A simple random sample of 1000 American adults found that the average number of hours spent
watching television during a typical week was 13.8. A simple random sample of 500 Canadians
yielded an average of 12.5 hours per week of television viewing. Assume that the American and
Canadian distributions for weekly television, viewing times have the same standard deviations.
The sampling variability associated with the sample means (described above) is
A.) smaller for the sample of Canadians, since the population of Canada is less than half that of
the United States.
B.) larger for the sample of Canadians because the sample size is smaller.
C.) smaller for the sample of Canadians, because their population is smaller.
D.) larger for the sample of Canadians, since Canadian citizens are more widely dispersed
throughout their country than American citizens are in the United States. Hence, Canadians have
more variable views.
Solution
b. remember that sample standard deviation is given by sigma/sqrt(n).
A social worker is involved in a study about family structure. She o.pdfagrawalsales2
A social worker is involved in a study about family structure. She obtains information regarding
the number of children per family for a certain community from the census data. Identify the
variable of interest, determine whether it is discrete or continuous and list some possible values.
Solution.
A social psychologist interested in cultural differences compared wo.pdfagrawalsales2
A social psychologist interested in cultural differences compared women of two ethnic groups on
a Role Approval Index on which high scores indicated high degrees of approval of one\'s own
social role. The results were as follows.
Ethnic Group A: N=15, M=55, S2=6.5
Ethnic Group B: N=23, M=51, S2=4.5
If the standard deviation of the distribution of the difference between means is .76, what is the t
score?
Solution
d.
A small town has been hit hard by the flu this winter. It is estimat.pdfagrawalsales2
A small town has been hit hard by the flu this winter. It is estimated that 20% of the town is
contagious. If we randomly select 4 people from the town (assuming the estimation is correct),
what is the probability that exactly two of them will be contagious?
Solution
binompdf(4,.2,2) = .1536 answer = .1536.
A skeptical paranormal researcher claims the proportion of Americans.pdfagrawalsales2
A skeptical paranormal researcher claims the proportion of Americans that have seen a UFO, p,
is less than 1 in every one thousand. Identify the type 2 error for the test.
Please show all work on how you got this answer.
Solution
Type II error is when you do not reject the null hypothesis even though it is false.
Ho: p = 1/1000 H1: p < 1/1000 (the researches claim) In this example the type II error would be
finding that the proportion of Americans that have seen a UFO is 1/1000 or higher when in fact,
the true proportion is less than 1 in 1000..
A skeptical paranormal researcher claims that the proportion of Amer.pdfagrawalsales2
A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO
is less than 1 in every 1,000. State the null hypothesis and t alternative hypothesis for a test of
significance.
Solution
p0=1/1000 = 0.001
H0: p = 0.001
Ha: p < 0.001.
A situation in which the Fed holds interest rates constant (at a low.pdfagrawalsales2
A situation in which the Fed holds interest rates constant (at a low level) in order to increase the
effectiveness of fiscal stimulus is known as ___________.
A. monetary policy that is non-cooperative and stringent
B. non-accommodating monetary policy action
C. non-accommodating monetary policy action
D. accommodating monetary policy action
Solution
D. accommodating monetary policy action.
A simple random sample of 100 observations was taken from a large po.pdfagrawalsales2
A simple random sample of 100 observations was taken from a large population and the sample
mean and the standard deviation were determined to be 80 and 12 respectively. The standard
error of the mean is 0.12.
A. True
B. False
Solution
standard error of the mean = s/n =12/10=1.2
Answer: False.
A shipment of 11 printers contains 2 that are defective. Find the pr.pdfagrawalsales2
A shipment of 11 printers contains 2 that are defective. Find the probability that a
sample of 4 printers, drawn from the 11, will not contain a defective printer.
Solution
There are 9 nondefective printers out of the 11 total, so: (9/11)(8/10)(7/9)(6/8) =
21/55 = 0.38181818.
A social worker is involved in a study about family structure. She o.pdfagrawalsales2
A social worker is involved in a study about family structure. She obtains information regarding
the number of children per family for a certain community from the census data. Identify the
variable of interest, determine whether it is discrete or continuous and list some possible values.
Solution.
A social psychologist interested in cultural differences compared wo.pdfagrawalsales2
A social psychologist interested in cultural differences compared women of two ethnic groups on
a Role Approval Index on which high scores indicated high degrees of approval of one\'s own
social role. The results were as follows.
Ethnic Group A: N=15, M=55, S2=6.5
Ethnic Group B: N=23, M=51, S2=4.5
If the standard deviation of the distribution of the difference between means is .76, what is the t
score?
Solution
d.
A small town has been hit hard by the flu this winter. It is estimat.pdfagrawalsales2
A small town has been hit hard by the flu this winter. It is estimated that 20% of the town is
contagious. If we randomly select 4 people from the town (assuming the estimation is correct),
what is the probability that exactly two of them will be contagious?
Solution
binompdf(4,.2,2) = .1536 answer = .1536.
A skeptical paranormal researcher claims the proportion of Americans.pdfagrawalsales2
A skeptical paranormal researcher claims the proportion of Americans that have seen a UFO, p,
is less than 1 in every one thousand. Identify the type 2 error for the test.
Please show all work on how you got this answer.
Solution
Type II error is when you do not reject the null hypothesis even though it is false.
Ho: p = 1/1000 H1: p < 1/1000 (the researches claim) In this example the type II error would be
finding that the proportion of Americans that have seen a UFO is 1/1000 or higher when in fact,
the true proportion is less than 1 in 1000..
A skeptical paranormal researcher claims that the proportion of Amer.pdfagrawalsales2
A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO
is less than 1 in every 1,000. State the null hypothesis and t alternative hypothesis for a test of
significance.
Solution
p0=1/1000 = 0.001
H0: p = 0.001
Ha: p < 0.001.
A situation in which the Fed holds interest rates constant (at a low.pdfagrawalsales2
A situation in which the Fed holds interest rates constant (at a low level) in order to increase the
effectiveness of fiscal stimulus is known as ___________.
A. monetary policy that is non-cooperative and stringent
B. non-accommodating monetary policy action
C. non-accommodating monetary policy action
D. accommodating monetary policy action
Solution
D. accommodating monetary policy action.
A simple random sample of 100 observations was taken from a large po.pdfagrawalsales2
A simple random sample of 100 observations was taken from a large population and the sample
mean and the standard deviation were determined to be 80 and 12 respectively. The standard
error of the mean is 0.12.
A. True
B. False
Solution
standard error of the mean = s/n =12/10=1.2
Answer: False.
A shipment of 11 printers contains 2 that are defective. Find the pr.pdfagrawalsales2
A shipment of 11 printers contains 2 that are defective. Find the probability that a
sample of 4 printers, drawn from the 11, will not contain a defective printer.
Solution
There are 9 nondefective printers out of the 11 total, so: (9/11)(8/10)(7/9)(6/8) =
21/55 = 0.38181818.
A set of waveforms applied to two D flip-flops is shown in Figure 4-.pdfagrawalsales2
A set of waveforms applied to two D flip-flops is shown in Figure 4-56. These waveforms are
applied to the flip-flops shown along with the values of their timing parameters.
(a) List the time(s) at which there are timing violations in signal D1 for flip-flop 2.
(b) List the time(s) at which there are timing violations in signal D2 for flip-flop 2.
Solution
flip flop 1 is a positive edge triggered flop.
The inputs should be seen at the rising edge of the clock
for signal D1, at the fourth rising edge i.e t = 28ns, violation occurs.
D1 is rising at the same edge, it should be stable for Ts = 1ns before and Th =0.5ns after the
edge.
2)
flipflop 2 is a negative edge triggered.
At second negative edge i.e t = 16ns and third negative edge i.e t = 24ns, D2 signal viloates the
timing requirements of flop2.
A significant advantage of internal funding is the fact that retaine.pdfagrawalsales2
A significant advantage of internal funding is the fact that retained earnings do not incur flotation
cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s
shareholders. True False
Solution
A significant advantage of internal funding is the fact that retained earnings do not incur flotation
cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s
shareholders. TRUE
UNLIKE OTHER SOURCES OF FINANCEING THE RETAINED EARNNINGS DOES
NOOT REQUIRE ANY ISSUE RELATED COST..
A short quiz has two truefalse questions and one multiple choice qu.pdfagrawalsales2
A short quiz has two true/false questions and one multiple choice question with four choices. A
student guesses at each question. Assuming the choices are all equally likely, what is the
probability that the student gets all three correct?
Solution
The events are independent, so P(A and B) = P(A)*P(B) --- P(2 correct on true-
false) = (1/2)(1/2) = 1/4 P(1 correct mult-choice) = 1/4 ---- P(this and that) = (1/4)(1/4) = 1/16.
A share of common stock has just paid a dividend of $2.00 (D0 = 2.00.pdfagrawalsales2
A share of common stock has just paid a dividend of $2.00 (D0 = 2.00). The long-term growth
rate for this stock is 15% (g = .15) and investors require a 19% return (r = .19). What is the price
of the stock today?
Solution
Price of stock = D0 (1+g) /(R- g)
= 2 (1+.15) /(.19-.15)
= 2*1.15 /.04
= $57.50 per share.
A set of 50 data values has a mean of 27 and a variance of 16. I. .pdfagrawalsales2
A set of 50 data values has a mean of 27 and a variance of 16.
I. Find the standard score (z) for a data value = 19.
II. Find the probability of a data value < 19.
III. Find the probability of a data value > 19.
Show all work.
Solution
I am assuming X ~ N(27, 16) to find the probability of greater than 19, just take the
fact that the probabilities add to 1 and 1 - the above probability is the probability that x is greater
than 19.
A set of 50 data values has a mean of 21 and a variance of 16. I. .pdfagrawalsales2
A set of 50 data values has a mean of 21 and a variance of 16.
I. Find the standard score (z) for a data value = 31.
II. Find the probability of a data value > 31.
III. Find the probability of a data value < 31.
Show all work.
Solution
If the variance is 4, sigma is the sqrt(4) = 2 z = (x - mu) / sigma z = (31 - 28) / 2 z =
1.5 So from there, look at a z-table. P(z>1.5) = 1 - .9332 = .0668 P(z<1.5) = .9332.
A set of 50 data values has a mean of 16 and a variance of 4. I. F.pdfagrawalsales2
A set of 50 data values has a mean of 16 and a variance of 4.
I. Find the standard score (z) for a data value = 12.
II. Find the probability of a data value > 12.
Solution
n=50
=16 = population mean
2 = 4, so standard devation =2
A) z = (x - )/ = (12 - 16)/2 = 2
B) P(x > 12) = P( (x - )/ > (12-16)/2 ) = P( z > 2) = .0228
I did not use the standard error of the sampling distribution [ (x - )/(/n) ] because that is only used
when dealing with a sample. The problems clearly state \"data value\" which I understood as it
being a single observation rather than the sample mean..
A set of 10 data values has a mean and a median of $1425 and ranges .pdfagrawalsales2
A set of 10 data values has a mean and a median of $1425 and ranges from a min value of $987
to a max value of $1945. Later, we discover that the $1945 value was mid-recorded and should
have been $2945. Find the corrected values for th mean and median.
Solution
Total sum of data = 10*1425 =14250 New sum of data =14250-1945+2945 =15250
mean = 1525 median remains unchanged = $1425.
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.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
A set of waveforms applied to two D flip-flops is shown in Figure 4-.pdfagrawalsales2
A set of waveforms applied to two D flip-flops is shown in Figure 4-56. These waveforms are
applied to the flip-flops shown along with the values of their timing parameters.
(a) List the time(s) at which there are timing violations in signal D1 for flip-flop 2.
(b) List the time(s) at which there are timing violations in signal D2 for flip-flop 2.
Solution
flip flop 1 is a positive edge triggered flop.
The inputs should be seen at the rising edge of the clock
for signal D1, at the fourth rising edge i.e t = 28ns, violation occurs.
D1 is rising at the same edge, it should be stable for Ts = 1ns before and Th =0.5ns after the
edge.
2)
flipflop 2 is a negative edge triggered.
At second negative edge i.e t = 16ns and third negative edge i.e t = 24ns, D2 signal viloates the
timing requirements of flop2.
A significant advantage of internal funding is the fact that retaine.pdfagrawalsales2
A significant advantage of internal funding is the fact that retained earnings do not incur flotation
cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s
shareholders. True False
Solution
A significant advantage of internal funding is the fact that retained earnings do not incur flotation
cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s
shareholders. TRUE
UNLIKE OTHER SOURCES OF FINANCEING THE RETAINED EARNNINGS DOES
NOOT REQUIRE ANY ISSUE RELATED COST..
A short quiz has two truefalse questions and one multiple choice qu.pdfagrawalsales2
A short quiz has two true/false questions and one multiple choice question with four choices. A
student guesses at each question. Assuming the choices are all equally likely, what is the
probability that the student gets all three correct?
Solution
The events are independent, so P(A and B) = P(A)*P(B) --- P(2 correct on true-
false) = (1/2)(1/2) = 1/4 P(1 correct mult-choice) = 1/4 ---- P(this and that) = (1/4)(1/4) = 1/16.
A share of common stock has just paid a dividend of $2.00 (D0 = 2.00.pdfagrawalsales2
A share of common stock has just paid a dividend of $2.00 (D0 = 2.00). The long-term growth
rate for this stock is 15% (g = .15) and investors require a 19% return (r = .19). What is the price
of the stock today?
Solution
Price of stock = D0 (1+g) /(R- g)
= 2 (1+.15) /(.19-.15)
= 2*1.15 /.04
= $57.50 per share.
A set of 50 data values has a mean of 27 and a variance of 16. I. .pdfagrawalsales2
A set of 50 data values has a mean of 27 and a variance of 16.
I. Find the standard score (z) for a data value = 19.
II. Find the probability of a data value < 19.
III. Find the probability of a data value > 19.
Show all work.
Solution
I am assuming X ~ N(27, 16) to find the probability of greater than 19, just take the
fact that the probabilities add to 1 and 1 - the above probability is the probability that x is greater
than 19.
A set of 50 data values has a mean of 21 and a variance of 16. I. .pdfagrawalsales2
A set of 50 data values has a mean of 21 and a variance of 16.
I. Find the standard score (z) for a data value = 31.
II. Find the probability of a data value > 31.
III. Find the probability of a data value < 31.
Show all work.
Solution
If the variance is 4, sigma is the sqrt(4) = 2 z = (x - mu) / sigma z = (31 - 28) / 2 z =
1.5 So from there, look at a z-table. P(z>1.5) = 1 - .9332 = .0668 P(z<1.5) = .9332.
A set of 50 data values has a mean of 16 and a variance of 4. I. F.pdfagrawalsales2
A set of 50 data values has a mean of 16 and a variance of 4.
I. Find the standard score (z) for a data value = 12.
II. Find the probability of a data value > 12.
Solution
n=50
=16 = population mean
2 = 4, so standard devation =2
A) z = (x - )/ = (12 - 16)/2 = 2
B) P(x > 12) = P( (x - )/ > (12-16)/2 ) = P( z > 2) = .0228
I did not use the standard error of the sampling distribution [ (x - )/(/n) ] because that is only used
when dealing with a sample. The problems clearly state \"data value\" which I understood as it
being a single observation rather than the sample mean..
A set of 10 data values has a mean and a median of $1425 and ranges .pdfagrawalsales2
A set of 10 data values has a mean and a median of $1425 and ranges from a min value of $987
to a max value of $1945. Later, we discover that the $1945 value was mid-recorded and should
have been $2945. Find the corrected values for th mean and median.
Solution
Total sum of data = 10*1425 =14250 New sum of data =14250-1945+2945 =15250
mean = 1525 median remains unchanged = $1425.
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.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
A simple random sample of 1000 American adults found that the averag.pdf
1. A simple random sample of 1000 American adults found that the average number of hours spent
watching television during a typical week was 13.8. A simple random sample of 500 Canadians
yielded an average of 12.5 hours per week of television viewing. Assume that the American and
Canadian distributions for weekly television, viewing times have the same standard deviations.
The sampling variability associated with the sample means (described above) is
A.) smaller for the sample of Canadians, since the population of Canada is less than half that of
the United States.
B.) larger for the sample of Canadians because the sample size is smaller.
C.) smaller for the sample of Canadians, because their population is smaller.
D.) larger for the sample of Canadians, since Canadian citizens are more widely dispersed
throughout their country than American citizens are in the United States. Hence, Canadians have
more variable views.
Solution
b. remember that sample standard deviation is given by sigma/sqrt(n)
