Ois-Quiz Study Guide Chapter 8-9 KFUPM STAT 211 Quiz

Ois-Quiz Study for Chapter 8 and 9
KFUPM |Term 041 |Date: 25/12/2004 | |
|Mathematical Sciences |STAT 211 |Duration: 10 minutes |
| |Quiz# 7 | |
|Name: |ID#: |Section#: |Serial#: |
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1. A 95 percent confidence interval estimate will have a margin of error that is approximately + 95
percent of the size of the population mean.
Answer: False ... Show more content on Helpwriting.net ...
Given this information, what sample size was used to arrive at this estimate? a. About 344. b.
Approximately 1,066. c. Just under 700. d. Can't be determined without more information.
Answer: B
With My Best Wishes
|KFUPM |Term 041 |Date: 25/12/2004 |
|Mathematical Sciences |STAT 211 |Duration: 10 minutes |
| |Quiz# 7 | |
|Name: |ID#: |Section#: 1 2 4 |Serial#: |
Show your work in detail and write neatly and eligibly
1. All other factors held constant, the higher the confidence level, the closer the point estimate for
the population mean will be to the true population mean.
Answer: False
2. In estimating a population mean, increasing the confidence level will result in a higher margin of
error for a given sample
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WK 4 DQ1: Confidence Intervals
BUS 308 WK 4 DQ1 – Confidence Intervals As pointed out in the course text, point estimates only
provide information about the difference or association between groups. It is not possible to
determine the level of accuracy of these estimates because bias may occur in the estimation. In
particular, errors may arise in research due to sampling, an inadequate sampling frame, improperly
designed research instruments or non–response. While a researcher can minimize bias in research by
enhancing the design, sampling errors are inevitable as they result from chance (Explorable.com,
2009). Therefore, the likelihood that a point estimate is accurate is minimal. In this context,
confidence intervals suffice to provide a range of values within which one is likely to find the
population parameter with a specified probability level. The confidence interval are very useful,
particularly in a business context ... Show more content on Helpwriting.net ...
However, due to the impact of various factors on the business, point estimates are not useful.
Confidence intervals allow for a variety of point estimates of the population parameter. For instance,
suppose a manager intends to estimate the mean sales of a product for a specified period. In this
case, the manager will chose a level of confidence, say 95%, and calculate the interval using sample
statistics. As a result, the manager can conclude with 95% certainty that the mean sales of the
product is within the calculated confidence interval. Unlike a point estimate, the manager can
calculate many confidence intervals by adjusting the level of confidence accordingly. On a recent
chat with a manager at a local boutique, I asked the manager whether he prefers point estimates or
confidence intervals. The manager affirmed that point estimates are useful but confidence intervals
are more insightful because they provide a wide range of possible outcomes that conform to the
varied nature of the business

The Rate Ratio And Irr
providing a more complete picture of the relationships between variables missed by other regression
methods" (Cade, 2003, pg. 412). So, quantile regression is focusing on change from minimum
responses.
B research binomial, the group in question being repeatedly tested. The incidence rate ratio or IRR
can be defined as, "...a measure of the frequency with which a case occurs in a population over a
period of time" (Statistical, n.d., pg. 1). The IRR is an event that takes place in a group over the
course of a period. When the researchers are implementing Q25 and Q95, they are comparing it to
the linear regression. They are looking at regression parameters estimates the change and response
of a variable. With the confidence levels close together it shows that there is a close relationship,
which shows a very close gap and that in turn means more accuracy. All of the confidence levels for
this chart show a very close relationship. That means that there is more accuracy within the figures
and table. Since the numbers are so close, the graph for this will very tall and tight. Also, the mean
or average falls into the middle of the confidence intervals. We have to look at the 95 percent
confidence interval. That means that the sample size was taken, that 95 percent of them would
include the true population.
Statistical conclusions
What are the statistical results?
The results of this study should be surprising. The results of this study found that the impairment

The T-Distribution and T-Test
The T–Distribution and T–Test "In probability and statistics, Student 's t–distribution (or simply the
t–distribution) is a continuous probability distribution that arises when estimating the mean of a
normally distributed population in situations where the sample size is small" (Narasimhan , 1996).
Similar to the normal distribution, the t–distribution is symmetric and bell–shaped, but has heavier
tails, meaning that it is more likely to produce values far from its mean. This makes the t–
distribution useful for understanding statistical behaviors of random quantities. It plays a role in a
number of widely–used statistical analyses, including the Student 's t–test for assessing the
statistical significance of the difference between two ... Show more content on Helpwriting.net ...
An important property of a test statistic is that we must be able to determine its sampling
distribution under the null hypothesis, which allows us to calculate p–values.
For example, suppose we wish to test whether a coin is fair. If we flip the coin 100 times and record
the results, the raw data can be represented as a sequence of 100 Heads and Tails. If our interest is in
the probability of obtaining a head, we only need to record the number T out of the 100 flips that
produced a head, and use T0 = 50 as our null value. In this case, the exact sampling distribution of T
is the binomial distribution, but for larger sample sizes the normal approximation can be used.
Using one of these sampling distributions, it is possible to compute either a one–tailed or two–tailed
p–value for the null hypothesis that the coin is fair.
The two–tailed test is when a given statistical hypothesis, H0 (the null hypothesis), will be rejected
when the value of the test statistic is either sufficiently small or sufficiently large. This contrasts
with a one–tailed test, in which only one of the rejection regions "sufficiently small" or "sufficiently
large" is preselected according to the alternative hypothesis being selected, and the hypothesis is
rejected only if the test statistic satisfies that criterion. (UCLA, & , 2009)
Among the most frequently used t–tests are: a one–sample location test of whether the mean of a
normally distributed population has a value

Game Theory
PS II 2013 Problem Set II
1. Reading newspapers: A survey on student behavior in some major B–schools in India asked a
randomly selected group of 240 students on the number of hours a week he/she reads a newspaper.
The sample mean was 4.1 and standard deviation 3. We can assume that the underlying distribution
(of number of hrs reading newspaper) is approximately normal.
a) What will be a 99% confidence interval of the population mean (number of hours a B–school
student read newspapers/week)?
b) Suppose you perform a similar survey at IIMA with 24 randomly selected students; the sample
mean and the sample standard deviation were 4.5 and 2.8 respectively. What will be a 99%
confidence interval of the population mean ... Show more content on Helpwriting.net ...
Similarly, the number of defectives among the ten televisions of brand B is 2M. In order to test the
null hypothesis H0: M = 2 against the alternative Ha: M > 2, the following procedure is adopted:
(i) From the six televisions of brand A, draw a random sample of size two without replacement.
(ii) From the ten televisions of brand B, draw a random sample of size two without replacement.
(iii) Reject if and only if both the televisions in at least one of the two samples are defectives.
a) What is the probability of type I error for the above test procedure?
b) Find the probability of type II error for this procedure when M equals 3.
5. Anorexia in teenage girls: Anorexia is an eating disorder that can cause a person to be
dangerously underweight. A recent study analysed the effect of cognitive behavioural therapy in
aiding weight gain on 29 teenage girls affected with Anorexia. Each girl's weight was measured
before and after the therapy and the weight change (positive: weight gain; negative: weight loss)
was noted. The weight changes for the 29 girls had a sample mean of = 3 kilos and a standard
deviation of s = 7.32 kilos. Let µ be the population mean weight change. It can be assumed that
weight change follow a normal distribution in the population.
a) What would be a reasonable set of hypotheses (null and alternative) for this problem?
b) How many standard errors separate the sample average weight change of the 29 girls and the null
value

Confidence Intervals
Consider the following question: someone takes a sample from a population and finds both the
sample mean and the sample standard deviation. What can he learn from this sample mean about the
population mean?
This is an important problem and is addressed by the Central Limit Theorem. For now, let us not
bother about what this theorem states but we will look at how it could help us in answering our
question.
The Central Limit Theorem tells us that if we take very many samples the means of all these
samples will lie in an interval around the population mean. Some sample means will be larger than
the population mean, some will be smaller. The Central Limit Theorem goes on to state that 95% of
the sample means ... Show more content on Helpwriting.net ...
A brief note: Statistical software often gives slightly different but more accurate values. Let us stick
with the consensus of the majority of people working with elementary statistics and use the critical
values given above. Summarized:
|Confidence Interval |Critical Value |
|90% |–1.65 |
|95% |–1.96 |
|99% |–2.58 |
Due to the perfect symmetry of the curve we will find the same but this time positive z–values for
the right tail. [pic]
Now we can start calculating confidence intervals.
Sample Mean and Standard Deviation are known, in the case of Sample Size [pic]
The confidence interval for the population mean μ in this situation is given by [pic].
Example 1:
We are looking at a very large group of tests. Forty five tests were selected from this group. The
mean of the sampled tests is 58 and the standard deviation is 13. Find the 95% confidence interval
for the mean grade for the whole class.

