Question 1

A researcher was interested in comparing the salaries of female and male employees of a particular company. Independent random samples of female employees (sample 1) and male employees (sample 2) were taken to calculate the mean salary, in dollars per week, for each group. A $95 \%$ confidence interval for the difference, $\mu _ { 1 } - \mu _ { 2 }$, between the mean weekly salary of all female employees and the mean weekly salary of all male employees was determined to be $( - \$ 180 , \$ 60 )$.

Accepted Answer

A)  We know that $95 \%$ of female employees at this company make between $\$ 180$ less and $\$ 60$ more than the male employees. 
B)  We know that $95 \%$ of all random samples done on the employees at this company will show that the average female salary is between $\$ 180$ less and $\$ 60$ more per week than the average male salary. 
C)  We are $95 \%$ confident that a randomly selected female employee at this company makes between $\$ 180$ less and $\$ 60$ more per week than a randomly selected male employee. 
D)  Based on these data, with $95 \%$ confidence, female employees at this company average between $\$ 180$ less and $\$ 60$ more per week than the male employees. 
E)  Based on these data, with $95 \%$ confidence, male employees at this company average between $\$ 180$ less and $\$ 60$ more per week than the female employees. D

Question 2

Data were collected on annual personal time (in hours) taken by a random sample of 16 women (group1) and 7 men (group 2) employed by a medium sized company. The women took an average of $24.75$ hours of personal time per year with a standard deviation of $2.84$ hours. The men took an average of $21.89$ hours of personal time per year with a standard deviation of $3.29$ hours. The Human Resources Department believes that women tend to take more personal time than men because they tend to be the primary child care givers in the family. The correct null and alternative hypotheses to test this belief are

Accepted Answer

A)  $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ and $H _ { A } : \mu _ { 1 } - \mu _ { 2 }  0$. 
C)  $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } = 0$ and $H _ { A } : \mu _ { 1 } - \mu _ { 2 } \neq 0$. 
D)  $H _ { 0 } : \mu _ { 1 } - \mu _ { 2 } > 0$ and $H _ { A } : \mu _ { 1 } - \mu _ { 2 } < 0$. 
E)  None of the above. B

Question 3

A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to construct a $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ where $\mu _ { 1 }$ and $\mu _ { 2 }$ represent the mean for the treatment group and the control group respectively.
Treatment Group: $n _ { 1 } = 85 , \bar { Y } _ { 1 } = 189.1 , s _ { 1 } = 38.7$
Control Group: $n _ { 2 } = 75 , \bar { Y } _ { 2 } = 203.7 , s _ { 2 } = 39.2$

Accepted Answer

A)  $( - 29.0 , - 0.2 )$ 
B)  $( - 1.5,30.7 )$ 
C)  $( - 30.7,1.5 )$ 
D)  $( - 26.8 , - 2.4 )$ 
E)  $( - 1.3,30.5 )$ C

Question 4

A two-sample $z$-test for two population proportions is to be performed using the $P$-value approach. The null hypothesis is $H _ { 0 } : p _ { 1 } = p _ { 2 }$ and the alternative is $H _ { A } : p _ { 1 } \neq p _ { 2 }$. Use the given sample data to find the $p$-value for the hypothesis test. Give an interpretation of the $p$-value.
$$n _ { 1 } = 200 , x _ { 1 } = 11 , n _ { 2 } = 100 \text {, and } x _ { 2 } = 8$$

Accepted Answer

A)  $P$-value $= 0.402$; If there is no difference in the proportions, there is a $40.2 \%$ chance of seeing the exact observed difference by natural sampling variation. 
B)  $P$-value $= 0.1011$; There is about a $10.11 \%$ chance that the two proportions are equal. 
C)  $P$-value $= 0.0201$; There is about a $2.01 \%$ chance that the two proportions are equal 
D)  $P$-value $= 0.402$; If there is no difference in the proportions, there is about a $40.2 \%$ chance of seeing the observed difference or larger by natural sampling variation. 
E)  $P$-value $= 0.0201$; If there is no difference in the proportions, there is about a $2.01 \%$ chance of seeing the observed difference or larger by natural sampling variation.

Question 5

Results of a small experiment show that people are likely to offer a different amount for used exercise equipment when bargaining with a friend than when bargaining with a stranger. The $p$-value from testing the difference in mean offers was equal to $0.00162$. At an $\alpha = 0.05$, the correct conclusion is to

Accepted Answer

A)  reject the null hypothesis. 
B)  fail to reject the null hypothesis. 
C)  conclude that there is no difference between bargaining with a stranger versus a friend. 
D)  Both A and C. 
E)  Both $\mathrm { B }$ and $\mathrm { C }$.

