 2
MANAGEMENT OF CONGESTIVE HEART FAILURE THROUGH MO 8
Managing Congestive Heart Failure through Motivational Education
Rosaline Hicks
Chamberlain University
Dr. Sheryl Cator
March 26, 2020
The purpose of this paper is to discuss how motivation can improve outcomes in congestive heart failure (CHF). CHF is a chronic progressive condition that affects the pumping ability of the heart muscles. This paper will cover CHF as a practice problem, the role of evidence to in regard to CHF, and the role of the DNP practice scholar in the translation of evidence.
Addressing issues related to CHF management through education program is important in the improvement of self-management. Most of the reported readmission cases, morbidity, and mortality are associated with poor self-care and self-management of the diseases. The focus of most healthcare facilities when it comes to the management of the CHF is focused on an identified medication regimen, and little to no attention is given to the importance of patient education to improve self-management of CHF.
A study by Bader et al (2018) revealed that an advanced heart failure program helped in the improvement of disease awareness and self-care behaviors when the patients were led by well-trained heart failure nurses. Another study by Howie-Esquivel et al (2015) used the approach of TEACH-HF intervention to manage CHF patients. The study outcome revealed a significantly lower hospital re-admission rate and decrease in the length of stay.
DNP practice scholar play a key role in the translation of evidence. The DNP practice scholar is instrumental in the initiation of projects that focus on the standardized educational process for CHF patients. The initiation is done through the development of new education tools and clinician documentation of evidence-based heart failure care (Myslenski, 2018). Practice Problem and Question
Patient education is becoming an effective process of managing CHF at home. Patient education aids in the improvement of knowledge and self-care behaviors, thereby, reducing the incidence of readmissions cases (Bader, et al., 2018).
Heart failure is a common, high-risk condition that is characterized with high reports hospitalization and sometimes death. This disease affects more than 6.5 million Americans and in 2012 the CDC reported that it cost approximately 30.7 billion dollars to care for CHF patients and wages lost due to hospitalization. Unlike other cardiovascular illnesses, CHF appears to be the most common one and nearly 1 million new cases are being reported annually internationally. This, therefore, makes it the fastest growing cardiovascular disorder (Savarese & Lund, 2017).
This study is guided by the following Picot question: Does the multidisciplinary educational approach work effectively towards the prevention of hospital re-admission for patients diagnosed with congestive heart .
Running head: MALWARE 1
MALWARE 2
Student’s name:
Professor' name:
Topic:
Institution:
Date:
Malware-Trojan horse virus
Malware can be defined as any file or program that is introduced to a computer with the intention of harming the user. The harm to the user can be through interfering with his use of the compute, unauthorized access to his data, locking the user out of his computer and also spying on the user’s activity. There are several types of malware and they include ransom ware, Trojan horses, computer viruses, worms and spyware (White, Fisch & Pooch, 2017). For this particular assignment, I will focus on Trojan horse virus. The name Trojan horse comes from the famous Greek story, where Greek soldiers were able to take down the city of Troy after they sneaked into the city inside a wooden horse that was guised as a gift to the people of Troy. Just like the story the Trojan horse virus disguises itself as a legitimate program however the program provides unauthorized access into the system most of the time to hackers.
Most of the time, Trojan horses gain access to a secured system through social engineering. Most of the time, Trojan horse viruses are introduced into a system by duping a user into executing an attachment on an email guised to be unsuspicious. They can also be introduced via social media where users are tricked into clicking on fake advertisements or advertisements that offer fake rewards. Once the links or attachments are clicked on, a Trojan horse virus is introduced. Trojan horse viruses can allow an attacker to have access to a user’s personal information and other forms of data. Trojan horse viruses can affect other devices on the network through infection caused by the introduction of the first Trojan horse; most ransom ware is introduced through Trojan horse viruses (Wang, Lorch & Parno, 2016). In addition, through the use of Trojan horse viruses, attackers can modify data, copy data, block data, delete data and generally disrupt or distort the performance and operations of targeted computers or devices in a network.
Steps of mitigating a Trojan horse virus attack
The first step in mitigating a Trojan horse virus attack is the installation of effective anti-malware software or what is commonly referred to as an anti-virus. The anti-malware will detect as well as prevent any Trojan horse virus attack on a computer or a network. The second step in mitigating Trojan horse virus attacks is the installation of the latest available patches of the operating system in use. The third step is proper scanning of all external devices that are introduced to a computer or a network (Rader & Rahman, 2015). The fourth step is through the cautioning on the execution of any program th.
Running head LOS ANGELES AND NEW YORK BUDGETARY COMPARISON .docxwlynn1
Running head: LOS ANGELES AND NEW YORK BUDGETARY COMPARISON 1
LOS ANGELES AND NEW YORK BUDGETARY COMPARISON
3
Los Angeles and New York budgetary comparison
Vibert Jacob
South University
Los Angeles and New York budgetary comparison
The cities for comparison in this assignment are the city of New York and the city of Los Angeles. These two are major cities in the United States that have large population and play a crucial role both locally and internationally. The cities have major infrastructural, social, and economic burdens to bear. They also have huge finances to budget for the management of their cities. In the financial year 2017, the city of New York budgeted for an expenditure of $84 billion (The City of New York, 2017). Los Angeles has a budget of $9.2 billion (City of Los Angeles, 2017). The New York City budget is larger than some of the states in the USA. Both cities are required to ensure they have a balanced budget each year with clear information about the sources of the funds, use of the fund and ensure that the budgetary deficits are clearly financed in each year.
The city of Los Angeles budgets is prepared with several underlying principles that must adhered. The city has a reserve fund, which equals to 5% of the city’s general fund revenues. The capital improvements fund for the city is equal to 1% of the city’s general fund revenue. The city holds that all the funds from one-time sources must be used to finance the one-time expenditures. The city of New York has also established several reserves to take care of uncertainties in the city (City of Los Angeles, 2017). These reserves include the Retiree health benefit trust funds, a general reserve as well as a capital stabilization reserve fund for the city.
Sources of funds
The two cities have almost similar sources of funds for their budgets. These sources of funds, however, have differing contributions to the city’s finances. The table below presents the proportional sources of incomes to the cities.
Los Angeles
New York
Source
%
%
property taxes
21.9
29
allocation from other government agencies
6.5
27
utility user tax
7.1
7
business occupation
8.6
4
licenses and other fees
24.5
8
sales tax
5.7
8
proprietary
5.3
13
miscellaneous
20.4
4
100
100
In the two cities, the property taxes account for the largest source of incomes. In New York, the allocation and distribution from other government and government agencies is the second largest source of income. This is due to the international nature of the city, which hosts major national and international offices. The city of Los Angeles has large commercial enterprises within its jurisdiction that contributed large amount of incomes in form of licenses, fees, and permits compared to New York’s city income from license and fees amounting to only 8% of the overall incomes. The miscellaneous sources of finance include the transfers from the reserve transfers, the special funds .
Running head MAJOR PROJECT1MAJOR PROJECT9Initial Ou.docxwlynn1
Running head: MAJOR PROJECT
1
MAJOR PROJECT
9
Initial Outline
Chicago
University
(The Working Title of this Major Paper Should Go Here Exactly as on the Title Page)
Foreclosure is a scary word for homeowners, but it is
not all that common today (citation needed). Bortz (2017) reported that the foreclosure rate (meaning the percentage of loans in foreclosure) currently hovers just under 1%. During economic downturns, like the housing crisis of 2011, foreclosure rates rose as high as 3.6% in United State (Bortz, 2017).
Research question
The phenomenon as mentioned above and literature background lead to the overriding research question, “what are the lived experiences of management executives whose companies face foreclosure?” The subareas of exploration for this question are:
i. The manager’s self-care practices
ii. The manager’s relationship with immediate relatives
iii. The manager’s business practices
iv. The manager’s relationships with subordinates
Methodology
In order to investigate the lived experiences of management executives, a phenomenological qualitative method will be employed. The relationships and practices of managers facing company foreclosure are the core of this research. Creswell (2013) discussed that the purpose of a phenomenological qualitative method is to …….
Proposed population
1. The homogenous group for the study is former management executives strictly from the operations department. The selected executives will have a background of having undergone company foreclosure at least once in the past 20 years.
2. Participants will be solicited through enticing advertisements online for filling surveys to participate in a study interview.
3. The number of participants will be restricted to 16 executives aged 35 years or more. Their former positions will be limited to operations management.
Data collection
1. The type of data to be accrued will be unstructured and semi-structured interviews.
2. Participants will be asked to participate in at least two rounds of one-on-one interviews spanning anywhere from 50 to 60 minutes each. Interviews will be conducted in person, by phone, or through an internet source such as Zoom.
3.
Bracketing
I am especially interested in this research question because my research showed scarce primary literature about the impact of company foreclosure on the personal and professional lives of executives’ manager and their families. With many companies facing foreclosure around the globe every year, it is surprising that very little research has been conducted on how they affected the lives of the involved executives. I suspect I may find it useful to know the real potential consequences of organizational shutdown in case I become a manager in the future. Even though one works hoping for the best, preparing for the worst is also a very rational route for any organizational management model.
(Do you have any first or third-party experience and/or knowledge of a.
Running Head MAJOR CONCERNS OF CLIMATE CHANGE IN CHINA 1MAJO.docxwlynn1
Running Head: MAJOR CONCERNS OF CLIMATE CHANGE IN CHINA 1
MAJOR CONCERNS OF CLIMATE CHANGE IN CHINA 10
Major Concerns of Climate Change in China
Student’s Name:
Course Title:
Course Number:
Professor’s Name:
Date:
Major Concerns of Climate Change in China
Introduction
China is one of the critical countries in the world, which are considered to significantly contribute to the issue of climate change. Research indicates that China produces over 6.000 megatons of carbon dioxide every year. The increased concentration of carbon dioxide in the atmosphere is associated with increase in global warming, which perpetrates the climate change. To this end, China is regarded as the largest emitter of greenhouse gases across the globe based on absolute terms, contributing to about 22 percent of the total amount of emissions (Held, Nag & Roger, 2011). At the moment, the emissions of the greenhouse gases by China have exceeded the global per capita average, following the growth in the emissions by over 200 percent from 1990 to 2008. The concern of increased greenhouse gases emissions in China is largely associated with the countries appetite for economic growth. The historical growth of the Chinese economy has been tremendously effected through the use of fossil fuels as a major source of energy in industries. Despite the increased desire from the global community to mitigate the impacts of climate change, there is fear that the emission of greenhouse gases in the country may rise by between 55 and 75 by 2025 (Held, Nag & Roger, 2011). Therefore, it is important to discuss the different concerns presented by China regarding the issue of climate change that is tremendously perpetrated by increase in emission of carbon dioxide and other greenhouse gases.
Overview of the Issue of Climate Change in China
The Chinese government has established policies that are aimed at adopting effective governance of climate change, improved domestic capacity of effectively governing the energy use and emissions, as well as supporting the commitments that positively impact decline in future international emissions. China acknowledges the need to lower the emission of greenhouse gases as well as mitigating the impacts of climate change, which is a critical solution towards obtaining a healthier international environment (Lipin, 2016). As a matter of fact, numerous multinational negotiations have been advanced so as to develop a global climate regime that governs the efforts of reducing the emission of carbon dioxide and other greenhouse gases. Being among the world’s largest polluters, China has received increase attention from the global community. The country, which has the highest population of over 1.3 billion, has been steadfastly reluctant to comply to the suggestions by international organizations such as the United Nations Framework Convention on Climate Change (UNFCCC) (Held, Nag & Roger, 2011). These organizations have been engaged in pushing for .
Running Head LOGISTICS1Running Head LOGISTICS7.docxwlynn1
Running Head: LOGISTICS 1
Running Head: LOGISTICS 7
Logistics and Supply Chain Operations
Stanley Thompson Jr.