Solution:
[pic]. This means that we are 95% confident that the population mean for all these tests is between
[pic] and [pic] or more simply said, between 54.2 and 61.8.
The calculator keystrokes on the TI 30XIIB for that part of the calculation that follows the ( in
example 1 are:
1.96 x 13
(
√
45
)
=
Please note that taking a sample gives us information

Essay on Polling America
NORTHCENTRAL UNIVERSITY
ASSIGNMENT COVER SHEET
Student: Emanuel Parker
THIS FORM MUST BE COMPLETELY FILLED IN
Follow these procedures: If requested by your instructor, please include an assignment cover sheet.
This will become the first page of your assignment. In addition, your assignment header should
include your last name, first initial, course code, dash, and assignment number. This should be left
justified, with the page number right justified. For example:
|DoeJXXX0000–1 1 |
Save a copy of your assignments: You may need to re–submit an assignment at your instructor's
request. Make sure you save your files in accessible location. ... Show more content on
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president. Polls can give us a general idea of the Nation's preference in terms of who will be
selected as the next president. This article uses the Bloomberg National Poll which showed Obama
having a thirteen point lead over Romney while the Associated Press Poll resulted in Obama having
a three point lead while the Pew Research Center had the president up by four points. These polls
ranged from a three point to a thirteen point lead which means that different sampling will yield
different results. Location, demographic, and whether the state is democratic or republican plays an
important part in the results you will get from the polls. According to the article, small differences in
methods between popular polls can result in large differences when voter commitment is still
uncertain (Tackett, 2012). During this election, polls were a great source of information for voters to
use in their decision making process. Bloomberg's poll surveyed 1,002 adults, with a margin of error
of plus or minus 3.1 percentage points for the full sample. This means that if the polls were
conducted 100 times, the data would be within 3 points above or below the reported percentage in
95 of 100 surveys (Tackett, 2012). The confidence interval is used as a type of interval estimate of a
sample population to indicate the reliability of the

Case Study National Century Bank
TABLE OF CONTENT 1 | Introduction | 2 | 2 | Question 1 | 3 | 3 | Question 2 | 4 | 4 | Question 3 | 6 |
5 | Question 4 | 9 | 6 | Question 5 | 11 | 7 | Conclusion | 12 | INTRODUCTION Century National
Bank has offices in several cities in the Midwest and the southeastern part of the United States. Mr.
Dan Selig, president and CEO, would like to know the characteristics of his checking account
customer. To better understand the customers, Mr. Selig asked Ms. Wendy Lamberg, director of
planning, to select a sample of customers and prepare a report. To begin, she has appointed a team
from her staff and the team has selected a random sample of 60 customers. All the information
gathered is tabulated in the table below: X1 = ... Show more content on Helpwriting.net ...
For many years the mean checking balance has been $1600. Does the sample data indicate that the
mean account balance has declined from this value? SOLUTION µ (Checking balance) = $1600.00
Number of customers, n = 60 In this case, level of significance, α was not provided. Therefore, the
analysis will be evaluated based on two α values which are: α = 0.05; Zα = – 1.65 α = 0.01; Zα = –
2.33 Hypotesis testing 1) H0 : µ ≥ 1600 H1 : µ < 1600 2) α = 0.05 α = 0.01 3) Left–tail test
Z(0.05) = – 1.65 Z(0.01) = – 2.33 4) Z calculated :– –1.65 –2.33 α=0.05 α=0.01 Z= –1.2994 5)
From calculations, computed z value is more than –1.65 and falls within Ho not rejected region. Ho
is not rejected at α = 0.05 & α = 0.01 significance levels. It is concluded that the mean
checking balance is still similar or more than $1600 QUESTION 3 Recent years have also seen an
increase in the use of ATM machines. When Mr. Selig took over the bank, the mean number of
transactions per month per customer was 8; now he believes it has increased to more than 10. In
fact, the advertising agency that prepares TV commercial for Century would like to use this on the
new commercial being designed. Is there sufficient evidence to conclude that the mean number of
transactions per customer is more than 10 per month? Could the advertising agency say the mean is
more than 9 per month? SOLUTION

Barron's Ap Statistics
I read Barron's How to Prepare for the AP Statistics Exam. A very educational book helped a lot on
the AP test. It clarified ideas that I was uncertain on. It helped me to understand when to use each
test and the assumptions needed for each test. Type I and Type II errors were explained in such a
way that they became crystal clear to me instead of muddy. Computer and Minitab outputs were
thoroughly explained, and I became comfortable with them after reading this book. The Barron's
guide also formatted equations in the same manner as the AP equation sheet, which helped me
become familiar with this format before going into the AP test. I feel that the Barron's guide helped
me to review all the Statistics concepts and refreshed my ... Show more content on Helpwriting.net
...
The Barron's guide more effectively explained the necessary steps for a confidence interval, as well.
The Brase text and Barron's guide also differ in their treatment of formulas, specifically the
Binomial Formula for Probability. In the Brase text, the binomial formula is: P(r) = n! (pr qn–r) r!
(n–r)!
While in the Barron's guide it is: This equation is given to us on the AP formula sheet, so since
Barons actually teaches you how to use this specific formula it makes it much easier to understand
and was very helpful on the AP Exam. Another topic the Brase text did not cover very thoroughly
was the Least Squares Line. Brase simply defines the line as, "The least–squares line devolved with
x as the explanatory variable and y as the response variable can be used only to predict y values
from specific x values. Baron 's on the other hand goes into detail in explaining how to find the line,
its slope, and the standard deviation. It gives the equation ŷ=y + b1 (x–x). It also goes onto define ß
as the slope of the true regression line, which can be found using a t score with degrees of freedom
n–2. The equation for standard deviation is the sum of the squared residuals divided by the sum of
the squared deviations of the mean. Overall, this AP study book was a very helpful tool in preparing
for the exam. It explained difficult concepts in a much

Mat 510 Week 10 Discussion Question
MAT 510 WEEK 10 DISCUSSION QUESTION
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"Hypothesis Test"
Note: Online students, please select one of the two subjects to discuss.
Use the Internet or Strayer Library to research articles on hypothesis test and its application in
business. Select one (1) company or organization which utilized hypothesis test technique for its
business process (e.g., whether or not providing flexible work hours improve employee produMAT
510 WEEK 10 DISCUSSION QUESTION
To purchase this visit following link: ... Show more content on Helpwriting.net ...
Select one (1) company or organization which utilized hypothesis test technique for its business
process (e.g., whether or not providing flexible work hours improve employee productivity.) Give
your opinion as to whether or not the utilization of such a technique improved business process for
the selected company or organization. Justify your response.
Select one (1) project from your working or educational environment that you would use the
hypothesis test technique. Next, propose the hypothesis structure (e.g., the null hypothesis, data
collection process, confidence interval, test statistics, reject or not reject the decision, etc.) for the
business process of the selected project. Provide a rationale for your response.
Course Home Work aims to provide quality study notes and tutorials to the students of MAT 510
Week 10 Discussion Question in order to ace their studies.
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discussion–question/ Contact us at:
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"Hypothesis Test"
Note: Online students, please select one of the two subjects to discuss.
Use the Internet or Strayer Library to research articles on hypothesis test and its application in
business. Select one (1) company or organization which utilized hypothesis test

Sample Testing Is Mandatory Against A Static Groundwater...
Statistical methods can be useful in compliance assessment and corrective action monitoring, where
single–sample testing is mandatory against a static groundwater protection standard. These criteria
health and also those derived from related as a fixed standard. The primary tool for monitoring is the
confidence interval. Numerous ranges of confidence intervals, including confidence intervals around
means, medians. The goal is on planning tests with reasonable statistical performance in terms of
per–test false positive rates.
Confidence interval are designed to estimate statistical characteristics of some parameter of sampled
population. Given a statistical parameter of interest such as the population mean (µ), the lower and
upper limits of a confidence interval define the most probable concentration range in which the true
parameter should be.
Like any estimate, the true parameter many not be located within the confidence interval. The
frequency with which this error tends to occur is denoted α, while its complement (1– α) is known
as the confidence level. The confidence level represents the percentage of cases where a confidence
interval constructed according to a fixed algorithm or equation will actually contain its intended
target, e.g., the population mean.
A point worth clarifying is the distinction between α as the complement of the confidence level
when constructing a confidence interval and the significance level (α) used in hypothesis testing.
Confidence

The Baseline Study Of The Elderly Study Sample Essay
RESULTS
The baseline study of the elderly study sample are as shown in the given table. It can be observed
that three–fifths of the total sample were composed of women, and also an expression of the high
prevalence of diabetes, antihypertensive treatment, and CVD are evident.
Table 1: Baseline characteristics of the study sample
Characteristic Value
Clinical features Women (%) 62.3 Age (years) 79.0 ± 4.5 Systolic blood pressure (mmHg) 143 ± 21
Diastolic blood pressure (mmHg) 72 ± 11 BMI (kg/m2) 26.8 ± 4.7 Antihypertensive treatment (%)
48.9 Current smoking (%) 8.3 Diabetes (%) 27.4 History of atrial fibrillation (%) 10.6 Prevalent
heart failure (%) 5.3 History of CVD (%) 35.0
Biochemical features Total–to–HDL cholesterol ratio (mg/dl) 4.54 ± 1.51 Leptin (ng/ml) 12.8 (7.0,
22,7) Log–Leptin 2.51 ± 0.81
The Clinical Correlation of Leptin and its Relation to the Antecedent BMI
Leptin was found to strongly correlate to the BMI at the index of examination where it was found
that r = 0.67, P < 0.0001. In the case of multivariable analyses testing the clinical correlates of the
log–Leptin, BMI with a partial R2 = 0.26 and sex with a partial R2 = 0.28 were found to be the
strongest correlates. For the women with a median of 17.4 and the quartiles 1 and 3 to be 10.6 and
28.7 respectively; this demonstrated a higher Leptin level than in the men with a median of 7.2 and
the first and third quartiles of 4.5 and 11.4

Teacher Essay examples
StatCrunch Assignment 2
First save this file to your computer. Answer each of the following questions, then resave the file
along with your answers and turn it in using the assignment link in Module 3, Activity 4. the first
four problems are worth 10 points each. Problems 5 and 6 are worth 30 points each.
1. If the original sample is 48, 55, 43, 61, 39, which of the following would not be a possible
bootstrap sample? Explain why it wouldn't be.
a) 48, 55, 43, 61, 39
b) 43, 39, 56, 43, 61
c) 55, 48, 55, 48, 61
d) 39, 39, 39, 39, 39
The answer is be because the number "56" is not part of the original data so it cannot be used in
bootstrapping.
2. The following bootstrap output is for mileage of a random sample of 25 mustang ... Show more
content on Helpwriting.net ...
Use the StatCrunch data file you created to complete the remainder of this assignment.
ID Gender Class Hours Work Loans CC Debt
226 Male 1 17 15.5 0 1257
341 Male 3 15 10 12766 3254
1139 Male 1 15 21 2575 1369
5347 Female 4 13 0 0 2113
6435 Male 1 9 36 4240 1843
6790 Female 4 11 29 13855 5336
6972 Female 1 15 15.5 0 960
7060 Female 1 16 0 4241 1166
8065 Male 3 15 0 0 4115
10190 Male 1 11 26.5 3919 1350
10720 Female 3 14 16 13088 3970
12934 Female 2 15 0 0 1786
17199 Male 3 15 0 12479 3423
21650 Female 4 11 22 14605 6695

21813 Male 1 11 23 2615 1443
23858 Male 3 13 0 0 0
27197 Male 4 15 13.5 16306 7785
27856 Female 4 12 17 15530 4992
29490 Male 2 12 0 0 2941
30229 Female 3 15 0 0 3311
30298 Male 4 18 0 0 2593
30895 Female 1 19 0 2763 0
31670 Male 3 16 0 11227 4698
33476 Female 4 3 36.5 0 3712
34151 Male 1 9 29 3426 1657
34543 Male 2 13 0 6831 3414
38144 Female 4 16 0 0 2518
38200 Female 4 15 17.5 0 8496
39541 Male 2 8 28 7499 3515
45931 Female 4 7 0 0 4976
5. Construct a bootstrap distribution of the credit card debt data from your sample using 3000
resamples.
a) Paste a copy of your distribution here. (With a PC, you can press the control–alt–fn–F11 keys to
copy the window showing the distribution.)
b) What is the mean of your original sample? 3146.01%
c) What is the 95% confidence interval estimate of the population

Final Projedct
The Final Project is worth 105 points. Please download this document to your computer and save it
using the naming convention specified in the course syllabus. For the Final Project you will be using
the MM207 Student Data Set, the survey codebook, and StatCrunch as necessary. You should enter
your answers/responses directly after the question. There is no need to retype the project.
After completing and saving the project, submit your project in the Final Drop Box.
In the course, go to Unit 9 –> Instructor Graded Project –> StatCrunch to access the MM207
Student Data Set. When the page loads you will need to click on Data Set on the left side of the
page. You do not need a StatCrunch ID or a password to access the data set; simply ... Show more
Be sure to show the data from your sample and the data to support your estimate.
8. Assume that the MM207 Student Data Set is a random sample of all Kaplan students; estimate the
proportion of all Kaplan students who are male using a 90% level of confidence.
9. Assume you want to estimate with the proportion of students who commute less than 5 miles to
work within 2%, what sample size would you need?
10. A professor at Kaplan University claims that the average age of all Kaplan students is 36 years
old. Use a 95% confidence interval to test the professor 's claim. Is the professor 's claim reasonable
or not? Explain.
NOTE: Project problems should not be posted to the Discussion threads. Questions on the project
problems should be addressed to the instructor by sending an email or by attending office hours.
Would you like a Math Center tutor to review your project?
Students in MM207 may submit their projects to the Math Center for review. Tutors will not grade
or correct the project, but they will provide guidance for improvement.
Students should submit assignments early enough to receive feedback and make corrections before
the project due date (24 hour turn–around times Monday–Thursday and 48 hour turn–around times
on weekends are typical).
Email projects to: kumc@kaplan.edu. Please put "project review" in the subject line of the message.
Please email the project using your Kaplan student account as messages sent from outside

Statistics
STATISTICS – Lab #6
Statistical Concepts: Data Simulation Discrete Probability Distribution Confidence Intervals
Calculations for a set of variables
Open the class survey results that were entered into the MINITAB worksheet.
We want to calculate the mean for the 10 rolls of the die for each student in the class. Label the
column next to die10 in the Worksheet with the word mean. Pull up Calc > Row Statistics and select
the radio–button corresponding to Mean. For Input variables: enter all 10 rows of the die data. Go to
the Store result in: and select the mean column. Click OK and the mean for each observation will
show up in the Worksheet.
We also want to calculate the median for the 10 rolls of the die. Label the ... Show more content on
Helpwriting.net ...
Either show work or explain how your answer was calculated.
Mean: Summation xP(x) = 1(1⁄6) +2(1⁄6) + 3(1⁄6) + 4(1⁄6) + 5(1⁄6) +6(1⁄6) = 21⁄6= μ 3.5
Standard deviation: sq. root ((1–3.5)^2 (1⁄6) + (2–3.5)^2(1⁄6) + (3–3.5)^2(1⁄6) + (4–3.5)^2(1⁄6) + (5–
3.5)^2(1⁄6) + (6–3.5)^2(1⁄6))= sq. root2.916=σ 1.707
3.) Give the mean for the mean column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Mean: 3.560. Yes, this is very closely centered around the parameter of interest (3.5)
4.) Give the mean for the median column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)?
μ of Median: 3.600. Yes, this too is also centered around the parameter of interest (3.5).
5.) Give the standard deviation for the mean and median column. Compare these and be sure to
identify which has the least variability?
σ of Mean: .0476 σ of Median: .0754

The standard deviation of the Means is smaller, thus having less variability than the Median,
meaning the data for the Means is grouped closer

Ops 561- Week 5
Process Improvement Plan
OPS 571
April 22, 2013
Richard Franchetti, Facilitator
Process Improvement Plan
In week one of this class, I was tasked to design a flowchart for a process in my daily life that I can
improve. I chose my morning routine before work. My goal is to get out of the house by 7:38am
Monday–Friday in order to catch my bus for work at 7:41am. I have trouble keeping track of time,
usually missing the bus about one time every two weeks. I have discovered through the flowchart in
week one, that my entire process has serious flaws. All of the bottlenecks were identified and
improved upon as necessary. After collecting data over the past four weeks, I see changes in the
process, using time as the metric. This paper will ... Show more content on Helpwriting.net ...
The final table calculates the upper and lower control limits for the entire sample. Since I wake up
every day at 6:30am and must leave the house by 7:40am, I have 70 minutes total to complete all of
my tasks in the morning. According to my calculations on the final table, I need to leave the house
anywhere between 7:34am and 7:39am. Doing so means that I have made my bus and my process
has been a success for the day.
Seasonal Factors Seasonal factors are defined as "the amount of correction needed in a time series to
adjust for the season of the year" (Chase, Jacobs, & Aquilano, 2006, Chapter 13). The word seasonal
does not always mean weather. Seasonal in business terms is usually classified as an activity of
some sort. In my situation, I don't have any seasonal factors that can affect my morning routine. My
husband has already left the house and I am alone in a stable environment (my home).
Confidence Intervals Confidence intervals allow us to pinpoint data to a degree of confidence. The
intervals are used to estimate the reliability of an estimate. Usually, the confidence levels that are
calculated are 90%, 95%, and 99%. The confidence intervals for my particular situation are as
follows:
|Confidence Intervals |Lower Confidence Interval (mins.) |Upper Confidence Interval (mins.) |

The Royal Theatre And Its Consortium
Executive Summary The following highlights the main issues affecting the Royal Theatre and its
consortium: since the issues is in regard to a claim exists on the basis of fraudulent
misrepresentation against the surrounding theatres and the Royal Theatre, a statistical analysis was
required to collect data from movie–goers to lead to an ethical conclusion. The analysis performed
includes inquiries in regard to the legality of Tommy's claim and whether it is justifiable under the
rule of misrepresentation; both fraud and innocent misrepresentation have been considered. In
addition, surveys were collected from groups of movie–goers, and the analysis concludes the
percentage of patronage preferences. It is recommended that Mr. Plex, and the ... Show more
Tommy indicates that the duration of time in which commercials were playing was 20 minutes.
Tommy was unaware that the theater would be showing commercials. Although he was unhappy
with the commercials, he stayed until he was discontent with the movie that he had paid for.
Ultimately, Tommy was denied a refund, as it is a policy of the Royal theatre; therefore, Tommy has
expressed his intent to initiate a class action lawsuit believing that his rights were violated. Analysis
Liability for Fraud: What Tommy Must Prove to Win Assuming that a contract existed at the time of
the event, between Tommy and the Theater, Tommy may use the following argument to prove his
fraudulent claim. Tommy may argue that (1) a representation was made and that was that one of the
terms of the contract provides that the movie will begin at 1 PM. Thus, he was induced by a
fraudulent claim because the movie started 20 minutes after commercials ended. Tommy may also
suggest (2) the representation was known to be false because the ticket stub indicated the movie
would start at 1 p.m. Therefore, (2) the plaintiff was expected to rely on the fact that the movie was
starting at 1 p.m.; moreover, (3) it was known to be false because the newspaper ad stated the time,
(4) and it was made with the intention that the plaintiff rely on it because the cashier at the ticket–
stub affirmed it, (5) the

Presidential Pestel Analysis
Presidential Knowledge In today's society most individuals are more concerned with modern
popular culture transmitted via mass media but those individuals sometimes do not have simple
knowledge about the United States presidential history. Conducting this survey I gained a
perspective on the knowledge of Chaffey College students in regards to the history of the past
president of the country they reside in. The survey I am conducting will contain four questions, two
will be modern popular culture and two will be about the history of the past presidents. Out of the
four questions I will only be focusing on one particular question on the past presidential history, "
Four of the country's most honored Presidents have their likenesses carved into ... Show more
I got the formula for the margin of error and applied it, 1.644853626*√62/100*(1–62/100)/100=
.0798389652 which will be the margin of error. I applied the formula of p̂ –E ≤ p ≤ p̂ +E since we
know p̂ = 62/100 and we calculated the margin of error (E) 1.644853626 then I applied the formula
62/100–1.644853626 ≤ p ≤ 62/100+1.644853626 and calculate it to be .5401610348 ≤ p ≤
.6998389652. I will calculate the same process for the two other confidence intervals.
.025
.95
.025
I did the diagram for the 95% confidence interval which then added the left side of .95 and .025 to
get .975. I used InvNorm(.975) and calculated the z score of 1.959963986 and applied the margin of
error formula 1.959963986*√62/100*(1–62/100)/100= .0951339949. Once finding the margin of
error I applied the formula 62/100–.0951339949 ≤ p ≤ 62/100+.09513399496 and I calculated
.5248660051 ≤ p ≤ .7151339949.
.005
.99

Advantages Of Portfolio Optimization
Robust Optimization Approach to Multi–Period Portfolio Selection Progress Report
Introduction
Investors always seek for a way that they can get back greatest return while enjoying minimized
risk. Instead of investing in a single asset, holding a portfolio is obviously less risky. However, how
to select the best portfolio among tens of thousands of assets in today's financial market? The
stringent need of investors promote the raising of modern portfolio theory. In 1952, Harry
Markowitz [1] established the fundamental model of modern portfolio theory: the Markowitz
model, also called the mean–variance model. This model aimed to achieve a tradeoff between the
expected return and the risk of return. As shown in Figure 1, among all efficient portfolios, the
efficient frontier consisted of all those with highest return at each given risk level. C_1,C_2,and C_3
were the investors indifference curves which showed that traders prefer portfolios with high return
or low risk. The tangent point R of the highest indifference curve and the efficient frontier gave the
optimized portfolio. Figure 1: Efficient Portfolio and Trader's Indifference Curve in ... Show more
One direction among them is the robust portfolio optimization. It was carried out to compensate the
instability and sensitivity of the classical optimal model due to the uncertainty of the coefficients of
variables. This paper aims to use the robust optimization techniques to take input uncertainties into
consideration. Moreover, the model achieved should be applicable to multi–period portfolio
selection problems. In the next section, previous studies on this topic were reviewed. The following
section introduced the method used to get the final result. Section 4 discussed the achievement until
now and the problems that have arisen. The final section summarized the current research progress
and possible direction in

Ops 571 Statistical Process Control
Chase, Jacobs and Aquilano pose questions such as, "How many paint defects are there in the finish
of a car? [and] Have we improved our painting process by installing a new sprayer?" These
questions are meant to investigate and apply different techniques that we can use to improve the
quality of life. Quality control not only applies to manufacturing techniques, it can also be applied to
everyday life. This discussion will focus on a specific method of quality control called statistical
process control that will ensure my morning process is effective.
One method of quality control can be pursued through process control procedures like statistical
process control or SPC. SPC "involves testing a random sample of output from a process to ... Show
more content on Helpwriting.net ...
The more data that is available the stronger your confidence intervals are.
UCL = p + z Sp
UCL = p + 3Sp
UCL = .08333 + 3(.05050) = .23483
LCL = p – z Sp
LCL = p – 3Sp
LCL = .08333 − 3(.05050) = –.06817
In the control chart, the data from the sample stays in between the controls. This means that my
process in the morning is working properly and is effective. Now, it is important to look to the
future trends in order to predict seasonal factors. "A seasonal factor is the amount of correction
needed in a time series to adjust for the season of the year." (Chase, Jacobs & Anquilano, 533)
Seasonal factors may affect the samples by taking into consideration factor based on seasons or time
periods. The alarm clock that is used to wake me up in the morning is not dependent on any factors
of time or season.
Statistical process control is one way to control quality and make sure goals are attained. Statistical
methods show that the samples taken can create visual representations that conclude my alarm clock
is an effective method to starting my morning process. This ensures that it is operating at its fullest
potential.
REFERENCES
Chase, R. B., Jacobs, F. R., Aquilano, N.J. Operations management for competitive advantage (11th
ed). New York: McGraw Hill/Irwin.

Green Jr. K, Toms L, Stinson T. STATISTICAL PROCESS CONTROL APPLIED WITHIN AN
EDUCATION SERVICES ENVIRONMENT. Academy Of Educational Leadership Journal [serial
online]. June 2012;16

1) Various Federal Agencies Impact Marketing...
1) Various federal agencies impact marketing activities. Research each agency below, discuss the
elements of marketing that are impacted by that agency, and briefly present a recent marketing case
or issue on which each agency has focused (1 paragraph, 100 words per agency).
Federal Trade Commission:
The Federal Trade Commission is an agency of the United States government that works to prevent
fraudulent, deceptive, and unfair business practices. They also provide information to help
consumers' stop, and avoid scams and fraud. The FTC has an impact in the economy, because its
purpose is to help the US maintain a more stable market and economy by helping customers and
keeping a more accurate information of product, in aspects such ... Show more content on
Helpwriting.net ...
(FDA, 2016)
Consumer Product Safety Commission:
The mission of CPSC is protecting the public from unreasonable risks of injury or death associated
with the use of the thousands of types of consumer products under the agency 's jurisdiction. Deaths,
injuries, and property damage from consumer product incidents cost the nation more than $1 trillion
annually. The CPSC is committed to help and protect consumers and families to products that
threatened one's life, such as fire, electrical, chemical or mechanical hazard. A recent CPSC cases
was against TOY R US, they were accused for a "Wooden Coloring Cases" due to violation of lead
paint standard. About 27,000 cases have excess of lead, which violates the federal lead paint
standard. (CPSC, 2017)
2) Do you know Danica from the Philippines, Peter from London, Nargis from India, Marina from
Russia, Chieko from Japan, or Miran from the United States? These are some of the babies whose
parents claimed they were the 7th billion human born into the world. The world population
continues to grow, even though women are having fewer children than before. Markets are made up
of people, and to stay competitive, marketers must know where populations are located and where
they are going. The fertility rate in the United States is declining and the population is aging,
creating opportunities as well as threats for marketers. That is why tracking and predicting
demographic trends are so important in

Confidence Interval
34. Information from the American Institute of Insurance indicates the mean amount of life
insurance per household in the United States is $110,000. This distribution follows the normal
distribution with a standard deviation of $40,000. a. If we select a random sample of 50 households,
what is the standard error of the mean? b. What is the expected shape of the distribution of the
sample mean? c. What is the likelihood of selecting a sample with a mean of at least $112,000? d.
What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the
likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. a) The
standard error of the mean, S.E. = /Sqrt(n) = 40,000/Sqrt(50) = 5656.8542 ... Show more content on
Helpwriting.net ...
Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to
conclude that less than 5 percent of the employees are not able to pass the random drug test? a) Here
the sample proportion, p = 14/220 = 0.06364 SE = Sqrt[p(1–p)/n] = Sqrt[0.06364*(1–0.06364)/220]
= 0.01646 The z– score for 99% confidence is z = 2.576 The 99 percent confidence interval for the
proportion of applicants that fail the test is given by (p – z*SE, p – z*SE) = (0.06364 –
2.576*0.01646, 0.06364 + 2.576*0.01646) =(0.06364 – 0.04239, 0.06364 + 0.04239) =(0.02124,
0.10603) =(2.12%, 10.6%) Since 10% falls within the 99% confidence interval but near the upper
limit of 10.6%, it may not be reasonable to conclude that more than 10% are now failing the test. (b)
Here the sample proportion, p = 14/400 = 0.035 SE = Sqrt[p(1–p)/n] = Sqrt[0.035*(1–0.035)/400] =
0.0092 The z– score for 99% confidence is z = 2.576 The 99 percent confidence interval for the
proportion of applicants that fail the test is given by (p – z*SE, p – z*SE) = (0.035 – 2.576 * 0.0092,
0.035 + 2.576 * 0.0092) = (0.035 – 0.0237, 0.035 + 0.0237) = (0.0113, 0.0587) = (1.13%, 5.87%)
Since 5% falls within the 99% confidence interval, we may conclude that less than 5 percent of the
employees are not able to pass the random drug

Statisitical report Essay
BU1007 Business Data Analysis and Interpretation Singapore Campus, Study Period SP53, 2013
Statistical Report Analysis of Case study 3: Heavenly Chocolates website transactions Prepared for
Dr Tjong Budisantoso Done by Mr. Keung, Tseung Student ID : 12776910 20/12/2013 Table of
contents Introduction
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Question a
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Question b ... Show more content on Helpwriting.net ...
Vice versa, the less time spent and page viewed, the less payment made. The total amount of
transaction have reached two peak points recorded $758.86 on Monday and $925.43 on Friday.
From Tuesday to Thursday, has a slight decline from $414.86 to $294.03. Furthermore, on both
Saturday and Sunday, the transactions have a relative low amount compared to the rest of the days,
recorded $394.4 and $222.15, respectively. c). You have to develop a scatter diagram and compute a
simple correlation coefficient to explore the relationship between the number of pages viewed and
the amount ($) spent (use horizontal axis for the number of pages viewed). Correlations Analysis
Step 1: Scatter Diagram: the amount ($) spent and the number of pages viewed Independent
variable: Number of pages viewed Dependent variable: Amount Spent Step 2: Elaboration The
Scatter diagram illustrate a strong positive linear correlation and a direct relationship between the
two variables. To apply the Coefficient of Correlation (r) formula: Correlation coefficient =
0.9(Solved by Excel) When r is close to 1, which proves it is a positive correlation, so there is a
direct relationship can be seen between the number of pages viewed and the amount spent , and the
value of 0.9 is rather close to 1.00, so it can be concluded that the

Yellow Dung Fly Breeding
Success of male yellow dung fly breeding.
Introduction
Looking at the yellow dung fly, Scathophaga stercoraria, it is seen that the species experiences
sexual dimorphism in the form of both size and colour. This gives changes between the males and
females that are specially adapted to give them the best chance of producing offspring. Males are
considerably larger than the females and have clearly defining yellow hair along the abdomen,
whereas females are smaller and are more of an olive green colour. Copulation for this species
occurs on large mammal faeces as it acts as a source of nutrition for larvae once eggs are hatched.
The right to breed with a female is highly competitive among yellow dung flies and only the most
suited mate or the most opportunistic mate will be able to pass on its sperm. Assessing the size of
males is key when looking at the mating probability of the yellow dung fly ... Show more content on
Helpwriting.net ...
Mean HTL of the 4 different male yellow dung fly categories. The mean size for males found on the
faeces is clearly larger than those off the pat, and the mean HTL is the largest in paired males on the
pat.
Looking at Fig 1. it can be speculated that larger males are more successful at copulating, however
with the data presented there is not a large enough sample size over the two sites to confirm this.
Males on the pat are larger on both instances compared to males off the pat but due to overlapping
of error bars, it cannot be stated with confidence that they are indeed more successful. The only
statement that can be made with confidence is that paired males on the pat are considerably larger
than solo males off the pat, this can be stated as the error bars do not overlap. With only one sample
of paired males off the pat, there is no way to know if error bars would overlap with any other group
if the sample size was larger. The paired males off the pat then can be only taken as an estimation
because of this.

holla Essay
Introduction to Confidence Intervals (page 248)
In chapter 7 we discussed how to make inferences about a population parameter based on a sample
statistic. While this can be useful, it has severe limitations. In Chapter 8, we expand our toolbox to
include Confidence Intervals. Instead of basing our inference on a single value, a point estimate, a
Confidence Interval provides a range of values, an interval, which – at a certain level of confidence
(90%, 95%, etc.) – contains the true population parameter. Having a range of values to make
inferences about the population provides much more room for accuracy than making an inference
off of only one value.
When we worked with probabilities based on sample means, we learned that there is ... Show more
Assume that the population standard deviation is fairly stable at 1.8 hours.
Calculate the 95% confidence interval for the population mean weekday sleep time of all adult
residents of this Midwestern town.
95% = 1.96 6.4 +(–) 1.96 x (1.8/sqrt80) = [6.01, 6.79]
Can we conclude with 95% confidence that the mean sleep time of all adult residents in this
Midwestern town is not 7 hours?
Yes, we can conclude with 95% confidence that the mean sleep in this Midwestern town is not 7
hours because the value 7 does not fall within the confidence interval.
Confidence Interval for the Population Mean when Sigma is Unknown
While it is possible that we could know enough about our population to make an assumption about
what the population standard deviation is, it is much more likely that if we do not know the
population mean, then we do not know the population standard deviation. In this case, we can't use
the standard normal distribution, and we use a different distribution, the Student's t distribution.
Instead of , we use s, the sample standard deviation.
The formula is: x±ta/2, df sn .
/2 is still defined the same way, and df is degrees of freedom, calculated as n–1, where n is the
sample size. Degrees of freedom determine the extent of the broadness of the tails of the
distribution; the fewer the degrees of freedom, the broader the tails.
Solve the following problems:
Find ta/2, df for the following confidence levels: T–SCORE

Quantitative Outcomes For Nursing Students Essay
Title
The title of the article critiqued is, "Quantitative Outcomes for Nursing Students in a
Flipped Classroom". Authors of this study are listed as Susan Ann Harrington, Melodee Vanden
Bosch, Nancy Schoofs, Cynthia Beel–Bates and Kirk Anderson. While this nurse was researching
for Quantitative studies, this article drew interest to this nurse. While this nurse begins to advance
education in nursing, the article was interesting to read. Pedagogy or the art of teaching; education;
instructional methods (dictionary.reference.com). The concept of the flipped classroom is
comparable to the advanced learning techniques this nurse is experiencing. Hurt (2013), stated
"flipping isn't really a new concept, individual instructors in various disciplines have used variations
of it for years" (Hart, 2013, p. 1).
Authors
The article was written by five authors. No credentials were stated behind each authors names at the
beginning of the article, however, at the end of the article credentials of each author was discussed.
Susan Ann Harrington, PhD, RN, assistant professor at Kirkhof College of Nursing, Grand Valley
State University. Melodee Vanden Bosch, PhD, RN, assistant professor. Nancy Schoofs, PhD, RN,
associate professor at Kirkhof College of Nursing. Cynthia Beel–Bates, PhD, RN, FGSA associate
professor at Kirkhof College of Nursing. Kirk Anderson, PhD, associate professor of statistics at
Allendale, Michigan. With the authors of the study all having their PhD, the

Web Calculater Exercise 2 Essay
Web Calculator Exercise 2
Daniel Alvarado
Liberty University
Z test & One sample t–test
1. A researcher is interested in whether students who attend private high schools have higher
average SAT Scores than students in the general population. A random sample of 90 students at a
private high school is tested and and a mean SAT score of 1030 is obtained. The average score for
public high school student is 1000 (σ= 200).
a. Is this a one– or two tailed test? This test is a one – tailed test, because the researcher wants to
know that students who attend private high school will have a higher Sat score, which makes the
alternative hypothesis a prediction in favor of the private school over the public high school. ...
Show more content on Helpwriting.net ...
He measured their resting pulses. Their pulses were 45, 45, 64, 50, 58, 49, 47, 55, 50, 52 beats per
minute. The average resting pulse of athletes in the general population is normally distributed with a
pulse rate of 60 beats per minute. a. What statistical test should be used to analyze the data?
For this statistical test we can use the single sample research t–test, predicting that the runner pulse
will be slower than the other athlete pulses.
b. Is this a one– or two– tailed test?
This is a one tailed test, because the researcher predicts that the result will be slower than other
athletes.
c. What are H0 and Ha for this study?
H0 for the study will be that the runners pulse will be the same as the other althete, and the Ha for
the study is that runners pilse is slower than other athletes.
d. Find tcv from appendix A in Jackson's text.
Df = –1 Df 10 –1 = Df = 9

Using alpha 0.05
Tcv = 1.833 e. Compute t obt Tobt = –4.4799 http://in–silico.net/tools/statistics/ttest
f. Should H0 be rejected? What should the researcher conclude?
The H0 Should not be rejected because the runner pilse is slower than the one from other athletes.
3. A researcher hypothesizes that people who listen to music via headphones have greater hearing
loss and will thus score lower on a

Ops/571 Week 5 Paper
Statistical Process Control
OPS/571
Over the past five weeks, data has been collected from the process of getting my daughter, Sophie,
ready for daycare in the morning. I have tracked six key areas, or steps, in the process: The time it
takes to wake her up, The time it takes to get her to go to the bathroom, The time it takes to get her
stuff ready, The time it takes to get her dressed, The time it takes to brush her teeth and hair, and The
time it takes to get her into the car. In this paper, I will discuss what I have discovered based on this
data, I will identify roadblocks to the process and recommend strategies to overcome them, and I
will discuss the variables which affect the steps in the process. Finally I will discuss ... Show more
The following step is getting her dressed. This is also the step that has been identified as the main
bottleneck in the process. This step has a mean time of 16.5 minutes and a standard deviation of
5.16. The extreme deviation in this step is based on a single primary factor: fussiness. On some
mornings, Sophie is happy with the clothes I have chosen and eager to get dressed. On other
mornings she does not like what I have chosen and/or does not want to get dressed. On good days, it
generally takes from ten to fifteen minutes to get her dressed, on days when she doesn't want to get
ready, it can take up to twenty–seven minutes. Since this step is the primary bottleneck in the
process, this is the step I have focused on in regards to improving efficiency. The strategy I have
recommended in order to improve efficiency is simple. On the night before, prior to Sophie's
bedtime, I will pick out her clothes for the following day, with her help. This will allow her to have
some level of control over what she wears and she will also know ahead of time what outfit she will
be putting on. Doing this should eliminate the problem of her not liking the clothes I have picked
out for her and it should improve the mean time and decrease the standard deviation of the time it
takes to get her dressed. Brushing her teeth and combing her hair is generally an easy process as she
enjoys

Confidence Interval
Elon poll Understand the problem: The problem is asking (a) what is the relationship between
confidence interval, point estimate, and margin of error for poll results, and (b) how one would
design a poll to make the margin of error smaller. The information given is the confidence interval
and the sample size. Solve the problem: (a) given a confidence interval, determine the point estimate
and the margin of error. Confidence interval is symmetric, so the point estimate has to be the
midpoint of the interval. Confidence interval is 46.4% to 59.2%, point estimate is 52.8%. The
margin of error is defined as the half–width of the confidence interval the distance between either
end of the confidence interval and the point estimate. Margin of error is 6.4%. (b) the margin of
error is related to sample size for a given % confidence interval. Where Z1–Î± is the critical value
for the confidence interval 1–Î±. For a 95% confidence interval, Î±=0.05 and Z = 1.96 SE is the
standard error p is the proportion n is the sample size This equation shows that margin of error is
inversely related to the square root of the sample size. In order to get a smaller margin of error, one
needs to have a larger sample size. In this problem, for Z=1.96, p=0.528, and n=125 people, the
margin of error is 6.4%. If the sample size were changed to 1000, then the margin of error would be
3.1%. Reflect on the problem: The crux of this problem is the concept of confidence interval. This

Math: Normal Distribution and Confidence Interval
Statistics Math 1342 Final Exam Review
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the
question. Provide an appropriate response. 1) A card is drawn from a standard deck of 52 playing
cards. Find the probability that the card is an ace or a heart. 17 7 3 4 A) B) C) D) 52 52 13 13
Answer: D 2) The events A and B are mutually exclusive. If P(A) = 0.7 and P(B) = 0.2, what is P(A
or B)? A) 0.5 B) 0.9 C) 0.14 D) 0 Answer: B 3) A coin is tossed. Find the probability that the result
is heads. A) 1 B) 0.5 C) 0.1 Answer: B 4) Suppose you are using
= 0.01 to test the claim that µ = 950 using a P–value. You are given the
1)
2)
D) 0.9
3)
4) ... Show more content on Helpwriting.net ...
Round your answer to three decimal places. A) –2.617 B) –3.010 C) –3.186 Answer: D 17) Find the
area of the indicated region under the standard normal curve. 17)
= 0.05 if
15)
16)
D) –2.189
A) 0.0968 Answer: C
B) 0.0823

C) 0.9032
D) 0.9177
18) A sample of candies have weights that vary from 2.35 grams to 4.75 grams. Use this information
to find the upper and lower limits of the first class if you wish to construct a frequency distribution
with 12 classes. A) 2.35–2.65 Answer: B 19) Assume that the heights of men are normally
distributed. A random sample of 16 men have a mean height of 67.5 inches and a standard deviation
of 1.4 inches. Construct a 99% confidence interval for the population standard deviation, . A) (0.8,
2.1) B) (1.0, 2.6) C) (0.9, 2.5) D) (1.0, 2.4) Answer: C B) 2.35–2.55 C) 2.35–2.54 D) 2.35–2.75
18)
19)
3
20) In a survey of 2480 golfers, 15% said they were left–handed. The survey 's margin of error
was 3%. Construct a confidence interval for the proportion of left–handed golfers. A) (0.12, 0.15) B)
(0.12, 0.18) C) (0.11, 0.19) D) (0.18, 0.21) Answer: B 21) For the stem–and–leaf plot below, what is
the maximum and what is the minimum entry?
Key : 11 7 = 11.7 11 3 7 12 4 6 6 7 8 9 13 0 1 1 2 3 6 6 7 8 8 14 3 4 6 6 8 9 9 9 15 0 1 1 2 3 7 7 8 9
16 2 2 5 7 8 8 9 9 17 1 7 A) max: 177; min: 113 C) max: 17.7; min: 11.7
20)
21)
B) max: 17.7; min: 11.3 D) max: 17.1; min: 11.3

Campus Safety Analysis: Capella University
MBA 6018
Ryan Smith
Unit 6 Assignment 1
7/3/15
Campus Safety Analysis, Minnesota – Prepared for Capella University
Due to recent concerns from the Capella University Security team, their recommendation is to
analyze data from crime rates on Minnesota College campuses, to understand crime trends, and
reduce occurrences on campus. To gain insight on crime rates, data was pooled from the U.S. Dept.
of Education's website on Campus Crime. This data will be utilized to examine the following
questions from the security team:
What crimes were most commonly committed on Minnesota campuses between 2009 and 2011?
Based on the data, would you say the crime rates decreased or increased from 2009 to 2011?
The campus security ... Show more content on Helpwriting.net ...
Although the standard deviation is large between numbers in the data set, which increases the
standard error rate, The Test shows at all levels of α, evidence that supports the client's belief that
Public Institutions have a mean crime rate that is larger than the mean crime rate at Private
Institutions.
3. A confidence interval was requested, at a 95% level for the difference in total campus crime rates
between public and private institutions in Minnesota. Utilizing the information in Table 2,
summarized from the Sample Data for all colleges in Minnesota, a Confidence interval for the crime
rates for Public, and for Private institutions was calculated below.
a. Confidence Interval – Public Institutions
Input Variables –– Public Institutions
Sample Mean (x–bar)
10.67
Sample standard deviation (s):
21.45
Sample Size (n)
67
Confidence Level:

0.95
Intermediate Calculations ––
Degrees of freedom:
66
Standard Error of the Estimate:
2.6205346
Prob. in the tails for this Conf Level:
0.05
t–Multiple:
1.9965644
Lower limit:
5.4379339
We can be 95% confident that the population mean
Upper limit:
15.90
lies between 5.44 and 15.9
Margin of error:
5.23
b. Confidence Interval – Private Institutions
Input Variables ––
Sample Mean (x–bar)
6.04
Sample standard

Mat 540 Ash Course Tutorial/Tutorialrank Essay
MAT 540 Entire Course Click Here to Buy the Tutorial
http://www.tutorialrank.com/MAT/ASHFORD–MAT–540/product–7758–ashford–mat–540–entire–
course For more course tutorials visit www.tutorialrank.com Tutorial Purchased: 5 Times, Rating:
A+ MAT 540 Week 1 DQ 1 Gallup Poll MAT 540 Week 1 DQ 2 Qualitative vs. Quantitative MAT
540 Week 2 DQ 1 Scatter Plot MAT 540 Week 2 DQ 2 Correlation and Causation MAT 540 Week 3
DQ 1 Actuaries MAT 540 Week 3 DQ 2 Probability Video Analysis MAT 540 Week 4 DQ 1 Normal
or Bell–Shaped Curve MAT 540 Week 4 DQ 2 Standardized Testing MAT 540 Week 5 DQ 1
Repeated Sampling MAT 540 Week 5 DQ 2 Confidence Intervals MAT 540 Week 6 DQ 1 Statistical
Concept MAT 540 Week 6 Final Paper ... Show more content on Helpwriting.net ...
These can be found in the Ashford Online Library by going to a research database such as JSTOR or
ProQuest and searching for your topic. If available, narrow your search category to scholarly
journals, including peer–reviewed and full–text documents only. At least three of your sources must
be from peer–reviewed scholarly journals. Introduction Describe, in third person, the issue you've
selected, why it was selected, the perspective of the approach, and the scope of the paper. The
introduction describes what will be covered in the paper, and serves to engage the reader. Be
specific and to the point. Statement of the Problem Describe, in third person, why the topic is
relevant, and a problem or issue associated with the topic. It is important to provide scholarly
sources in support of your discussion of the problem or issue. Literature Review In the literature
review, process and develop a conceptual framework of the issue or problem that is being
researched. This section should present a comprehensive review of the historical and current
literature on the topic. The literature review should: Identify the topic, and describe specific research
related to the topic (describe the study, sample, findings, important points from the discussion in the
research – describe any variables that may influence the findings of the research). Address any key
issues such as political, social, legal, and/or ethical implications the

Ethical Behavior At Rocky University
Ethical Behavior at Rocky University Ethical behavior should be expected from college students;
however, it is not always present. There is a moral obligation for students, earning a degree, that
they are in fact the ones doing the homework and tests. The prevalence of online institutions has
helped enable a multitude of working individuals earn a degree. Without the classroom setting, it is
arduous for a professor to ensure who is on the other end of the keyboard and monitor. In fact,
Gopala, Paswan, and Qin (2015) stated that regarding online classes and degree programs, "as a
component of distant–learning initiatives, have attracted a lot of attention from educational
institutions, administrators, policymakers, and society at large. According to results of Pew
Research Center 2011 Surveys, about 23 percent of college graduates have taken online courses" (p.
67). Rocky University has conducted a survey asking 90 of its students whether or not he or she
may have cheated on any assignments. Rocky University is interested in comparing itself with other
Universities as well as setting a benchmark of its own situation. The following will enable the reader
to understand just where Rocky stands after the results of the survey are tallied. Overview of the
Rocky University Survey The data analyzed herein, were gathered from 90 of Rocky University
students. There were 46 female respondents and 44 male respondents. Each was asked a series of
three questions: have you ever copied

Completely Randomized Factorial Anova
Chapter 16
Completely Randomized Factorial ANOVA
This tutorial describes the procedures for computing F tests for a completely randomized factorial
analysis of variance design. The reading–speed data in Table 16.4–2 of the textbook are used to
illustrate the procedures.
1. Enter a description of the data in the SPSS Data Editor following steps 1–4 described in the
Frequency Distribution tutorial for Chapter 2. Use rows 1, 2, and 3 of the SPSS Data Editor Variable
View window to describe the two independent variables and the dependent variable. There are two
levels of room illumination, Illumination Level, denoted by a1 and a2. You identify the illumination
levels in the Values cell of the Variable View window. When you click on ... Show more content on
Helpwriting.net ...
Next, click on Tukey to select this multiple comparison procedure. The preferred procedure, Fisher–
Hayter statistic, is not an option in SPSS. When the n's are equal, the Tukey and Fisher–Hayter
statistics are equal. You can compute the Fisher–Hayter statistic from the information in the ANOVA
and Multiple Comparisons tables given later. The Fisher–Hayter statistic can be referred to the
Studentized Range table (Table D9) in your textbook to obtain a slightly more powerful test. Click
on the Continue button to return to the Univariate window.
11. In the Univariate window, click on the Options button to bring up the Univariate: Options
window shown here.
[pic]
12. Select (Overall) in the Factor(s) and Factor Interactions box and click on the arrow beside the
Display Means for box. This moves (Overall) into the Display Means for box. Repeat the procedure
for I_level, T_size, and I_level*T_size. Next, click on the Descriptive Statistics box, Estimates of
effect size box, and the Homogeneity tests box. Then click on the Continue button to return to the
Univariate window. Click on the OK button in the Univariate window to obtain the ANOVA output
shown here.
[pic]
[pic]

[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
13. The Between–Subjects Factors window displays the number of observation in each level of the
two independent variables.
The Descriptive Statistics window displays the mean, standard deviation, and sample size for

Confidence Intervals Consider the following question: someone takes a sample from a population
and finds both the sample mean and the sample standard deviation. What can he learn from this
sample mean about the population mean? This is an important problem and is addressed by the
Central Limit Theorem. For now, let us not bother about what this theorem states but we will look at
how it could help us in answering our question. The Central Limit Theorem tells us that if we take
very many samples the means of all these samples will lie in an interval around the population
mean. Some sample means will be larger than the population mean, some will be smaller. The
Central Limit Theorem goes on to state that 95% of the sample means will lie ... Show more content
on Helpwriting.net ...
Find the 98% confidence interval for the population proportion. Solution: We first find zc for the
98% confidence interval. Consulting the table above we find it to be –2.33. Also [pic] Thus [pic].
We can be 98% confident that the population proportion, who thought that the homeless are not
adequately assisted by the government, is between 0.819 and 0.855. Example 5: In a study of 150
accidents that required treatment in an emergency room, 36% involved children less than 6 years of
age. Find the 90% confidence interval of the true proportion of accidents that involve children less
than 6 years old who require treatment in an emergency room. Solution: [pic] Thus we can be 90%
confident that the true proportion of accidents that involve children less than 6 years old, who
require treatment in an emergency room, is between 0.295 and 0.425. Sample Mean and Standard
Deviation are known, in the case of Sample Size [pic] When working with samples whose size is
less than 30, we often have to work with a new and different concept, that of "degrees of freedom".
Let us simply not bother with its origin but state that in the case of confidence intervals with sample

Statistics: Normal Distribution and Confidence Interval
Study Set for Midterm II, Chapters 7 & 8
ESSAY. Write your answer in the space provided or on a separate sheet of paper.
1)
The average score of all pro golfers for a particular course has a mean of 70 and a standard
deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average
score of the 36 golfers exceeded 71. 2) At a computer manufacturing company, the actual size of
computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1
centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample
mean will be between 0.99 and 1.01 centimeters? MULTIPLE CHOICE. Choose the one alternative
that best ... Show more content on Helpwriting.net ...
19) Referring to Table 7–7, the population mean of all possible sample proportions is ________. 19)
_____________ 20) Referring to Table 7–7, the standard error of all possible sample proportions is
________. 20) _____________ 21) Referring to Table 7–7, ________% of the samples are likely to
have between 35% and 40% who take advantage of online customer service. 21) _____________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the
question.
22)
If you were constructing a 99% confidence interval of the population mean based on a sample of
n=25 where the standard deviation of the sample s = 0.05, the critical value of t will be: 22) ______
A)
2.7874 B) 2.7969 C) 2.4851 D) 2.4922 23) A confidence interval was used to estimate the
proportion of statistics students that are female. A random sample of 72 statistics students generated
the following 90% confidence interval: (0.438, 0.642). Using the information above, what size
sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95%
confidence? 23) ______
A)
150 B) 105 C) 420 D) 597 24) A major department store chain is interested in estimating the average
amount its credit card customers spent on their first visit to the chain's new store in

Mat 300 Statistics Bottling Company Case Study Essay
Assignment 1: Bottling Company Case Study
MAT 300 Statistics
Strayer University
December 15, 2013
Calculate the mean, median, and standard deviation for ounces in the bottles.
Bottle Number Ounces Bottle Number Ounces Bottle Number Ounces
1 14.5 11 15 21 14.1
2 14.6 12 15.1 22 14.2
3 14.7 13 15 23 14
4 14.8 14 14.4 24 14.9
5 14.9 15 15.8 25 14.7
6 15.3 16 14 26 14.5
7 14.9 17 16 27 14.6
8 15.5 18 16.1 28 14.8
9 14.8 19 15.8 29 14.8
10 15.2 20 14.5 30 14.6
Total: 446.1/ 30 = 14.87
Mean: (Avg) 14.87
Median: (14.87 + 14.87) / 2 = 14.8
Standard Deviation: 0.55033
Construct a 95% Confidence Interval for the ounces in the bottles.
With a mean ... Show more content on Helpwriting.net ...
Provide the following discussion based on the conclusion of your test: If you conclude that there are
less than sixteen (16) ounces in a bottle of soda, speculate on three (3) possible causes. Next,
suggest the strategies to avoid the deficit in the future.
I conclude that there are less than sixteen (16) ounces in some of the bottles of soda. Some causes
that may contribute to the issues may include a mechanical defect with the company's equipment,
employees failed to notice or inspect the bottles before packaging, or the product evaporates or goes
flat after a period time has passed. As manager, I recommend that there will be a monthly
maintenance check on the equipment and documented. Also, employees will be monitored in how

thorough the product is inspected before leaving the assembly line and packaged. Lastly, I will
stress to consumers of the possibility that the product may have experienced evaporation and gone
flat depending on how long it stays on the shelf unopened.
Resources
Lane, D. M. (n.d.). Steps in hypothesis testing. Retrieved from
http://onlinestatbook.com/2/logic_of_hypothesis_testing/steps.html Sauro, J. (2006, February 06).
Graph and calculator for confidence intervals for task

Biophysical Ecology and Pattern Recognition Essay
Biophysical Ecology and Pattern Recognition
Abstract:
This study was undertaken to investigate behavioral adaptations of a lizard, Lacertilia, to its
environment. Twelve peeps, representing the lizards, were placed in a habitat with two
microhabitats of different temperatures. Six peeps were placed in one microhabitat, and six in the
other. The internal temperature of these "lizards" was measured over a period of 20 minutes to see if
their body temperatures matched that of their environment and to make inferences about the
behavioral adaptations the organism might acquire to maintain its body temperature. One
microhabitat was on a tree and under the branches; the other was at the base of the tree. We
hypothesized that the microhabitat in ... Show more content on Helpwriting.net ...
Groups decided on which organism to experiment with and a hypothesis about this organism with
testable predictions to explain their theory in terms of temperature differences.
Conducting the experiment
I–buttons were used to test predictions and to carry out an experiment. Once an experiment was
planned based on predictions, manipulative experiments to test predictions, independent variables,
dependent variables, and controls, groups obtained I–buttons and peeps, which were used to
simulate the organism. Groups chose one habitat and put 6 peeps in each microhabitat, which
differed in temperature. The I–buttons were inserted into each peep and groups waited 20 minutes
until taking the I–buttons out. Observations about each microhabitat were recorded. For our
experiment, we tried to find a habitat that had microhabitats of very different temperatures so that
we could clearly see how the organism reacts to different temperatures. Our microhabitats were on a
tree. One of them was on the actual tree, near the leaves. The other was at the base of the tree on the
soil, and we used a lizard, Lacertilia, as our organism.
Data Analysis
After conducting the experiment, statistical analysis was done to see how the observed results
compared to expected results (hypotheses). ). The data was obtained from blackboard and put it into
Excel along minute intervals of 1–20. A table was set up in excel to include 5 replicates, average
temperatures,

Wentworth Medical Center
Results for the Good Health Survey Florida: Confidence Level at 95% is 1.001191951; Upper Limit
is 6.551; Lower Limit is 4.548. The expected value has a 95% chance of being in the confidence
range. In the case of the above problem, it has a 95% chance of being between 6.551 and 4.548.
New York: Confidence Level at 95% is 1.0298559; Upper Limit is 9.0299; Lower Limit is 7.0299.
The expected value has a 95% chance of being in the confidence range. In the case of the above
problem, it has a 95% chance of being between 9.0299 and 7.0299. California: Confidence Level at
95% is 1.3278753; Upper Limit is 8.3779; Lower Limit is 5.7221. The expected value has a 95%
chance of being in the confidence range. In the case of the above problem, ... Show more content on
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Some body that has heart ailment will probably be more depressed than somebody with hyper
tension. Two–way ANOVA: 1–What is the response variable? The levels of Depression 2–What are
the factors? Graphical Location and Health. 3–How many levels do the factors have? Factor1 has
two levels (Good Health and Chronic Health), and Factor2 has three levels (Florida, North Carolina,
and New York) 4–How many treatments are there, and what are they? Six treatments: GH–F, GH–
NY, GH–NC, CH–F, CH–NY–CH–NC. These are sometimes referred to as factor–levels. 5–How
many replicates are there in each treatment? 19 replicate each treatment. Null hypothesis: H0: All
treatment means are equal Alternative hypothesis: Ha: at least one treatment mean is not equal
Conclusion, at the .05 significance level, there is a difference between Good Health and Bad Health,
between Florida, North Carolina and New York, and between Heath and States. Tukey Analysis
First, we would reject the null hypothesis of no interraction, p = .2620 suggests that there is no
interraction between health and States. Second, we can reject the null hypothesis concerning no
differences between health, p = .7.11E–27 is Very strong evidence of a difference between Good
Healty and Chronic Health. Third, we can reject the null hypothesis concerning states, p = .05 is
evidence that there is differences

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Ois-Quiz Study Guide Chapter 8-9 KFUPM STAT 211 Quiz