Question 6

A consumer advocate decided to investigate the average wait time for a table for one at two local restaurants. Eighteen customers were sent to each restaurant at the same randomly selected times and the time they waited for a table was recorded in minutes. The following sample data was obtained.
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline \text { A } & 10 & 15 & 7 & 9 & 20 & 5 & 15 & 5 & 7 & 0 & 10 & 12 & 19 & 6 & 0 & 5 & 22 & 18 \
\hline \text { B } & 12 & 17 & 5 & 12 & 8 & 2 & 10 & 25 & 6 & 35 & 10 & 14 & 9 & 22 & 20 & 18 & 5 & 13 \
\hline
\end{array}$

Find a $90 \%$ confidence interval for the difference, $\mu _ { A } - \mu _ { B }$, between the mean wait time for restaurant $\mathrm { A }$ and the mean wait time for restaurant $\mathrm { B }$.

Accepted Answer

A)  $( 10.28,13.5 )$ 
B)  $( - 7.45 , - 1 )$ 
C)  $( - 1,7.45 )$ 
D)  $( - 7.45,1 )$ 
E)  $( - 13.5 , - 10.28 )$

Question 7

Suppose the proportion of women who favor stricter gun control legislation is $p _ { w }$ and the proportion of men who favor stricter gun control legislation is $p _ { m }$. The survey found a $90 \%$ confidence interval for $p _ { w } - p _ { m }$ is $( - 0.08,0.24 )$. Give an interpretation of this confidence interval.

Accepted Answer

A)  We know that $90 \%$ of all random samples done on the population will show that the proportion of men who favor stricter gun control legislation is between $8 \%$ lower and $24 \%$ higher than the proportion of women who favor stricter gun control legislation. 
B)  We know that $90 \%$ of all random samples done on the population will show that the proportion of women who favor stricter gun control legislation is between $8 \%$ lower and $24 \%$ higher than the proportion of men who favor stricter gun control legislation. 
C)  We are $90 \%$ confident that the proportion of men who favor stricter gun control legislation is between $8 \%$ lower and $24 \%$ higher than the proportion of women who favor stricter gun control legislation. 
D)  We are $90 \%$ confident that women favor gun control legislation that is between $8 \%$ less strict and $24 \%$ more strict than legislation favored by men. 
E)  We are $90 \%$ confident that the proportion of women who favor stricter gun control legislation is between $8 \%$ lower and $24 \%$ higher than the proportion of men who favor stricter gun control legislation.

Question 8

Suppose the proportion of women who follow a regular exercise program is $p _ { w }$ and the proportion of men who follow a regular exercise program is $p _ { m }$. A study found a $98 \%$ confidence interval for $p _ { w } - p _ { m }$ is $( - 0.024,0.115 )$. Give an interpretation of this confidence interval.

Accepted Answer

A)  We are $98 \%$ confident that the proportion of women who follow a regular exercise program is between $2.4 \%$ less and $11.5 \%$ more than the proportion of men who follow a regular exercise program. 
B)  We are $98 \%$ confident that the proportion of men who follow a regular exercise program is between $2.4 \%$ less and $11.5 \%$ more than the proportion of women who follow a regular exercise program. 
C)  We know that $98 \%$ of men exercise between $2.4 \%$ less and $11.5 \%$ more than women. 
D)  We know that $98 \%$ of all random samples done on the population will show that the proportion of women who follow a regular exercise program is between $2.4 \%$ less and $11.5 \%$ more than the proportion of men who follow a regular exercise program. 
E)  We know that $98 \%$ of women exercise between $2.4 \%$ less and $11.5 \%$ more than men.

Question 9

Suppose the proportion of sophomores at a particular college who purchased used textbooks in the past year is $p _ { s }$ and the proportion of freshmen at the college who purchased used textbooks in the past year is $p _ { f }$. A study found a $90 \%$ confidence interval for $p _ { s } - p _ { f }$ is $( 0.234,0.421 )$. Give an interpretation of this confidence interval.