DB 8035
24 May 2020
INTRODUCTION
Amazon is one of the fastest growing online retailer company in the United States of America that has been able to overhaul its business structure by using innovative strategies in supply chain management. Amazon has left most of its competitors have a hard time trying to catch up. The firm has made huge investments in the management of its inventory to include recent forms of technology to beat its competition. The firm has optimized every link in its supply chain to ensure its customers are satisfied and well attended to (Leblanc, 2019). This paper hence seeks to discuss Amazons supply chain operation factors such as; transport and security, procurement and inventory management, technology and information management, and articulate some of the global risk factors affecting the firm. Comment by TJS: Paragraphs need to be left justified Comment by TJS: Great point here. Amazon is dominating the industry Comment by TJS: Anthropomorphisms should not be utilized. An anthropomorphism is the attribution of human characteristics or behavior to a good, animal, or object.
TRANSPORTATION AND SECURITY
Transportation cost structures, modes, and distribution centers, inventory control systems, and inventory costs reduction strategies
Amazon initially launched a two-day delivery program for its customers to ensure that its customers had fast delivery of products but soon other competitors started catching on. Amazon hence had to make another adjustment in its freight services and now offers a two-hour delivery service to Amazon Prime customers. For product freight, Amazon has equally sub-contracted firms such as the United Parcel Service to transport its products to its customers. Amazon has been relying on third-party couriers to make their deliveries as they have a better-established delivery route and path that they can leverage for efficient delivery services (Leblanc, 2019). Comment by TJS: Yes. They set a new industry standard
However, due to the consideration of numerous factors involved in using third-party carriers for deliveries, Amazon has developed its privately-owned freight service. Amazon hence uses its privately-owned vehicles to carry products to its clients specifically for same-day deliveries. In recent times, Amazon has been developing cargo freight service in certain specific areas where the firm uses drones to carry items straight to their clients who are within a 10-mile radius from their warehouses. This has cut product deliveries to half an hour or less. Amazon is progressively incorporating newer technologies in its supply chain that systems can hence run without human supervision. This strategy has been articulated to be efficient so far as there are has been reduced inventory management costs over the last few years since the acquisition of Kiva Systems (Leblanc, 2.
Running head LOGIC MODELLOGIC MODEL 2Logic modelStu.docxwlynn1
Running head: LOGIC MODEL
LOGIC MODEL
2
Logic model
Student’s name
University affiliation
Date
References
Blue-Howells, J., McGuire, J., & Nakashima, J. (2008). Co-location of health care services for homeless veterans: a case study of innovation in program implementation. Social work in health care, 47(3), 219-231.
Output
Integrating patient care
Communication and collaboration between workers hence resulting to communities of practicing clinicians
Attracting new patients to GLA
Funding a two-year pilot grant
Effective process for psychiatric screening for homeless patients
Outcomes
Homeless project were integrated
The issues of homeless veterans were addressed due to institutional barriers
There was creation of coalition and linking the project to legitimate VA-wide goals
Good sustained program maintenance, process evaluation and encouraging development of communities.
Activities
Building a coalition of decision makers
Introduction of a new integrated program
Inputs
The decision to implement
Initial implementation
Sustained maintenance
Termination or transformation
Running head: PROGRAM EVALUATION 1
PROGRAM EVALUATION 2
Program Evaluation
Institutional Affiliation
Insert the student’s name
Instructor’s name
Course
Date
Introduction
Evaluation of the program is usually done to in order to determine the quality of the program, how effective the program is and how the program is performing. This can help to know if the program is making a significant difference among the targeted people. It can also assist to know if the program is functioning or not. This paper therefore seeks to evaluate the program which is assisting the homeless people within the community.
The two program evaluation questions are: what is the reach of the program? And what has been the impact of the program on the homeless people? The answers to these questions would elicit both qualitative and quantitative results. Therefore, the program evaluation will require both quantitative and qualitative data collection plan. This is because the use of mixed-method approach is convenient since the results and findings would be reliable (Creswell, 2017). After identifying the evaluation program questions, the next step will be to come up with plan of evaluating a program. The plan should consist of methods of collecting data, evidences, the person responsible and the duration.
Program Evaluation Question
Evidence
Methods and sources of collecting data
Person in charge
Duration
1. What is the reach of the program?
Number of building materials distributed
Records of the program
Robert
One month
2. What has been the impact of the program on the homeless people?
Number of people resettled
Number of people not yet re.
Running head LITERATURE REVIEW1MINORITY BOYS SCHOOL DROPOUT A.docxwlynn1
Running head: LITERATURE REVIEW 1
MINORITY BOYS SCHOOL DROPOUT AND CONTINUATION SCHOOL 2
Literature Review
Literature Review
It is expected that every student enrolled in high school works hard towards the completion of their high school diploma. However, research indicates there was a 5.4% drop out among the minority groups, in which 6.4% of the overall status dropout rate is that of the male youth. Among the Africans, Hispanics, and American Indian Natives, the dropout rates among the boys are 8%, 10%, and 11.6%, respectively (Musu-Gillette, De Brey, McFarland, Hussar, Sonnenberg, & Wilkinson-Flicker, 2017). These dropouts often join continuation schools later in life with the hope that they will get an equivalent of their high school diploma. The theoretical framework of this research is based on the phenomenological approach, in which the aim is to examine the occurrence of school dropout among minority boys and their performance after joining continuation school.
One of the theories that explain why minority boys drop out of school is the Critical Race Theory. The model argues that education opportunities are often affected by an individual’s race and racism (Colbert, 2017). Based on this theory, minority groups are often faced with issues such as poverty and racial discrimination in schools, which causes some of the male students to drop out of school. Racism victims in school feel inferior to the whites and sometimes feel like they do not deserve a quality education, and they end up falling behind in school.
Cultural production theory, on the other hand, explains why the dropouts choose to go back to school. The theory holds that the education system helps to level out the playing field so that people get equal opportunities to make their lives. The approach provides an essential perspective as to why minority boys dropouts join continuation schools and complete their learning process.
According to Bania, Lydersen, and Kvernmo (2016), non-completion of high school mostly results from different problems, most of which are health-related. In research in which the authors carried out among the youths in the Arctic, they found out that dropout rates were higher among males. Additionally, minority males often drop out due to mental issues. Based on the article, education affects an individual’s employment opportunities and income, as well as the quality of life, which explains why the dropouts choose to join continuation schools later in life.
Hernandez and Ortez (2019) undertake research in which they analyze the experiences of some Latinas who are enrolled in continuation school. Based on the writers’ claims, continuation schools have put in place strategies that enable the students to cope and realize that they have an opportunity to succeed just like any other individual. Additionally, due to the improvement in the prospects for quality education presented to the marginalized groups, the article indicates that there are .
Running head LIVING WITH CHRONIC ILLNESS1Living with Chroni.docxwlynn1
Running head: LIVING WITH CHRONIC ILLNESS 1
Living with Chronic Illnesses 2
Living with chronic illnesses: How are those with a chronic illness treated by their families since their diagnosis?
Maura K. Little
University of West Florida
Abstract
This study aims to figure out what the relationship and meaning of the ways that a family treats a family member with a chronic mental or physical illness. The exploration of the way those with a chronic illness are treated since their diagnosis is important to understand the perceptions, behaviors, and communication that surrounds illness. Chronic mental illness will be analyzed against chronic physical illness to assess similarities and differences in family behaviors. Participants included individuals selected from local support groups based on their illness as well as family structure. An ethnographic study would be used to compare both the verbal and nonverbal relationship between the ill family member and the rest of the family.
Introduction
This study aimed to focus on both physical chronic illnesses and mental chronic illnesses and their effects on family communication, particularly surrounding the diagnosis of the illnesses.
Family has a large impact on the perceptions of illness. In recent times, the publicity around individuals with chronic illnesses, both mental and physical, has increased dramatically in the media. From the production of films about those with physical chronic illnesses to celebrity diagnosis of a mental illness, illness is something our society is beginning to talk about more frequently. However there are certain stigmas attached to these illnesses that make it harder for patients and their families to cope with their situation. Most often because of the portrayals of chronic illness that romanticize illnesses and do not necessarily show all of the effects of these illnesses on the patient or their family.
Both mental and physical chronic illnesses are much more complex than how they are portrayed in the media. These illnesses often produce copious amounts of side effects that bring a whole new level of challenges to the patient's struggle through their daily life and readjustment after diagnosis. One effect that is often not publicized as much as others is the relationships that exist between the patient and their family. These family relationships may change drastically with the diagnosis of and grappling with a chronic illness, changing how family members perceive one another, how they act, and even how they communicate. All of these things depend upon the nature of the family, and the illness and produce different changes. However, through all different types of families and illnesses, communication in situations like these is essential to understanding one another. According to Rosland (2009), several interviews and focus groups showed that family members lowered stress, and are central to patient success. In most instances, the family i.
Running Head LITERATURE REVIEW2LITERATURE REVIEW 2.docxwlynn1
Running Head: LITERATURE REVIEW 2
LITERATURE REVIEW 2
Effect of Tobacco Use
Gideon Aryertey
Embry Riddle Aeronautical University
Introduction
Over decades, many individuals have been using tobacco without being aware of its harmful effects. For instance, in the U.S., the rate of cigarette smoking increased immensely in the early twentieth century. This was due to the invention of the cigarette rolling machine as well as an increase in the advertisement of tobacco products. As a result, cigarette smoking expanded regardless of the opposition of religious leaders or other members. Tobacco consumption reaches its peak especially between the ages of 20 to 40 in both females and males although statistically males consume more than females. Furthermore, the smoking rate amongst African-Americans (16.7%) are higher than the national average in comparison to Caucasians (16.6%). In fact, mixed race individuals and American Indian/Alaska Natives have higher smoking rates than African-Americans. As a result, this shows that there’s a big issue with the use of tobacco. Tobacco has led to many diseases such as lung cancer, diabetes, heart disease, stroke. It also leads to addiction. However, it is significant for one to overcome the addiction of tobacco use to improve their health status. Educating people about the harmful effect of tobacco consumption and making tobacco less affordable will correspond to a gradual decrease in its use.
Tobacco use has caused numerous deaths amongst individuals despite their socioeconomic backgrounds. For instance, approximately 30 percent of people who perish due to cancer in the United States; 80 percent of these deaths are caused by lung cancer. lung cancer is the main cause of cancer related deaths in the youth and adults. (Addicott, Sweitzer & McClernon, 2018). Lung cancer attacks both genders and the treatment process can be very complex. Consumers of tobacco are affected by this disease because it exterminates the cells responsible for fighting against the disease. Also, the use of tobacco affects the proper functioning of all the organs in the body. Other than lung cancer, tobacco consumption can also lead to mouth, esophagus, larynx, liver, kidney, bladder, cervix, pharynx, stomach, myeloid leukemia, pancreas and colon cancers (Ebbert, Elrashidi & Stead, 2015).
In fact, about 7300 nonsmokers die from lung disease every year according to the International Agency that is responsible for Research on Cancer (IARC) after being exposed to tobacco. Additionally, a 2009 survey that was conducted in China indicated that about 38 percent of smokers were aware that smoking contributes to attack of coronary heart disease while 27 percent were aware that it can lead to a stroke (Ambrose, et. al, 2017). However, individuals who smoke about five cigarettes a day showed signs of various diseases and damages to the blood vessels (Gilreath, et. al, 2016). In fact, blood vessels are thickened and then become narrow.
Running head LOGIC MODELLOGIC MODEL 4Situ.docxwlynn1
Running head: LOGIC MODEL
LOGIC MODEL
4
Situation: due to language barrier, patients are unable to receive adequate healthcare
Inputs
Outputs
Outcomes – Impact
Activities
Participation
ShortMediumLong
-Funding
-Staff
-Technology
-Trainers
-Software
-Facilitators
-Computer devices
In order to measure the effectiveness of these inputs, a comprehensive program evaluation may be done through interviews, questionnaires etc
-Training of staff
-Use of technology
-Use of professional interpreter
-Use of multiple languages
-Use of visuals like graphs and pictures
-Interview patients and healthcare
- Assessing the language barrier
-Improving staff ability to communicate using different languages
-Developing ways that can be used in eradicating the issue of language barrier
-50% of healthcare providers trained within three months.
75% of patients reporting greater satisfaction in healthcare services
-70% increase in number of patient comeback.
-Training completed
-100% effective communication between healthcare providers and patients
-Improved patient satisfaction
-Increase number of community patients
-Improved quality of patient quality.
Project assumptions
There will be enough funding for the training and equipments.
Healthcare providers/staff will be open to participation
References
Chou, C. & Cooley, L. (2018). Communication Rx : transforming healthcare through relationship-centered communication. New York: McGraw-Hill Education.
Jacobs, E. & Diamond, L. (2017). Providing health care in the context of language barriers : international perspectives. Bristol, U.K. Blue Ridge Summit, PA: Multilingual Matters.
.
Thinking of getting a dog? Be aware that breeds like Pit Bulls, Rottweilers, and German Shepherds can be loyal and dangerous. Proper training and socialization are crucial to preventing aggressive behaviors. Ensure safety by understanding their needs and always supervising interactions. Stay safe, and enjoy your furry friends!
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...NelTorrente
In this research, it concludes that while the readiness of teachers in Caloocan City to implement the MATATAG Curriculum is generally positive, targeted efforts in professional development, resource distribution, support networks, and comprehensive preparation can address the existing gaps and ensure successful curriculum implementation.
The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
How to Add Chatter in the odoo 17 ERP ModuleCeline George
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
1. Running head: INTERNATIONAL BUSINESS LAW
1
INTERNATIONAL BUSINESS LAW
6
International business law case
Name: Abdullah Alshaya
Voest-Alpine Trading USA Corp v. Bank of China
Introduction
This paper addresses the case of Voest-Alpine Trading
USA Corp v. Bank of China, case number 167F.Supp.2d
940(2000). The case deals with issues in international business
transactions as it is between Voest-Alpine Trading USA Corp,
an American-based company and Jiangyin Foreign Trade
Corporation, a Chinese-based company. The relationship
between the plaintiff (Voest-Alpine Trading USA Corp) and
Jiangyin Foreign Trade Corporation emerges from a contractual
agreement dated June 23, 1995. Voest-Alpine was to supply or
sell 1000 metric tons of styrene monomer at total of price of $
1.2 million. The defendant Bank of China entered the picture
when JFTC applied for a letter of credit to it in order to fund
the transaction with the plaintiff upon delivery to Zhangjiagang.
After a series of delays and miscommunication, the Bank of
China did not honor the letter of credit presented by the
plaintiff. However, the court ruled the case in favor of the
plaintiff Voest-Alpine after assessing the list of discrepancies
presented by the Bank of China.
The companies
The companies involved in the case include Voest-Alpine
Trading USA Corp, the Jiangyin Foreign Trade Corporation, the
Bank of China and the Texas Commerce Bank. The Voest
Alpine Trade Corporation deals in technology and capital
2. goods. The company supplies steel and other metals to
companies in the aerospace, automotive and consumer goods
market. The Jiangyin Foreign Trade Corporation deals in
imports and exports of products within the light industry and
thus sourcing of the 1000 metric tons of styrene monomer was
for the purpose of production. The Bank of China which is the
defendant in the case is a large commercial bank in China
owned by the government and deals in financial services. The
bank’s headquarters is in Beijing, China. It was tasked with
allocating payment to Voest-Alpine Trading USA Corp upon
end of transaction by honoring the letter of credit applied by
Jiangyin Foreign Trade Corporation. The case took root, when
JFTC asked for a price concession after the significant drop in
the market price of styrene monomer; however, Voest Alpine
Trade USA Corp declined and instead shipped the styrene
monomer on July 18 1995. The Texas Commerce bank had the
role of assessing the letter of credit and documents presented by
Voest –Alpine but identified some discrepancies. Even so,
Voest-Alpine ordered the bank to forward the documents and
letter of credit to the Bank of China. However, the Bank of
China declined to pay the letter of credit and highlighted to the
Texas Bank, “informing them of seven alleged discrepancies
between the letter of credit and the documents Voest-Alpine
presented, six of which are the subject of this action”(DiMatteo
& Dhooge, 2006, pg.377).
The Facts
FACTS: Voest-Alpine agreed to ship 1000 metric tons of
styrene monomer to JFTC valued at $ 1.2 million after
negotiations. JFTC applied for a letter of credit to the Bank of
China which would be honored upon the shipment of goods to
Zhangjiagang, China and that Voest-Alpine would have
presented the required paperwork to Bank of China. Before
shipment, JFTC applied for a price concession after the price of
styrene monomer had drastically dropped in the market but
Voest-Alpine declined the request and shipped the monomer on
3. July 18, 1995. On August 1, 1995, Vest-Alpine presented
documents specified in the letter of credit to the Texas
Commerce Bank. The Texas Commerce Bank, instead, identified
discrepancies between the documents presented and the letter of
credit and notified Voest –Alpine. However, Voest-Alpine did
not believe that discrepancies would disqualify the payment. On
August 3, 1995, the Texas Commerce Bank sent the documents
to the Bank of China. The Voest-Alpine exceeded the fifteen
days of the shipping dates it was required by the letter of credit.
The required date was August 2, 1995.On August 11. 1995, the
Bank of China declined to honor the letter of credit based on
identification of seven discrepancies between the letter of credit
and the documents. The reasons given were: a) the name of the
beneficiary was not similar to the name listed on the letter of
credit. b) Voest-Alpine failed to submit bills indicated original
and instead they were marked as duplicate and triplicate. c) The
invoice, packing list and certificate of origin were not indicated
as original. d) The dates between the survey report and bill of
lading did not match. e) The number indicated in the letter of
credit did not match the date on the beneficiary’s certified copy
of fax. f) The name of the destination was misspelled in the
certificate of origin and the beneficiary’s certificate.
Case significance
The case highlights the rules and policies followed when
international companies engage in international business deals.
It illuminates on whether a court has jurisdiction over a foreign
company similar to how the district court compelled the Bank of
China to honor the letter of credit to Voest-Alpine despite the
discrepancies. According to Frostestad (2000) under the
sovereign immunity framework, federal courts have jurisdiction
over foreign sates on any actions of commercial activity outside
the territories of the United States as long as the activity has a
direct effect on the United States. Furthermore, the case
signifies the importance of clear communication f terms and
agreements between foreign companies to ensure they do not
violate their respective laws in commercial activity. Trading
4. partners should be keen on the laws and regulations that
regulate international trade ( Azimuddin Law Associates, 2018).
Case brief
FACTS: Voest-Alpine agreed to ship 1000 metric tons of
styrene monomer to JFTC valued at $ 1.2 million after
negotiations. Before shipment, JFTC applied for a price
concession after the price of styrene monomer had drastically
dropped in the market but Voest-Alpine declined the request
and shipped the monomer on July 18, 1995. On August 3, 1995,
the Texas Commerce Bank sent the documents to the Bank of
China. On August 11, 1995, the Bank of China declined to
honor the letter of credit based on identification of seven
discrepancies between the letter of credit and the documents.
ISSUE: The bank of China failed to notify that it will not honor
the presentation documents within seven banking days of
receiving the documents
REASONS: The bank of China failed to present the notice of
refusal within seven banking days from receiving the
documents. b) In the notice, the Bank of China did not indicate
that it was rejecting the documents or would not honor the letter
of credit. c) The Bank of China’s telex sent on August 11, 1995
did not contain the list of discrepancies and the disposition of
the documents. d) the Bank of China seemed to indicate
acceptance when it notified the plaintiff they would contact
them to determine whether to waive the discrepancies.
CONCLUSION: The court ruled that the telex sent exceeded the
required date of August 18, 1995 and was instead sent on
August 19, 1995 as per the UCP 500 Article 14(d). Secondly,
the Bank of China did not forward a notice of refusal within
seven banking days upon receiving the documents as per UCP
500 Article 14(d). In conclusion, the court ruled the case in
favor of the plaintiff.
References
Azimmudin Law Associates.(2018). Legal Aspects of
International Business Transactions. Retrieved from
https://www.hg.org/legal-articles/legal-aspects-of-international-
5. business-transactins-30504
DiMatteo,L., & Dhooge, L.(2006). International Business Law:
A Transactional Approach. Thomson/Southwestern.
Frotestad, H.L. (2000). Voest-Alpine Trading v. Bank of China:
Can a Uniform Interpretation of a “Direct Effect” Be Attained
Under the Foreign Sovereign Immunities Act (FSIA) of 1976?
Valparaiso University Law Review, 34(3), 515-556.
35874 Topic: Discussion5
Number of Pages: 1 (Double Spaced)
Number of sources: 1
Writing Style: APA
Type of document: Essay
Academic Level:Master
Category: Psychology
Language Style: English (U.S.)
Order Instructions: Attached
I will attach the instruction
Please follow them carefully
6. Discussion: Please discuss, elaborate and give example on the
topic. Be careful with grammar and spelling. No running head
please. Please Use only the reference I will attach as the
professor will not be able to give grade.
Author: (Jackson, S. L. (2017). Statistics plain and simple. (4th
ed.). Boston, MA: Cengage Learning.)
Topic
What level of measurement can be used for this test for the
independent and dependent variables?
Reference:
Module 9: The Single-Sample z Test
The z Test: What It Is and What It Does
The Sampling Distribution
The Standard Error of the Mean
Calculations for the One-Tailed z Test
Interpreting the One-Tailed z Test
Calculations for the Two-Tailed z Test
Interpreting the Two-Tailed z Test
Statistical Power
Assumptions and Appropriate Use of the z Test
Confidence Intervals Based on the z Distribution
Review of Key Terms
Module Exercises
Critical Thinking Check Answers
Module 10: The Single-Sample t Test
The t Test: What It Is and What It Does
Student's t Distribution
Calculations for the One-Tailed t Test
The Estimated Standard Error of the Mean
Interpreting the One-Tailed t Test
Calculations for the Two-Tailed t Test
7. Interpreting the Two-Tailed t Test
Assumptions and Appropriate Use of the Single-Sample t Test
Confidence Intervals Based on the t Distribution
Review of Key Terms
Module Exercises
Critical Thinking Check Answers
Chapter 5 Summary and Review
Chapter 5 Statistical Software Resources
In this chapter, we continue our discussion of inferential
statistics—procedures for drawing conclusions about a
population based on data collected from a sample. We will
address two different statistical tests: the z test and t test. After
reading this chapter, engaging in the Critical Thinking checks,
and working through the problems at the end of each module
and at the end of the chapter, you should understand the
differences between the two tests covered in this chapter, when
to use each test, how to use each to test a hypothesis, and the
assumptions of each test.
MODULE 9
The Single-Sample z Test
Learning Objectives
•Explain what a z test is and what it does.
•Calculate a z test.
•Explain what statistical power is and how to make statistical
tests more powerful.
•List the assumptions of the z test.
•Calculate confidence intervals using the z distribution.
The z Test: What It Is and What It Does
The z test is a parametric statistical test that allows you to test
the null hypothesis for a single sample when the population
variance is known. This procedure allows us to compare a
sample to a population in order to assess whether the sample
differs significantly from the population. If the sample was
drawn randomly from a certain population (children in academic
8. after-school programs) and we observe a difference between the
sample and a broader population (all children), we can then
conclude that the population represented by the sample differs
significantly from the comparison population.
z test A parametric inferential statistical test of the null
hypothesis for a single sample where the population variance is
known.
Let's return to our example from the previous module and
assume that we have actually collected IQ scores from 75
students enrolled in academic after-school programs. We want
to determine whether the sample of children in academic after-
school programs represents a population with a mean IQ greater
than the mean IQ of the general population of children. As
stated previously, we already know μ (100) and σ (15) for the
general population of children. The null and alternative
hypotheses for a one-tailed test are:
H0:μ0≤μ1,orμacademicprogram≤μgeneralpopulationH0:μ0≤μ1, o
r μacademic program ≤μgeneral population
H0:μ0>μ1,orμacademicprogram>μgeneralpopulationH0:μ0>μ1,
or μacademic program >μgeneral population
In Module 6 we learned how to calculate a z score for a single
data point (or a single individual's score). To review, the
formula for a z score is:
z=X−μσz=X−μσ
where
X = each individual score
μ = the population mean
σ = the population standard deviation
Remember that a z score tells us how many standard deviations
above or below the mean of the distribution an individual score
falls. When using the z test, however, we are not comparing an
individual score to the population mean. Instead, we are
comparing a sample mean to the population mean. We therefore
cannot compare the sample mean to a population distribution of
individual scores. We must compare it instead to a distribution
of sample means, known as the sampling distribution.
9. The Sampling Distribution
If you are becoming confused, think about it this way.
A sampling distribution is a distribution of sample means based
on random samples of a fixed size from a population. Imagine
that we have drawn many different samples of some size (say
75) from the population (children whose IQ can be measured).
For each sample that we draw, we calculate the mean; then we
plot the means of all the samples. What do you think the
distribution will look like? Well, most of the sample means will
probably be similar to the population mean of 100. Some of the
sample means will be slightly lower than 100; some will be
slightly higher than 100; and others will be right at 100. A few
of the sample means, however, will not be similar to the
population mean. Why? Based on chance, some samples will
contain some of the rare individuals with either very high IQ
scores or very low IQ scores. Thus, the means for those samples
will be much lower than 100 or much higher than 100. Such
samples, however, will be few in number. Hence, the sampling
distribution (the distribution of sample means) will be normal
(bell-shaped), with most of the sample means clustered around
100 and a few sample means in the tails or the extremes.
Therefore, the mean for the sampling distribution will be the
same as the mean for the distribution of individual scores (100).
sampling distribution A distribution of sample means based on
random samples of a fixed size from a population.
The Standard Error of the Mean
Here is a more difficult question: Would the standard deviation
of the sampling distribution, known as the standard error of the
mean, be the same as that for a distribution of individual
scores? We know that σ = 15 for the distribution of individual
IQ test scores. Would the variability in the sampling
distribution be as great as it is in a distribution of individual
scores? Let's think about it. The sampling distribution is a
distribution of sample means. In our example, each sample has
75 people in it. Now, the mean for a sample of 75 people could
never be as low or as high as the lowest or highest individual
10. score. Why? Most people have IQ scores around 100. This
means that in each of the samples, most people will have scores
around 100. A few people will have very low scores, and when
they are included in the sample, they will pull the mean for that
sample down. A few others will have very high scores, and
these scores will raise the mean for the sample in which they
are included. A few people in a sample of 75, however, can
never pull the mean for the sample as low as a single
individual's score might be or as high as a single individual's
score might be. For this reason, the standard error of the mean
(the standard deviation for the sampling distribution) can never
be as large as σ (the standard deviation for the distribution of
individual scores).
standard error of the mean The standard deviation of the
sampling distribution.
How does this relate to the z test? A z test uses the mean and
standard deviation for the sampling distribution to determine
whether the sample mean is significantly different from the
population mean. Thus, we need to know the mean (μ) and the
standard error of the mean(σ¯¯¯X)(σX¯)for the sampling
distribution. We have already said that μ for the sampling
distribution is the same as μ for the distribution of individual
scores—100. How will we determine what σ¯¯¯XσX¯ is?
To find the standard error of the mean, we would need to draw a
number of samples from the population, determine the mean for
each sample, and then calculate the standard deviation for this
distribution of sample means. This is hardly feasible. Luckily
for us, there is a method of finding the standard error of the
mean without doing all of this. This is based on the central limit
theorem. The central limit theorem is a precise description of
the distribution that would be obtained if you selected every
possible sample, calculated every sample mean, and constructed
the distribution of sample means. The central limit
theorem states that for any population with mean μ and standard
deviation σ, the distribution of sample means for sample
size N will have a mean of μ and a standard deviation
11. of σ/√Nσ/N and will approach a normal distribution
as N approaches infinity. Thus, according to the central limit
theorem, in order to determine the standard error of the mean
(the standard deviation for the sampling distribution) we take
the standard deviation for the population (σ) and divide by the
square root of the sample size√N:N:
σ¯¯¯X=σ√NσX¯=σN
central limit theorem A theorem which states that for any
population with mean μ and standard deviation σ, the
distribution of sample means for sample size N will have a
mean of μ and a standard deviation of σ/√Nσ/N and will
approach a normal distribution as N approaches infinity.
We can now use this information to calculate the z test. The
formula for the z test is
z=¯¯¯X−μσ¯¯¯Xz=X¯−μσX¯
where
¯¯¯XX¯ = sample mean
μ = mean of the sampling distribution
σ¯¯¯XσX¯ = standard deviation of the sampling distribution, or
standard error of the mean
THE z TEST (PART I)
Concept
Description
use
Sampling Distribution
A distribution of sample means where each sample is the same
size (N)
Used for comparative purposes for z tests—a sample mean is
compared with the sampling distribution to assess the likelihood
that the sample is part of the sampling distribution
Standard Error of the Mean (σ¯¯¯X)(σX¯)
The standard deviation of a sampling distribution, determined
by dividing σ by √NN
Used in the calculation of z
z Test
12. Indication of the number of standard deviation units the sample
mean is from the mean of the sampling distribution
An inferential test that compares a sample mean with the
sampling distribution in order to determine the likelihood that
the sample is part of the sampling distribution
1.Explain how a sampling distribution differs from a
distribution of individual scores.
2.Explain the difference between σ¯¯¯XσX¯ and σ.
3.How is a z test different from a z score?
Calculations for the One-Tailed z Test
You can see that the formula for a z test represents finding the
difference between the sample mean (¯¯¯X)(X¯) and the
population mean (μ) and then dividing by the standard error of
the mean (σ¯¯¯X)(σX¯). This will tell us how many standard
deviation units a sample mean is from the population mean, or
the likelihood that the sample is from that population. We
already know μ and σ, so all we need is to find the mean for the
sample (¯¯¯X)(X¯) and to calculate σ¯¯¯XσX¯ based on a
sample size of 75.
Suppose we find that the mean IQ score for the sample of 75
children enrolled in academic after-school programs is 103.5.
We can calculate σ¯¯¯XσX¯ based on knowing the sample size
and σ.
σ¯¯¯X=σ√N=15√75=158.66=1.73σX¯=σN=1575=158.66=1.73
We now use σ¯¯¯XσX¯ (1.73) in the z test formula.
z=¯¯¯X−μσ¯¯¯X=103.5−1001.73=3.51.73=+2.02z=X¯−μσX¯=1
03.5−1001.73=3.51.73=+2.02
Instructions on using the TI-84 calculator to conduct this one-
tailed z test appear in the Statistical Software Resources section
at the end of this chapter.
Interpreting the One-Tailed z Test
Figure 9.1 represents where the sample mean of 103.5 lies with
respect to the population mean of 100. The z test score of 2.02
can be used to test our hypothesis that the sample of children in
the academic after-school program represents a population with
13. a mean IQ greater than the mean IQ for the general population.
To do this, we need to determine whether the probability is high
or low that a sample mean as large as 103.5 would be chosen
from this sampling distribution. In other words, is a sample
mean IQ score of 103.5 far enough away from, or different
enough from, the population mean of 100 for us to say that it
represents a significant difference with an alpha level of .05 or
less?
How do we determine whether a z score of 2.02 is statistically
significant? Because the sampling distribution is normally
distributed, we can use the area under the normal curve (Table
A.1 in Appendix A). When we discussed z scores in Module 6,
we saw that Table A.1 provides information on the proportion
of scores falling between μ and the z score and the proportion of
scores beyond the z score. To determine whether a z test is
significant, we can use the area under the curve to determine
whether the chance of a given score's occurring is 5% or less. In
other words, is the score far enough away from (above or
below) the mean that only 5% or less of the scores are as far or
farther away?
Using Table A.1, we find that the z score that marks off the top
5% of the distribution is 1.645. This is referred to as
the zcritical value, or zcv. For us to conclude that the sample
mean is significantly different from the population mean, then,
the sample mean must be at least ±1.645 standard deviations
(z units) from the mean. The critical value of 1.645 is
illustrated in Figure 9.2. The z we obtained for our sample mean
(zobt) is 2.02, and this value falls within the region of rejection
for the null hypothesis. We therefore reject H0 which states that
the sample mean represents the general population mean and
support our alternative hypothesis that the sample mean
represents a population of children in academic after-school
programs whose mean IQ is greater than 100. We make this
decision because the z test score for the sample is larger than
(further out in the tail than) the critical value of ±1.645. In APA
style, it would be reported as follows: z (N = 75) = 2.02, p < .05
14. (one-tailed).
critical value The value of a test statistic that marks the edge of
the region of rejection in a sampling distribution, where values
equal to it or beyond it fall in the region of rejection.
FIGURE 9.1 The obtained mean in relation to the population
meanFIGURE 9.2 The z critical value and the z obtained for
the z test example
Keep in mind that when a result is significant, the p value
(the a level, or probability of a Type I error) is reported as less
than (<) .05 (or some smaller probability) not greater than (>)—
an error commonly made by students. Remember the p value, or
alpha level, indicates the probability of a Type I error. We want
this probability to be small, meaning we are confident that there
is only a small probability that our results were due to chance.
This means it is highly probable that the observed difference
between the sample mean and the population mean is truly a
meaningful difference.
The test just conducted was a one-tailed test, because we
predicted that the sample would score higher than the
population. What if this were reversed? For example, imagine I
am conducting a study to see whether children in athletic after-
school programs weigh less than children in the general
population. Can you determine what H0 and Ha are for this
example?
H0:μ0≥μ1,orμweightofchildreninathleticporgrams≥μweightofchi
ldreningeneralpopulationH0:μ0≥μ1, or μweight of children in ath
letic porgrams≥μweight of children in general population
H0:μ0<μ1,orμweightofchildreninathleticporgrams<μweightofchi
ldreningeneralpopulationH0:μ0<μ1, or μweight of children in ath
letic porgrams<μweight of children in general population
Assume that the mean weight of children in the general
population (μ) is 90 pounds, with a standard deviation (σ) of 17
pounds. You take a random sample (N = 50) of children in
athletic after-school programs and find a mean
weight(¯¯¯X)(X¯) of 86 pounds Given this information, you can
now test the hypothesis that the sample of children in the
15. athletic after-school program represents a population with a
mean weight that is less than the mean weight for the general
population of children.
First, we calculate the standard error of the
mean (σ¯¯¯X)(σX¯).
σ¯¯¯X=σ√N=17√50=177.07=2.40σX¯=σN=1750=177.07=2.40
Now, we enter σ¯¯¯XσX¯ into the z test formula.
z=¯¯¯X−μσ¯¯¯X=86−902.40=−42.40=−1.67z=X¯−μσX¯=86−90
2.40=−42.40=−1.67
The z score for this sample mean is −1.67, meaning that it falls
1.67 standard deviations below the mean. The critical value for
a one-tailed test was 1.645 standard deviations. This means
the z score has to be at least 1.645 standard deviations away
from (above or below) the mean in order to fall in the region of
rejection. In other words, the critical value for a one-
tailed z test is ±1.645.
Is our z score at least that far away from the mean? It is, but
just barely. Therefore, we reject H0 and support Ha—that
children in the athletic after-school programs weigh
significantly less than children in the general population and
hence represent a population of children who weigh less. In
APA style, this would be written as z (N = 50) = −1.67, p <.05
(one-tailed). Instructions on using the TI-84 calculator to
conduct this one-tailed z test appear in the Statistical Software
Resources section at the end of this chapter.
Calculations for the Two-Tailed z Test
So far, we have completed two z tests, both one-tailed. Let's
turn now to a two-tailed z test. Remember that a two-tailed test
is also known as a nondirectional test—a test in which the
prediction is simply that the sample will perform differently
from the population, with no prediction as to whether the
sample mean will be lower or higher than the population mean.
Suppose that in the previous example we used a two-tailed
rather than a one-tailed test. We expect the weight of the
children in the athletic after-school program to differ from that
of children in the general population, but we are not sure
16. whether they will weigh less (because of the activity) or more
(because of greater muscle mass). H0 and Ha for this two-tailed
test appear next. See if you can determine what they would be
before you continue reading.
H0:μ0=μ1,orμathleticprograms=μgeneralpopulationH0:μ0=μ1, o
r μathletic programs=μgeneral population
H0:μ0≠μ1,orμathleticprograms≠μgeneralpopulationH0:μ0≠μ1, o
r μathletic programs≠μgeneral population
Let's use the same data as before: The mean weight of children
in the general population (μ) is 90 pounds, with a standard
deviation (σ) of 17 pounds; for children in the sample (N = 50),
the mean weight (¯¯¯X)(X¯) is 86 pounds. Using this
information, you can now test the hypothesis that children in
athletic after-school programs differ in weight from those in the
general population. Notice that the calculations will be exactly
the same for this z test. That is, σ¯¯¯XσX¯ and the z score will
be exactly the same as before. Why? All of the measurements
are exactly the same. To review:
σ¯¯¯X=σ√N=17√50=177.07=2.40σX¯=σN=1750=177.07=2.40
z=¯¯¯X−μσ¯¯¯X=86−902.40=−42.40=−1.67z=X¯−μσX¯=86−90
2.40=−42.40=−1.67
Interpreting the Two-Tailed z Test
If we end up with the same z score, how does a two-tailed test
differ from a one-tailed test? The difference is in the z critical
value (zcv). In a two-tailed test, both halves of the normal
distribution have to be taken into account. Remember that with
a one-tailed test, the zcv was ±1.645; this z score was so far
away from the mean (either above or below) that only 5% of the
scores were beyond it. How will the zcv for a two-tailed test
differ?
With a two-tailed test, the zcv has to be so far away from the
mean that a total of only 5% of the scores are beyond
it (both above and below the mean). A zcv of ±1.645 leaves 5%
of the scores above the positive zcv and 5% below the
negative zcv. If we take both sides of the normal distribution
into account (which we do with a two-tailed test because we do
17. not predict whether the sample mean will be above or below the
population mean), then 10% of the distribution will fall beyond
the two critical values. Thus, ±1.645 cannot be the critical value
for a two-tailed test because this leaves too much chance (10%)
operating.
To determine the zcv for a two-tailed test, then, we need to find
the z score that is far enough away from the population mean
that only 5% of the distribution—taking into account both
halves of the distribution—is beyond the score. Because Table
A.1 represents only half of the distribution, we need to look for
the z score that leaves only 2.5% of the distribution beyond it.
Then, when we take into account both halves of the distribution,
5% of the distribution will be accounted for (2.5% + 2.5% =
5%). Can you determine what zscore this would be, using Table
A.1?
If you concluded that it would be ±1.96, then you are correct.
This is the z score that is far enough away from the population
mean (using both halves of the distribution) that only 5% of the
distribution is beyond it. The critical values for both one- and
two-tailed tests are illustrated in Figure 9.3.
FIGURE 9.3 Regions of rejection and critical values for one-
tailed versus two-tailed testsFIGURE 9.4 The z critical value
and the z obtained for the two-tailed z test example
Okay, what do we do with this critical value? We use it exactly
the same way as we did the zcv for a one-tailed test. In other
words, the zobt has to be as large as or larger than the zcv in
order for us to reject H0. Is our zobt as large as or larger than
±1.96? No (this is illustrated in Figure 9.4). Our zobt was −1.67
and not in the region of rejection. We therefore fail to
reject H0 and conclude that the weight of children in the
athletic after-school program does not differ significantly from
the weight of children in the general population. Instructions on
using the TI-84 calculator to conduct this two-tailed z test
appear in the Statistical Software Resources section at the end
of this chapter.
With exactly the same data (sample size, μ, σ, ¯¯¯XX¯,
18. and σ¯¯¯XσX¯), we rejected H0 using a one-tailed test and
failed to reject H0 with a two-tailed test. How can this be? The
answer is that a one-tailed test is statistically a more powerful
test than a two-tailed test. Statistical power refers to the
probability of correctly rejecting a false H0. With a one-tailed
test, you are more likely to reject H0 because the zobt does not
have to be as large (as far away from the population mean) to be
considered significantly different from the population mean.
(Remember, the zcv for a one-tailed test is ±1.645, but for a
two-tailed test, it is ±1.96.)
statistical power The probability of correctly rejecting a
false H0.
Statistical Power
Let's think back to the discussion of Type I and Type II errors
in the previous module. We said that in order to reduce your
risk of a Type I error, you need to lower the alpha level—for
example, from .05 to .01. We also noted, however, that lowering
the alpha level increases the risk of a Type II error. How, then,
can we reduce our risk of a Type I error but not increase our
risk of a Type II error? As we just noted, a one-tailed test is
more powerful—you do not need as large a zcv in order to
reject H0. Here, then, is one way to maintain an alpha level of
.05 but increase your chances of rejecting H0. Of course,
ethically you cannot simply choose to adopt a one-tailed test for
this reason. The one-tailed test should be adopted because you
truly believe that the sample will perform above (or below) the
mean.
By what other means can we increase statistical power? Look
back at the z test formula. We know that the larger the zobt, the
greater the chance that it will be significant (as large as or
larger than the zcv) and that we can therefore reject H0. What
could we change in our study that might increase the size of
the zobt? Well, if the denominator in the z formula were a
smaller number, then the zobt would be larger and more likely
to fall in the region of rejection. How can we make the
denominator smaller? The denominator is σ¯¯¯XσX¯. Do you
19. remember the formula for σ¯¯¯XσX¯?
σ¯¯¯X=σ√NσX¯=σN
It is very unlikely that we can change or influence the standard
deviation for the population (σ). What part of
the σ¯¯¯XσX¯ formula can we influence? The sample size (N).
If we increase sample size, what will happen to σ¯¯¯XσX¯?
Let's see. We'll use the same example as before, a two-tailed
test with all of the same measurements. The only difference will
be in sample size. Thus, the null and alternative hypotheses will
be
H0:μ0=μ1,orμweightofchildreninathleticafter-
schoolprograms=μweightofchildreningeneralpopulationH0:μ0≠μ
1,orμweightofchildreninathleticafter-
schoolprograms≠μweightofchildreningeneralpopulationH0:μ0=μ
1, or μweight of children in athletic after-
school programs = μweight of children in general populationH0:
μ0≠μ1, or μweight of children in athletic after-
school programs ≠ μweight of children in general population
The mean weight of children in the general population (μ) is
once again 90 pounds, with a standard deviation (σ) of 17
pounds, and the sample of children in the after-school program
again has a mean (¯¯¯X)(X¯) weight of 86 pounds. The only
difference will be in sample size. In this case, our sample has
100 children in it. Let's test the hypothesis (conduct the z test)
for these data.
σ¯¯¯X=17√100=1710=1.70σX¯=17100=1710=1.70
z=86−901.70=−41.70=−2.35z=86−901.70=−41.70=−2.35
Do you see what happened when we increased sample size? The
standard error of the mean (σ¯¯¯X)(σX¯) decreased (we will
discuss why in a minute), and the zobt increased—in fact, it
increased to the extent that we can now reject H0 with this two-
tailed test because our zobt of −2.35 is further away from the
mean than the zcv of −1.96. Therefore, another way to increase
statistical power is to increase sample size.
Why does increasing sample size decrease (σ¯¯¯X)(σX¯)? Well,
you can see why based on the formula, but let's think back to
20. our earlier discussion about σ¯¯¯XσX¯ We said that it was the
standard deviation for a sampling distribution—a distribution of
sample means of a set size. If you recall the IQ example we
used in our discussion of σ¯¯¯XσX¯ and the sampling
distribution, we said that μ = 100 and σ = 15. We discussed
what σ¯¯¯XσX¯ would be for a sampling distribution in which
each sample mean was based on a sample size of 75. We further
noted that σ¯¯¯XσX¯ would always be smaller (have less
variability) than σ because it represents the standard deviation
of a distribution of sample means, not a distribution of
individual scores. What, then, will increasing sample size do
to σ¯¯¯XσX¯? If each sample in the sampling distribution had
100 people in it rather than 75, what do you think this would do
to the distribution of sample means?
As we noted earlier, most people in a sample will be close to
the mean (100), with only a few people in each sample
representing the tails of the distribution. If we increase sample
size to 100, we will have 25 more people in each sample. Most
of them will probably be close to the population mean of 100;
therefore, each sample mean will probably be closer to the
population mean of 100. Thus, a sampling distribution based on
samples of N = 100 rather than N = 75 will have less variability,
which means that σ¯¯¯XσX¯ will be smaller.
Assumptions and Appropriate Use of the z Test
As noted earlier in the module, the z test is a parametric
inferential statistical test for hypothesis testing. Parametric
tests involve the use of parameters or population characteristics.
With a z test, the parameters, such as m and s, are known. If
they are not known, the z test is not appropriate. Because
the z test involves the calculation and use of a sample mean, it
is appropriate for use with interval or ratio data. In addition,
because we use the area under the normal curve (Table A.1), we
are assuming that the distribution of random samples is normal.
Small samples often fail to form a normal distribution.
Therefore, if the sample size is small (N < 30), the z test may
not be appropriate. In cases where the sample size is small, or
21. where s is not known, the appropriate test would be the t test,
discussed in the next module.
THE z TEST (PART II)
Concept
Description
Examples
One-Tailed z Test
A directional inferential test in which a prediction is made that
the population represented by the sample will be either above or
below the general population
Ha:μ0<μ1orHa:μ0>μ1Ha:μ0<μ1orHa:μ0>μ1
Two-Tailed z Test
A nondirectional inferential test in which the prediction is made
that the population represented by the sample will differ from
the general population, but the direction of the difference is not
predicted
Ha:μ0≠μ1Ha:μ0≠μ1
Statistical Power
The probability of correctly rejecting a false H0
One-tailed tests are more powerful; increasing sample size
increases power
1.Imagine that I want to compare the intelligence level of
psychology majors to the intelligence level of the general
population of college students. I predict that psychology majors
will have higher IQ scores. Is this a one- or two-tailed test?
Identify H0 and Ha.
2.Conduct the z test for the previous example. Assume that μ =
100, σ = 15, ¯¯¯XX¯ = 102.75, and N = 60. Should we
reject H0 or fail to reject H0?
Confidence Intervals Based on the z Distribution
In this text, hypothesis tests such as the previously
described z test are the main focus. However, sometimes social
and behavioral scientists use estimation of population means
based on confidence intervals rather than statistical hypothesis
22. tests. For example, imagine that you want to estimate a
population mean based on sample data (a sample mean). This
differs from the previously described z test in that we are not
determining whether the sample mean differs significantly from
the population mean; rather, we are estimating the population
mean based on knowing the sample mean. We can still use the
area under the normal curve to accomplish this—we simply use
it in a slightly different way.
Let's use the previous example in which we know the sample
mean weight of children enrolled in athletic after-school
programs (¯¯¯XX¯ = 86), σ (17), and the sample size (N = 100).
However, imagine that we do not know the population mean (μ).
In this case, we can calculate a confidence interval based on
knowing the sample mean and s. A confidence intervalis an
interval of a certain width, which we feel “confident” will
contain μ. We want a confidence interval wide enough that we
feel fairly certain it contains the population mean. For example,
if we want to be 95% confident, we want a 95% confidence
interval.
confidence interval An interval of a certain width that we feel
confident will contain μ.
How can we use the area under the standard normal curve to
determine a confidence interval of 95%? We use the area under
the normal curve to determine the z scores that mark off the
area representing 95% of the scores under the curve. If you
consult Table A.1, you will find that 95% of the scores will fall
between ±1.96 standard deviations above and below the mean.
Thus, we could determine which scores represent ±1.96 standard
deviations from the mean of 86. This seems fairly simple, but
we must remember that we are dealing with a distribution of
sample means (the sampling distribution) and not with a
distribution of individual scores. Thus, we must convert the
standard deviation (σ) to the standard error of the mean
(σ¯¯¯XσX¯, the standard deviation for a sampling distribution)
and use the standard error of the mean in the calculation of a
confidence interval. Remember we calculate σ¯¯¯XσX¯ by
23. dividing σ by the square root of N.
σ¯¯¯X=17√100=1710=1.70σX¯=17100=1710=1.70
We can now calculate the 95% confidence interval using the
following formula:
CI=¯¯¯X±z(σ¯¯¯X)CI=X¯ ± z(σX¯)
where
¯¯¯X=thesamplemeanσ¯¯¯X=thestandarderrorofthemean,andz=t
hezscorerepresentingthedisiredconfidenceintervalX¯=the sample
meanσX¯=the standard error of the
mean, andz=the z score representing the disired confidence inter
val
Thus:
Cl=86±1.96(1.70)=86±3.332=82.668−89.332Cl=86 ±1.96(1.70)=
86±3.332=82.668−89.332
Thus, the 95% confidence interval ranges from 82.67 to 89.33.
We would conclude, based on this calculation, that we are 95%
confident that the population mean lies within this interval.
What if we wanted to have greater confidence that our
population mean is contained in the confidence interval? In
other words, what if we want to be 99% confident? We would
have to construct a 99% confidence interval. How would we go
about doing this?
We would do exactly what we did for the 95% confidence
interval. First, we would consult Table A.1 to determine
what z scores mark off 99% of the area under the normal curve.
We find that z scores of ±2.58 mark off 99% of the area under
the curve. We then apply the same formula for a confidence
interval used previously.
Cl=¯¯¯X±z(σ¯¯x)Cl=86±2.58(1.70)=86±4.386=81.614−90.386C
l=X¯±z(σx¯)Cl=86±2.58(1.70)=86±4.386=81.614−90.386
Thus, the 99% confidence interval ranges from 81.61 to 90.39.
We would conclude, based on this calculation, that we are 99%
confident that the population mean lies within this interval.
Typically, statisticians recommend using a 95% or a 99%
confidence interval. However, using Table A.1 (the area under
the normal curve), you could construct a confidence interval of
24. 55%, 70%, or any percentage you desire.
It is also possible to do hypothesis testing with confidence
intervals. For example, if you construct a 95% confidence
interval based on knowing a sample mean and then determine
that the population mean is not in the confidence interval, the
result is significant. For example, the 95% confidence interval
we constructed earlier of 82.67−89.33 did not include the actual
population mean reported earlier in the module (μ = 90). Thus,
there is less than a 5% chance that this sample mean could have
come from this population—the same conclusion we reached
when using the z test earlier in the module.
REVIEW OF KEY TERMS
central limit theorem (p. 142)
confidence interval (p. 151)
critical value (p. 144)
sampling distribution (p. 141)
standard error of the mean (p. 141)
statistical power (p. 148)
z test (p. 140)
MODULE EXERCISES
(Answers to odd-numbered questions appear in Appendix B.)
1.What is a sampling distribution?
2.Explain why the mean of a sampling distribution is the same
as the mean for a distribution of individual scores, but the
standard deviation for a sampling distribution differs from the
standard deviation of a distribution of individual scores.
3.Explain why the zcv for a one-tailed test differs from that for
a two-tailed test.
4.Explain what statistical power is and how we can increase
power.
5.A researcher is interested in whether students who attend
private high schools have higher average SAT scores than
students in the general population of high school students. A
random sample of 90 students at a private high school is tested
and has a mean SAT score of 1,050. The average for public high
school students is 1,000 (σ = 200).
25. a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute zobt.
d.What is zcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
6.The producers of a new toothpaste claim that it prevents more
cavities than other brands of toothpaste. In other words, those
who use the new toothpaste should have fewer cavities than
those who use other brands. A random sample of 60 people uses
the new toothpaste for six months. The mean number of cavities
at their next checkup is 1.5. In the general population, the mean
number of cavities at a six-month checkup is 1.73 (σ = 1.12).
a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute zobt.
d.What is zcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
CRITICAL THINKING CHECK ANSWERS
Critical Thinking Check 9.1
1.A sampling distribution is a distribution of sample means.
Thus, rather than representing scores for individuals, the
sampling distribution plots the means of samples of a set size.
2.σ¯¯¯XσX¯ is the standard deviation for a sampling
distribution. It therefore represents the standard deviation for a
distribution of sample means. σ is the standard deviation for a
population of individual scores rather than sample means.
3.A z test compares the performance of a sample to the
performance of the population by indicating the number of
standard deviation units the sample mean is from the mean of
the sampling distribution. A z score indicates how many
standard deviation units an individual score is from the
population mean.
26. CRITICAL THINKING CHECK 9.2
1.Predicting that psychology majors will have higher IQ scores
makes this a one-tailed test.
H0:μpsychologymajors≤μgeneralpopulationHa:μpsychologymajo
rs>μgeneralpopulationH0:μpsychology majors ≤μgeneral populat
ionHa:μpsychology majors >μgeneral population
2.σ¯¯¯X=15√60=157.75=1.94σX¯=1560=157.75=1.94
z=102.75−1001.94=2.751.94=+1.42z=102.75−1001.94=2.751.94
=+1.42
Because this is a one-tailed test, zcv = ±1.645.
The zobt = +1.42. We therefore fail to reject H0; psychology
majors do not differ significantly on IQ scores in comparison to
the general population of college students.
MODULE 10
The Single-Sample t Test
Learning Objectives
•Explain what a t test is and what it does.
•Calculate a t test.
•List the assumptions of the t test.
•Calculate confidence intervals using the t distribution.
The t Test: What It Is and What It Does
The t test for a single sample is similar to the z test in that it is
also a parametric statistical test of the null hypothesis for a
single sample. As such, it is a means of determining the number
of standard deviation units a score is from the mean (μ) of a
distribution. With a t test, however, the population variance is
not known. Another difference is that tdistributions, although
symmetrical and bell-shaped, are not normally distributed. This
means that the areas under the normal curve that apply for
the z test do not apply for the t test.
t test A parametric inferential statistical test of the null
hypothesis for a single sample where the population variance is
not known.
Student's t Distribution
27. The t distribution, known as Student's t distribution, was
developed by William Sealey Gosset, a chemist, who worked for
the Guinness Brewing Company of Dublin, Ireland, at the
beginning of the 20th century. Gosset noticed that when
working with small samples of beer (N < 30) chosen for quality-
control testing, the sampling distribution of the means was
symmetrical and bell-shaped, but not normal. Therefore, the
proportions under the standard normal curve did not apply. In
other words, with small sample sizes, the curve was
symmetrical, but it was not the standard normal curve. As the
size of the samples in the sampling distribution increased, the
sampling distribution approached the normal distribution and
the proportions under the curve became more similar to those
under the standard normal curve. He eventually published his
findings under the pseudonym “Student”; and with the help of
Karl Pearson, a mathematician, he developed a general formula
for the t distributions (Peters, 1987; Stigler, 1986; Tankard,
1984).
Student's t distribution A set of distributions that, although
symmetrical and bell-shaped, are not normally distributed.
We refer to t distributions in the plural because unlike
the z distribution, of which there is only one, the t distributions
are a family of symmetric distributions that differ for each
sample size. As a result, the critical value indicating the region
of rejection changes for samples of different sizes. As the size
of the samples increases, the t distribution approaches the z or
normal distribution. Table A.2 in Appendix A at the back of
your book provides the t critical values (tcv) for both one- and
two-tailed tests for various sample sizes and alpha levels.
Notice, however, that although we have said that the critical
value depends on sample size, there is no column in the table
labeled N for sample size. Instead, there is a column labeled df,
which stands for degrees of freedom. The degrees of freedom
are related to sample size. For example, assume that you are
given six numbers, 2, 5, 6, 9, 11, and 15. The mean of these
numbers is 8. If you are told that you can change the numbers as
28. you like, but that the mean of the distribution must remain at 8,
how many numbers can you change arbitrarily? You can change
five of the six numbers arbitrarily. Once you have changed five
of the numbers arbitrarily, the sixth number is determined by
the qualification that the mean of the distribution must equal 8.
Therefore, in this distribution of six numbers, five are free to
vary. Thus, there are five degrees of freedom. For any single
distribution, then, df= N − 1.
degrees of freedom (df) The number of scores in a sample that
are free to vary.
Look again at Table A.2, and notice what happens to the critical
values as the degrees of freedom increase. Look at the column
for a one-tailed test with alpha equal to .05 and degrees of
freedom equal to 10. The critical value is ±1.812. This is larger
than the critical value for a one-tailed z test, which was ±1.645.
Because we are dealing with smaller, nonnormal distributions
when using the t test, the t score must be farther away from the
mean in order for us to conclude that it is significantly different
from the mean. What happens as the degrees of freedom
increase? Look in the same column—one-tailed test, alpha =
.05—for 20 degrees of freedom. The critical value is ±1.725,
smaller than the critical value for 10 degrees of freedom.
Continue to scan down the same column, one-tailed test and
alpha=.05, until you reach the bottom, where df = ∞. Notice that
the critical value is ±1.645, the same as it is for a one-
tailed z test. Thus, when the sample size is large,
the t distribution is the same as the z distribution.
Calculations for the One-Tailed t Test
Let's illustrate the use of the single-sample t test to test a
hypothesis. Assume the mean SAT score of students admitted to
General University is 1,090. Thus, the university mean of 1,090
is the population mean (μ). The population standard deviation is
unknown. The members of the Biology Department believe that
students who decide to major in biology have higher SAT scores
than the general population of students at the university. The
null and alternative hypotheses are thus
29. H0:μ0≤μ1,orμbiologystudents≤μgeneralpopulationHa:μ0≤μ1,orμ
biologystudents>μgeneralpopulationH0:μ0≤μ1, or μbiology stud
ents ≤μgeneral populationHa:μ0≤μ1, or μbiology students >μgen
eral population
TABLE 10-1 SAT scores for a sample of 10 biology majors
x
1,010
1,200
1,310
1,075
1,149
1,078
1,129
1,069
1,350
1,390
ΣX = 11,760
¯¯¯X=ΣXN=11,76010=1,176X¯=ΣXN=11,76010=1,176
Notice that this is a one-tailed test because the researchers
predict that the biology students will perform higher than the
general population of students at the university. The researchers
now need to obtain the SAT scores for a sample of biology
majors. This information is provided in Table 10.1, which
30. shows that the mean SAT score for the sample is 1,176. This
sample mean represents our estimate of the population mean
SAT score for biology majors.
The Estimated Standard Error of the Mean
The t test will tell us whether this mean differs significantly
from the university mean of 1,090. Because we have a small
sample (N = 10) and because we do not know σ, we must
conduct a t test rather than a z test. The formula for the t test is
t=¯¯¯X−μS¯¯¯Xt=X¯−μSX¯
This looks very similar to the formula for the z test that we used
in Module 9. The only difference is in the denominator,
where S¯¯¯XSX¯ (the estimated standard error of the meanof
the sampling distribution) has been substituted forσ¯¯¯XσX¯.
We useS¯¯¯XSX¯ rather than σ¯¯¯XσX¯ because we do not
know σ (the standard deviation for the population) and thus
cannot calculate σ¯¯¯XσX¯. We can, however, determine s (the
unbiased estimator of the population standard deviation) and,
based on this, we can determine S¯¯¯XSX¯. The formula
for S¯¯¯XSX¯ is
estimated standard error of the mean An estimate of the
standard deviation of the sampling distribution.
S¯¯¯X=s√NSX¯=sN
We must first calculate s (the estimated standard deviation for a
population, based on sample data) and then use this to calculate
the estimated standard error of the mean (S¯¯¯X)(SX¯). The
formula for s, which we learned in Module 3, is
S=√Σ(X−¯¯¯X)2N−1S=Σ(X−X¯)2N−1
Using the information in Table 10.1, we can use this formula to
calculate s.
S=√156,3529=√17,372.44=131.80S=156,3529=17,372.44=131.8
0
Thus, the unbiased estimator of the standard deviation (s) is
131.80. We can now use this value to calculate S¯¯¯XSX¯, the
estimated standard error of the sampling distribution.
S¯¯¯X=S√N=131.80√10=131.803.16=41.71SX¯=SN=131.8010=
131.803.16=41.71
31. Finally, we can use this value for S¯¯¯XSX¯ to calculate t.
t=¯¯¯X−μS¯¯¯X=1,176−1,09041.71=8641.71=+2.06t=X¯−μSX¯
=1,176−1,09041.71=8641.71=+2.06
Interpreting the One-Tailed t Test
Our sample mean falls 2.06 standard deviations above the
population mean of 1,090. We must now determine whether this
is far enough away from the population mean to be considered
significantly different. In other words, is our sample mean far
enough away from the population mean that it lies in the region
of rejection? Because this is a one-tailed alternative hypothesis,
the region of rejection is in one tail of the sampling
distribution. Consulting Table A.2 for a one-tailed test with
alpha =.05 and df = N − 1 = 9, we see that tcv = ±1.833.
The tobt of 2.06 is therefore within the region of rejection. We
reject H0 and support Ha. In other words, we have sufficient
evidence to allow us to conclude that biology majors have
significantly higher SAT scores than the rest of the students at
General University. In APA style, this would be reported as t(9)
= 2.06, p < .05 (one-tailed). Figure 10.1illustrates the
obtained t with respect to the region of rejection. Instructions
on using Excel, SPSS, or the TI-84 calculator to conduct this
one-tailed single sample t test appear in the Statistical Software
Resources section at the end of this chapter.
FIGURE 10.1 The t critical value and the t obtained for the
single-sample one-tailed t test example
Calculations for the Two-Tailed t Test
What if the Biology Department had made no directional
prediction concerning the SAT scores of its students? In other
words, suppose the members of the department were unsure
whether their students' scores would be higher or lower than
those of the general population of students and were simply
interested in whether biology students differed from the
population. In this case, the test of the alternative hypothesis
would be two-tailed, and the null and alternative hypotheses
would be
H0:μ0=μ1,orμbiologystudents=μgeneralpopulationHa:μ0=μ1,orμ
32. biologystudents≠μgeneralpopulationH0:μ0=μ1, or μbiology stud
ents=μgeneral populationHa:μ0=μ1, or μbiology students≠μgene
ral population
Assuming that the sample of biology students is the
same, ¯¯¯XX¯, s, and S¯¯¯XSX¯ would all be the same. The
population at General University is also the same, so μ would
still be 1,090. Using all of this information to conduct the t test,
we end up with exactly the same t test score of + 2.06. What,
then, is the difference for the two-tailed t test? It is the same as
the difference between the one- and two-tailed z test—the
critical values differ.
Interpreting the Two-Tailed t Test
Remember that with a two-tailed alternative hypothesis, the
region of rejection is divided evenly between the two tails (the
positive and negative ends) of the sampling distribution.
Consulting Table A.2 for a two-tailed test with alpha = .05
and df= N − 1 = 9, we see that tcv = ±2.262. The tobt of 2.06 is
therefore not within the region of rejection. We do not
reject H0 and thus cannot support Ha. In other words, we do not
have sufficient evidence to allow us to conclude that the
population of biology majors differs significantly on SAT
scores from the rest of the students at General University. Thus,
with exactly the same data, we rejected H0 with a one-tailed
test, but failed to reject H0 with a two-tailed test, illustrating
once again that one-tailed tests are more powerful than two-
tailed tests. Figure 10.2 illustrates the obtained t for the two-
tailed test in relation to the regions of rejection. Instructions on
using Excel, SPSS, or the TI-84 calculator to conduct this two-
tailed single-sample t test appear in the Statistical Software
Resources section at the end of this chapter.
FIGURE 10.2 The t critical value and the t obtained for the
single-sample two-tailed test example
Assumptions and Appropriate Use of the Single-Sample t Test
The t test is a parametric test, as is the z test. As a parametric
test, the t test must meet certain assumptions. These
assumptions include that the data are interval or ratio and that
33. the population distribution of scores is symmetrical. The t test
is used in situations that meet these assumptions and in which
the population mean is known but the population standard
deviation (σ) is not known. In cases where these criteria are not
met, a nonparametric test is more appropriate. Nonparametric
tests are covered in Chapter 10.
THE t TEST
Concept
Description
Use/Examples
Estimated standard error of the mean (S¯¯¯X)(SX¯)
The estimated standard deviation of a sampling distribution,
calculated by dividing s by √NN
Used in the calculation of a t test
t test
Indicator of the number of standard deviation units the sample
mean is from the mean of the sampling distribution
An inferential statistical test that differs from the z test in that
the sample size is small (usually <30) and σ is not known
One-tailed t test
A directional inferential test in which a prediction is made that
the population represented by the sample will be either above or
below the general population
Ha:μ0<μ1orHa:μ0>μ1Ha:μ0<μ1orHa:μ0>μ1
Two-tailed t test
A nondirectional inferential test in which the prediction is made
that the population represented by the sample will differ from
the general population, but the direction of the difference is not
predicted
Ha:μ0≠μ1Ha:μ0≠μ1
1.Explain the difference in use and computation between
the z test and the t test.
2.Test the following hypothesis using the t test: Researchers are
interested in whether the pulse of long-distance runners differs
34. from that of other athletes. They suspect that the runners' pulses
will be lower. They obtain a random sample (N = 8) of long-
distance runners, measure their resting pulse, and obtain the
following data: 45, 42, 64, 54, 58, 49, 47, 55. The average
resting pulse of athletes in the general population is 60 beats
per minute.
Confidence Intervals Based on the t Distribution
You might remember from our discussion of confidence
intervals in Module 9 that they allow us to estimate population
means based on sample data (a sample mean). Thus, when using
confidence intervals, rather than determining whether the
sample mean differs significantly from the population mean, we
are estimating the population mean based on knowing the
sample mean. We can use confidence intervals with
the t distribution just as we did with the z distribution (the area
under the normal curve).
Let's use the previous example in which we know the sample
mean SAT score for the biology students (¯¯¯XX¯= 1,176), the
estimated standard error of the mean (S¯¯¯XSX¯= 41.71), and
the sample size (N= 10). We can calculate a confidence interval
based on knowing the sample mean and S¯¯¯XSX¯. Remember
that a confidence interval is an interval of a certain width,
which we feel “confident” will contain μ. We are going to
calculate a 95% confidence interval—in other words, an interval
that we feel 95% confident contains the population mean. In
order to calculate a 95% confidence interval using
the t distribution, we use Table A.2 (Critical Values for the
Student's t Distribution) to determine the critical value of t at
the .05 level. We use the .05 level because 1 minus alpha tells
us how confident we are, and in this case 1 − alpha is 1 −
.05 = 95%.
For a one-sample t test, the confidence interval is determined
with the following formula:
CI=¯¯¯X±tCV(S¯¯¯X)CI=X¯ ± tCV(SX¯)
We already know ¯¯¯XX¯ (1,176) and S¯¯¯XSX¯ (41.71), so
all we have left to determine is tcv. We use Table A.2 to
35. determine the tcv for the .05 level and a two-tailed test. We
always use the tcvfor a two-tailed test because we are
describing values both above and below the mean of the
distribution. Using Table A.2, we find that the tcv for 9 degrees
of freedom (remember df= N − 1) is 2.262. We now have all of
the values we need to determine the confidence interval.
Let's begin by calculating the lower limit of the confidence
interval:
CI=1,176−2.262(41.71)=1,176−94.35=1,081.65CI=1,176−2.262(
41.71)=1,176−94.35=1,081.65
The upper limit of the confidence interval is
CI=1,176+2.262(41.71)=1,176+94.35=1,270.35CI=1,176+2.262(
41.71)=1,176+94.35=1,270.35
Thus, we can conclude that we are 95% confident that the
interval of SAT scores from 1,081.65 to 1,270.35 contains the
population mean (μ).
As with the z distribution, we can calculate confidence intervals
for the t distribution that give us greater or less confidence (for
example a 99% confidence interval or a 90% confidence
interval). Typically, statisticians recommend using either the
95% or 99% confidence interval (the intervals corresponding to
the .05 and .01 alpha levels in hypothesis testing). You have
likely encountered such intervals in real life. They are usually
phrased in terms of “plus or minus” some amount, called
the margin of error. For example, when a newspaper reports that
a sample survey showed that 53% of the viewers support a
particular candidate, the margin of error is typically also
reported—for example, “with a ±3% margin of error.” This
means that the researchers who conducted the survey created a
confidence interval around the 53% and that if they actually
surveyed the entire population, μ would be within ±3% of the
53%. In other words, they believe that between 50% and 56% of
the viewers support this particular candidate.
REVIEW OF KEY TERMS
degrees of freedom (df) (p. 156)
estimated standard error of the mean (p. 157)
36. Student's t distribution (p. 155)
t test (p. 155)
MODULE EXERCISES
(Answers to odd-numbered questions appear in Appendix B.)
1.Explain how a t test differs from a z test.
2.Explain how S¯¯¯XSX¯ differs from σ¯¯¯XσX¯.
3.Why does tcv change when sample size changes? What must
be computed in order to determine tcv?
4.Henry performed a two-tailed test for an experiment in
which N = 24. He could not find his t table, but he remembered
the tcv at df = 13. He decided to compare his tobt to this tcv . Is
he more likely to make a Type I or a Type II error in this
situation?
5.A researcher hypothesizes that people who listen to music via
headphones have greater hearing loss and will thus score lower
on a hearing test than those in the general population. On a
standard hearing test, μ = 22.5. The researcher gives this same
test to a random sample of 12 individuals who regularly use
headphones. Their scores on the test are 16, 14, 20, 12, 25, 22,
23, 19, 17, 17, 21, 20.
a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute tobt.
d.What is tcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
6.A researcher hypothesizes that individuals who listen to
classical music will score differently from the general
population on a test of spatial ability. On a standardized test of
spatial ability, μ = 58. A random sample of 14 individuals who
listen to classical music is given the same test. Their scores on
the test are 52, 59, 63, 65, 58, 55, 62, 63, 53, 59, 57, 61, 60, 59.
a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute tobt.
37. d.What is tcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
CRITICAL THINKING CHECK ANSWERS
CRITICAL THINKING CHECK 10.1
1.The z test is used when the sample size is greater than 30 and
thus normally distributed, and σ
is known. The t test, on the other hand, is used when the sample
is smaller than 30 and bell-shaped but not normal, and s is not
known.
2.For this sample,
H0:μrunners≥μotherathletesH0:μrunners<μotherathletesH0:μrun
ners≥μother athletesH0:μrunners<μother athletes
¯¯¯X=51.75s=7.32μ=60s¯¯¯X=7.32√8=7.322.83=2.59t=51.75−6
02.59=−8.252.59=−3.19df=8−1=7tCV=±1.895tobt=−3.19X¯=51.
75s=7.32μ=60sX¯=7.328=7.322.83=2.59t=51.75−602.59=−8.25
2.59=−3.19df=8−1=7tCV=±1.895tobt=−3.19
Reject H0. The runners' pulses are significantly lower than the
pulses of athletes in general.
CHAPTER FIVE SUMMARY AND REVIEW
The z and t Tests
CHAPTER SUMMARY
Two parametric statistical tests were described in this chapter—
the z test and the t test. Each compares a sample mean to the
general population. Because both are parametric tests, the
distributions should be bell-shaped and certain parameters
should be known (in the case of the z test, μ and σ must be
known; for the t test, only μ is needed). In addition, because
they are parametric tests, the data should be interval or ratio in
scale. These tests use the sampling distribution (the distribution
of sample means). They also use the standard error of the mean
(or estimated standard error of the mean for the t test), which is
the standard deviation of the sampling distribution.
38. Both z and t tests can test one- or two-tailed alternative
hypotheses, but one-tailed tests are more powerful statistically.
CHAPTER 5 REVIEW EXERCISES
(Answers to exercises appear in Appendix B.)
Fill-in Self-Test
Answer the following questions. If you have trouble answering
any of the questions, restudy the relevant material before going
on to the multiple-choice self-test.
1.A ______________ is a distribution of sample means based on
random samples of a fixed size from a population.
2.The______________ is the standard deviation of the sampling
distribution.
3.A ______________ test is used when σ and μ are known and
the sample is 30 or larger.
4.The set of distributions that, although symmetrical and bell-
shaped, are not normally distributed is called the
______________.
5.The ______________ is a parametric statistical test of the null
hypothesis for a single sample where the population variance is
not known.
6.A ______________ test is used when μ is known but not σ and
the sample is 30 or less.
Multiple-Choice Self-Test
Select the single best answer for each of the following
questions. If you have trouble answering any of the questions,
restudy the relevant material.
1.The sampling distribution is a distribution of
a.sample means.
b.population means.
c.sample standard deviations.
d.population standard deviations.
2.A one-tailed z test, p = .05, is to ________ and a two-
tailed z test, p = .05, is to ________.
a.±1.645; ±1.96
b.±1.96; ±1.645
c.Type I error; Type II error
39. d.Type II error; Type I error
3.Which of the following is an assumption of the z test?
a.The data should be ordinal or nominal.
b.The population distribution of scores should be normal.
c.The population mean (μ) is known, but not the standard
deviation (σ).
d.The sample size is typically less than 30.
4.Which type of test is more powerful?
a.A directional test
b.A nondirectional test
c.A two-tailed test
d.All of the alternatives are equally powerful.
5.For a one-tailed test with df = 11, what is tcv?
a.1.796
b.2.201
c.1.363
d.1.895
6.Which of the following is an assumption of the t test?
a.The data should be ordinal or nominal.
b.The population distribution of scores should be skewed.
c.The population mean (μ) and standard deviation (σ) are
known.
d.The sample size is typically less than 30.
Self-Test Problems
1.A researcher is interested in whether students who play chess
have higher average SAT scores than students in the general
population. A random sample of 75 students who play chess is
tested and has a mean SAT score of 1,070. The population
average is 1,000 (σ = 200).
a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute zobt.
d.What is zcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
40. 2.A researcher hypothesizes that people who listen to classical
music have higher concentration skills than those in the general
population. On a standard concentration test, the overall mean is
15.5. The researcher gave this same test to a random sample of
12 individuals who regularly listen to classical music. Their
scores on the test were as follows:
16, 14, 20, 12, 25, 22, 23, 19, 17, 17, 21, 20
a.Is this a one- or two-tailed test?
b.What are H0 and Ha for this study?
c.Compute tobt.
d.What is tcv?
e.Should H0 be rejected? What should the researcher conclude?
f.Determine the 95% confidence interval for the population
mean, based on the sample mean.
CHAPTER FIVE
Statistical Software Resources
If you need help getting started with Excel or SPSS, please
see Appendix C: Getting Started with Excel and SPSS.
MODULE 9 Single-Sample z Test
The problems we'll be using to illustrate how to calculate the
single-sample z test appear in Module 9.
Using the TI-84 for the One-tailed z Test from Module 9
The z test will be used to test the hypothesis that the sample of
children in the academic after-school programs represents a
population with a mean IQ greater than the mean IQ for the
general population. To do this, we need to determine whether
the probability is high or low that a sample mean as large as
103.5 would be chosen from this sampling distribution. In other
words, is a sample mean IQ score of 103.5 far enough away
from, or different enough from, the population mean of 100 for
us to say that it represents a significant difference with an alpha
level of .05 or less?
1.With the calculator on, press the STAT key.
2.Highlight TESTS.
41. 3.1: Z-Test will be highlighted. Press ENTER.
4.Highlight STATS. Press ENTER.
5.Scroll down to μ0: and enter the mean for the population
(100).
6.Scroll down to σ: and enter the standard deviation for the
population (15).
7.Scroll down to X: and enter the mean for the sample (103.5).
8.Scroll down to n: and enter the sample size (75).
9.Lastly, scroll down to μ: and select the type of test (one-
tailed), indicating that we expect the sample mean to be greater
than the population mean (select >μ0). Press ENTER.
10.Highlight CALCULATE and press ENTER.
The z score of 2.02 should be displayed followed by the
significance level of .02. If you would like to see where
the z score falls on the normal distribution, repeat Steps 1-9,
then highlight DRAW, and press ENTER.
The z test score is 2.02, and the alpha level or significance level
is p= .02. Thus, the alpha level is less than .05, and the mean IQ
score of children in the sample differs significantly from that of
children in the general population. In other words, children in
academic after-school programs score significantly higher on IQ
tests than children in the general population. In APA style, it
would be reported as follows: z(N = 75) = 2.02, p < .05 (one-
tailed).
Using the TI-84 for the Two-Tailed z Test from Module 9
The previous example illustrated a one-tailed z test; however,
some hypotheses are two-tailed and thus the z test would also be
two-tailed. As an example, refer back to the study from Module
9 in which the researcher examined whether children in athletic
after-school programs weighed a different amount than children
in the general population. In other words, the researcher
expected the weight of the children in the athletic after-school
programs to differ from that of children in the general
population, but he was not sure whether they would weigh less
(because of the activity) or more (because of greater muscle
mass). Let's use the following data from Module 9: The mean
42. weight of children in the general population (μ) is 90 pounds,
with a standard deviation (σ) of 17 pounds; for children in the
sample (N = 50), the mean weight (¯¯¯X)(X¯) is 86 pounds.
Using this information, we can now test the hypothesis that
children in athletic after-school programs differ in weight from
those in the general population using the TI-84 calculator.
1.With the calculator on, press the STAT key.
2.Highlight TESTS.
3.1: Z-Test will be highlighted. Press ENTER.
4.Highlight STATS. Press ENTER.
5.Scroll down to μ0: and enter the mean for the population (90).
6.Scroll down to σ: and enter the standard deviation for the
population (17).
7.Scroll down to ¯¯¯XX¯: and enter the mean for the sample
(86).
8.Scroll down to n: and enter the sample size (50).
9.Lastly, scroll down to μ: and select the type of test (two-
tailed), indicating that we expect the sample mean to differ from
the population mean (select 2μ0). Press ENTER.
10.Highlight CALCULATE and press ENTER.
The z score of −1.66 should be displayed followed by the alpha
level of .096, indicating that this test was not significant. We
can therefore conclude that the weight of children in the athletic
after-school programs did not differ significantly from the
weight of children in the general population.
If you would like to see where the z score falls on the normal
distribution, repeat Steps 1−9, then highlight DRAW, and press
ENTER.
MODULE 10 The Single-Sample t Test
Let's illustrate the use of the single-sample t test to test a
hypothesis using the example from Module 10. Assume the
mean SAT score of students admitted to General University is
1,090. Thus, the university mean of 1,090 is the population
mean (μ). The population standard deviation is unknown. The
members of the Biology Department believe that students who
decide to major in biology have higher SAT scores than the
43. general population of students at the university.
Notice that this is a one-tailed test because the researchers
predict that the biology students will perform higher than the
general population of students at the university. The researchers
now need to obtain the SAT scores for a sample of biology
majors. This information is provided in Table 10.1 in Module
10, which shows that the mean SAT score for a sample of 10
biology majors is 1,176.
Using Excel
To demonstrate how to use Excel to calculate a single-
sample t test, we'll use the data from Table 10.1, which
represent SAT scores for 10 biology majors at General
University. We are testing whether biology majors have higher
average SAT scores than the population of students at General
University. We begin by entering the data into Excel. We enter
the sample data into the A column and the population mean of
1,090 into the B column. We enter the population mean next to
the score for each individual in the sample. Thus, you can see
that I've entered 1,090 in Column B ten times, one time for each
individual in our sample of 10 biology majors.
Next highlight the Data ribbon, and then click on Data
Analysis in the top right-hand corner. You should now have the
following pop-up window:
Scroll down to t-Test: Paired Two Sample for Means, which is
the procedure we'll be using to determine the single-
sample t test. Then click OK. You'll be presented with the
following dialog box:
With the cursor in the Variable 1 Range: box, highlight the data
from Column A in the Excel spreadsheet so that they appear in
the input range box. Move the cursor to
the Variable 2 Range: box and enter the data from Column B in
the spreadsheet into this box by highlighting the data. The
dialog box should now appear as follows:
44. Click OK to execute the problem. You will be presented with
the following output:
We can see the t test score of 2.063 and the one-tailed
significance level of p = .035. Thus, our sample mean falls 2.06
standard deviations above the population mean of 1,090. We
must now determine whether this is far enough away from the
population mean to be considered significantly different.
Because our obtained alpha level (significance level) is .035
and is less than .05, the result is significant. We reject H0 and
support Ha. In other words, we have sufficient evidence to
allow us to conclude that biology majors have significantly
higher SAT scores than the rest of the students at General
University. In APA style, this would be reported as t(9) =
2.06, p = .035 (one-tailed).
Using SPSS
To demonstrate how to use SPSS to calculate a single-
sample t test, we'll use the data from Table 10.1 in Module 10,
which represent SAT scores for 10 biology majors at General
University. We are testing whether biology majors have higher
average SAT scores than the population of students at General
University. We begin by entering the data into SPSS and
naming the variable (if you've forgotten how to name a variable,
please refer back to Appendix C).
Once the data are entered and the variable named, select
the Analyze tab and, from the drop-down menu, Compare
Means followed by One-Sample T Test. The following dialog
box will appear:
Place the SATscore variable into the Test Variable box by
utilizing the arrow in the middle of the window. Then let SPSS
know what the population mean SAT score is (1,090). We enter
this population mean in the Test Value box as in the following
window.
45. Then click OK, and the output for the single-sample t test will
be produced in an output window as follows:
We can see the t test score of 2.063 and the two-tailed
significance level. Thus, our sample mean falls 2.06 standard
deviations above the population mean of 1,090. We must now
determine whether this is far enough away from the population
mean to be considered significantly different. This was a one-
tailed test; thus when using SPSS you will need to divide the
significance level in half to obtain a one-tailed significance
level because SPSS reports only two-tailed significance levels.
Our alpha level (significance level) is .035 and is less than .05,
meaning the result is significant. We reject H0 and support Ha.
In other words, we have sufficient evidence to allow us to
conclude that biology majors have significantly higher SAT
scores than the rest of the students at General University. In
APA style, this would be reported as t(9) = 2.06, p = .035 (one-
tailed).
Using the TI-84
Let's use the data from Table 10.1 to conduct the test using the
TI-84 calculator.
1.With the calculator on, press the STAT key.
2.EDIT will be highlighted. Press the ENTER key.
3.Under L1 enter the SAT data from Table 10.1.
4.Press the STAT key once again and highlight TESTS.
5.Scroll down to T-Test. Press the ENTER key.
6.Highlight DATA and press ENTER. Enter 1,090 (the mean for
the population) next to μ0:. Enter L1 next to List (to do this
press the 2nd key followed by the 1 key).
7.Scroll down to μ: and select >μ0 (for a one-tailed test in
which we predict that the sample mean will be greater than the
population mean). Press ENTER.
8.Scroll down to and highlight CALCULATE. Press ENTER.
The t score of 2.06 should be displayed, followed by the
significance level of .035. In addition, descriptive statistics will
46. be shown. If you would like to see where the t score falls on the
distribution, repeat Steps 1−7, then highlight DRAW, and press
ENTER.