Your company sells exercise clothing and equipment on the Internet. To design the clothing, you collect data on the physical characteristics of your different types of customers. Here are the weights (in kilograms) for a sample of 25 male runners

This is a take-home final exam. Please complete the exam on your own! You will perform some
operations in Excel. Please make a pdf file with your answers to the following 5 questions
amounting to 30 points. Submit both the pdf file and your Excel file on Canvas. No late
submissions will be accepted. The grade will be evaluated based on both final answers and
calculation steps.
Question 1. (7pts)
Your company sells exercise clothing and equipment on the Internet. To design the clothing, you
collect data on the physical characteristics of your different types of customers. Here are the
weights (in kilograms) for a sample of 25 male runners (You can also see the data in Sheet 1 of
the excel file “Final”). Assume that these runners can be viewed as a random sample of your
potential male customers.
62.5 68.7 61.8 63.2 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 67.8
65.6 65.5 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 62.1
Suppose the standard deviation of the population is known to be σ = 4.5 kg. Answer Parts a) ~ d).
a) What is ̅
, the standard deviation of ̅?
b) Give a 95% confidence interval for , the mean of the population from which the sample is
drawn.
c) Will the interval contain the weights of approximately 95% of similar runners? Explain your
answer.
d) If you want the 95% margin of error to be 1 kg or less, what sample size would you use?
Suppose that the standard deviation of the population is unknown. Answer Parts e) ~ g).
e) What is ̅
, the standard error of ̅?
f) Give a 95% confidence interval for , the mean of the population from which the sample is
drawn. (Hint: at first find the degrees of freedom)
g) You want to test H0: = 61.3 kg versus Ha: ≠ 61.3 kg. Do you reject H0 at the 5% significance
level?
Question 2. (3pts)
The Survey of Study Habits and Attitudes (SSHA) is a psychological test designed to measure the
motivation, study habits, and attitudes toward learning of college students. These factors, along
with ability, are important in explaining success in school. Scores on the SSHA range from 0 to
200. A selective private college gives the SSHA to an SRS of both male and female first-year
students.
The scores for the women are as follows:
156 109 137 115 152 140 154 178 111
123 126 126 137 165 165 129 200 150
Here are the scores of the men:
118 140 114 91 180 115 126 92 169 139
121 132 75 88 113 151 70 115 187 114
You can also find these data in Sheet 2 of the excel file “Final”.
Most studies have found that the mean SSHA score for men is lower than the mean score in a
comparable group of women. Test this supposition here. That is, state the null and alternative
hypotheses, carry out the test and obtain a P-value, and give your conclusion. (Hint: you can use
the Excel function “t-Test: Two-Sample Assuming Unequal Variances” in “Data Analysis”)
Question 3. (4pts)
A cell phone manufacturer would like to know what proportion of its customers are dissatisfied
with the service received from their local distributor. The customer relations department will
survey a random sample of customers.
a) Find the sample size needed if the margin of error of the 95% confidence interval is to be 0.02
or less.
b) Suppose the sample size is 2500 and 12% of the sample say that they are dissatisfied. What is
the margin of error of the 95% confidence interval?
c) You want to test the null hypothesis that the population proportion is 0.10 using a two-sided test
with α = 0.05. Give the test statistic, P-value, and your conclusion.
Question 4. (7pts)
Did Utah Jazz have a home field advantage? In the 2018-2019 NBA regular season, Utah Jazz
played 41 games at home and 41 games away. They won 29 of their home games and 21 of the
games played away. We can consider these games as samples from potentially large populations
of games played at home and away. Please round any numerical answers to 4 decimal places.
a) Find the proportion of wins for the home games. Do the same for the away games.
b) Find the standard error needed to compute a confidence interval for the difference in the
proportions.
c) Compute a 99% confidence interval for the difference between the probability that the Jazz win
at home and the probability that they win when on the road.
d) Most people think that it is easier to win at home than away. State null and alternative
hypotheses to test this supposition.
e) Combining all of the games played, what proportion did the Jazz win?
f) Find the standard error needed for testing that the probability of winning is the same at home
and away.
g) Compute the z statistic and its P-value. What conclusion do you draw?
Question 5. (9pts)
We want to study whether there is a tradeoff between the time spent sleeping per week and the
time spent in paid work. The data are in Sheet 3 of the excel file “Final”. We will estimate the
following model:
= 0 + 1 +
where sleep is minutes spent sleeping at night per week and totwrk is total minutes worked during
the week.
a) What is the response variable and what is the explanatory variable?
b) Draw a scatterplot with the explanatory variable on the horizontal axis and the response variable
on the vertical axis.
c) Describe the relationship between sleep and totwrk. Is a linear relationship reasonable? Explain
your answer.
d) Run the regression in Excel. Remember to check the “Residual Plots”.
e) Does the residual plot suggest that the data fit a linear model? Why?
f) Is the slope statistically different from 0 at the 5% level? Why?
g) Is the intercept meaningful in this case? Why?
h) The standard error for estimating the mean response when the explanatory variable x takes the
value x
*
is
̂ = √
1

+
(
∗ − ̅)
2
∑( − ̅)
2
Explain why the prediction is most accurate near the mean of the explanatory variable.
i) Find the mean of totwrk. What is the predicted mean sleep time per week when to