Accepted Answer

A)  We are $90 \%$ confident that at this college the proportion of sophomores who bought used textbooks is between $23.4 \%$ and $42.1 \%$ higher than the proportion of freshmen who bought used textbooks. 
B)  We are $90 \%$ confident that at this college the proportion of freshmen who bought used textbooks is between $23.4 \%$ and $42.1 \%$ higher than the proportion of sophomores who bought used textbooks. 
C)  We are $90 \%$ confident that sophomores at this college bought between $23.4 \%$ and $42.1 \%$ more used textbooks than freshmen. 
D)  We know that $90 \%$ of all random samples done on the population will show that the proportion of sophomores who bought used textbooks is between $23.4 \%$ and $42.1 \%$ higher than the proportion of freshmen who bought used textbooks. 
E)  We know that $90 \%$ of sophomores bought between $23.4 \%$ and $42.1 \%$ more used textbooks than freshmen at this college.

Question 10

A survey of randomly selected college students found that 45 of the 99 freshmen and 57 of the 100 sophomores surveyed had purchased used textbooks in the past year. Construct a $98 \%$ confidence interval for the difference in the proportions of college freshmen and sophomores who purchased used textbooks.

Accepted Answer

A)  $( 0.317,0.593 )$ 
B)  $( 0.291,0.593 )$ 
C)  $( - 0.279,0.049 )$ 
D)  $( 0.291,0.619 )$ 
E)  $( - 0.253,0.593 )$

Question 11

A researcher is interested in the academic performance differences between individuals using an optimistic versus a pessimistic approach to their studies.If the researcher claims a
Significant difference between groups, when in fact none exists:

Accepted Answer

A) the research hypothesis was correctly rejected. 
B) the research hypothesis was correctly accepted. 
C) a Type 1 error is made. 
D) the null hypothesis was correctly accepted.

Question 12

Two types of flares are tested for their burning times (in minutes) and sample results are given below.
Brand A: $n = 35 , \bar { Y } = 19.4 , s = 1.4$
Brand B: $n = 40 , \bar { Y } = 15.1 , s = 0.8$
Construct a $95 \%$ confidence interval for the difference $\mu _ { A } - \mu _ { B }$ based on the sample data.

Accepted Answer

A)  $( 3.8,4.8 )$ 
B)  $( 3.6,5.0 )$ 
C)  $( 3.2,5.4 )$ 
D)  $( 3.5,5.1 )$ 
E)  $( - 4.7 , - 3.9 )$

Question 13

A two-sample $z$-test for two population proportions is to be performed using the $P$-value approach. The null hypothesis is $H _ { 0 } : p _ { 1 } = p _ { 2 }$ and the alternative is $H _ { A } : p _ { 1 } \neq p _ { 2 }$. Use the given sample data to find the $p$-value for the hypothesis test. Give an interpretation of the $p$-value.
$$n _ { 1 } = 50 , x _ { 1 } = 20 , n _ { 2 } = 75 , \text { and } x _ { 2 } = 15$$

Accepted Answer

A)  $P$-value $= 0.0032$; If there is no difference in the proportions, there is about a $0.32 \%$ chance of seeing the observed difference or larger by natural sampling variation. 
B)  $P$-value $= 0.0147$; If there is no difference in the proportions, there is about a $1.47 \%$ chance of seeing the observed difference or larger by natural sampling variation. 
C)  $P$-value $= 0.0292$; If there is no difference in the proportions, there is a $2.92 \%$ chance of seeing the exact observed difference by natural sampling variation. 
D)  $P$-value $= 0.0032$; There is about a $0.32 \%$ chance that the two proportions are equal. 
E)  $P$-value $= 0.0147$; There is about a $1.47 \%$ chance that the two proportions are equal.

Question 14

A survey of randomly chosen adults found that 38 of the 61 women and 46 of the 83 men follow regular exercise programs.Construct a 95% confidence interval for the difference in
The proportions of women and men who have regular exercise programs.

Accepted Answer

A) (-0.124, 0.786) 
B) (-0.093, 0.231) 
C) (-0.124, 0.815) 
D) (0.460, 0.786) 
E) (0.430, 0.815)

Question 15

A study was conducted to determine if patients recovering from knee surgery should receive physical therapy two or three times per week. Suppose $p _ { 1 }$ represents the proportion of patients who showed improvement after one month of therapy three times a week and $p _ { 2 }$ represents the proportion of patients who showed improvement after one month of therapy twice a week. A $90 \%$ confidence interval for $p _ { 1 } - p _ { 2 }$ is $( 0.12,0.38 )$. Give an interpretation of this confidence interval.

Accepted Answer

A)  We are $90 \%$ confident, based on the data, that patients who receive therapy three times per week show between $12 \%$ and $38 \%$ more improvement after one month than patients who receive therapy twice a week. 
B)  We know that $90 \%$ of all knee-surgery patients will show between $12 \%$ and $38 \%$ more improvement with therapy three times per week instead of twice a week. 
C)  We are $90 \%$ confident, based on the data, that the proportion of patients who show improvement with therapy three times per week is between $12 \%$ and $38 \%$ higher than the proportion who show improvement with therapy twice a week. 
D)  We know that $90 \%$ of all knee-surgery patients will show between $12 \%$ and $38 \%$ more improvement with therapy twice a week instead of three times per week. 
E)  We are $90 \%$ confident, based on the data, that the proportion of patients who show improvement with therapy three times per week is between $12 \%$ and $38 \%$ lower than the proportion who show improvement with therapy twice a week.

Question 16

A grocery store is interested in determining whether or not a difference exists between the shelf life of Tasty Choice doughnuts and Sugar Twist doughnuts. A random sample of 100 boxes of each brand was selected and the mean shelf life in days was determined for each brand. A $90 \%$ confidence interval for the difference of the means, $\mu _ { T C } - \mu _ { S T }$, was determined to be $( 1.3,2.5 )$.

Accepted Answer

A)  We know that $90 \%$ of Tasty Choice doughnuts last between $1.3$ and $2.5$ days longer than Sugar Twist doughnuts. 
B)  Based on this sample, we are $90 \%$ confident that Sugar Twist doughnuts will last on average between $1.3$ and $2.5$ days longer than Tasty Choice doughnuts. 
C)  We are $90 \%$ confident that a randomly selected box of Tasty Choice doughnuts will have a shelf life that is between $1.3$ and $2.5$ days longer than a randomly selected box of Sugar Twist doughnuts. 
D)  We know that $90 \%$ of all random samples done on the population will show that the mean shelf life of Tasty Choice doughnuts is between $1.3$ and $2.5$ days longer than the mean shelf life of Sugar Twist doughnuts. 
E)  Based on this sample, we are $90 \%$ confident that Tasty Choice doughnuts will last on average between $1.3$ and $2.5$ days longer than Sugar Twist doughnuts.

Question 17

A researcher is interested in the academic performance differences between individuals using an optimistic versus a pessimistic approach to their studies.If the researcher fails to find a
Significant difference, when in fact one exists in the population:

Accepted Answer

A) the null hypothesis was correctly accepted. 
B) the research hypothesis was correctly accepted. 
C) the null hypothesis was correctly rejected. 
D) a Type 2 error has been made. 
E) a Type 1 error has been made.

Question 18

A researcher was interested in comparing the number of hours of television watched each day by two-year-olds and three-year-olds. A random sample of 18 two-year-olds and 18 three-year-olds yielded the follow data.
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 2 y & 0.5 & 1.5 & 1.5 & 2.0 & 1.5 & 0.0 & 1.0 & 1.0 & 1.0 & 0.0 & 2.0 & 1.5 & 2.5 & 2.0 & 0.5 & 0.0 & 1.5 & 2.5 \
\hline 3 y & 2.0 & 3.0 & 1.5 & 1.5 & 1.5 & 2.0 & 1.0 & 0.0 & 0.0 & 1.5 & 1.5 & 2.0 & 2.5 & 2.0 & 3.0 & 1.0 & 1.5 & 0.5 \
\hline
\end{array}$

Find a $98 \%$ confidence interval for the difference, $\mu _ { 2 } - \mu _ { 3 }$, between the mean number of hours for two-year-olds $( 2 y )$ and the mean number of hours for three-year-olds ( $3 y$ ).

Accepted Answer

A)  $( 1.25,1.56 )$ 
B)  $( - 1.56 , - 1.25 )$ 
C)  $( - 0.98 , - 0.37 )$ 
D)  $( - 0.37,0.98 )$ 
E)  $( - 0.98,0.37 )$

Exam 20: Comparing Groups

A researcher is interested in the academic performance differences between individuals using an optimistic versus a pessimistic approach to their studies.If the researcher claims a Significant difference between groups, when in fact none exists:

A survey of randomly chosen adults found that 38 of the 61 women and 46 of the 83 men follow regular exercise programs.Construct a 95% confidence interval for the difference in The proportions of women and men who have regular exercise programs.

A researcher is interested in the academic performance differences between individuals using an optimistic versus a pessimistic approach to their studies.If the researcher fails to find a Significant difference, when in fact one exists in the population: