use the paired tinterval procedure to obtain the required
use the paired tinterval procedure to obtain the required.
1. Use the paired t–interval procedure to obtain the required confidence interval for the mean difference. Assume that the conditions and assumptions for inference are satisfied.
Ten families are randomly selected and their daily water usage (in gallons) before and after viewing a conservation video. Construct a 90% confidence interval for the mean of the difference of the “before” minus the “after” times if d (afterbefore) = 4.8 and Sd 5.2451
Before 33 33 38 33 35 35 40 40 40 31
After 34 28 25 28 35 33 31 28 35 33
 (1.5,8.1)
 (2.5,7.1)
 (1.8,7.8)
 (3.8,5.8)
 (2.1,7.5)
2.From the sample statistics, find the value of the pooled estimate Pcap used.
n_{1 }= 36 n_{2} = 418
x_{1} = 7 x_{2} = 132
Pcap =
3. Provide an appropriate response.
Do motivation levels between midlevel and upperlevel managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper Level 
Midlevel 
Sample size 
73 
109 
Mean score 
77.4 
79.71 
Standard Deviation 
10.6 
6.43 
Assuming equal population standard deviations, calculate the test statistic for determining whether the mean scores differ for upperlevel and midlevel managers.
 1.89c.63.69 e. none of these
 0.29 d. 1.74
4. Use the paired t–interval procedure to obtain the required confidence interval for the mean difference. Assume that the conditions and assumptions for inference are satisfied.
A test for abstract reasoning is given to a random sample of students before and after they complete a formal course in logic. The results are given below. Construct a 95% confidence interval for the difference in scores if d = afterbefore, Xbar_{d} = 3.7 and s_{d} = 4.945.
After 74 83 75 88 84 63 93 84 91 77
Before 73 77 70 77 74 67 95 83 84 75
 (4.4, 11.8)
 (0.2, 7.2)
 (1.2,5.7)
 (1.0,6.4)
 (0.8,6.6)
5. Find the appropriate test statistic/p–value.
Do motivation levels between midlevel and upperlevel managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper Level 
Midlevel 
Sample size 
73 
109 
Mean score 
77.4 
79.71 
Standard Deviation 
10.6 
6.43 
Calculate the appropriate test statistic and give your conclusion for testing
Ho: Uj=Ua
Ha: Uj<Ua
using a significance level of α = 0.05. Assume df = 100.
 t= 1.89; there is insufficient evidence to conclude that the mean scores differ for midlevel and upperlevel managers.
 t=1.89; reject the H0 and conclude that the mean scores differ for midlevel and upperlevel managers.
 t=1.31; there is insufficient evidence to conclude that the mean scores differ for midlevel and upperlevel managers.
 t=1.74; reject H0 and conclude that the mean scores differ for midlevel and upperlevel managers.
 t=1.74; there is insufficient evidence to conclude that the mean scores differ for midlevel and upperlevel managers.
6. Select the most appropriate answer.
For 12 pairs of females, the reported means are 24.8 on the wellbeing measure for the children of alcoholics and 29.0 for the control group. A t test statistic of 2.67 for the test comparing the means was obtained. Assuming that this is the result of a dependentsamples analysis testing for a difference between the group means, report the Pvalue.
 0.01 < Pvalue < 0.02
 0.005 < Pvalue < 0.01
 0.0076
 0.0152
 0.02 < Pvalue <0.05
7. A test for abstract reasoning is given to a random sample of students before and after they complete a formal course in logic. Calculate the test statistic for testing that the course improves the test scores assuming that d=afterbefore, Xbar_{d} = 3.7 and s_{d} = 4.945, n = 10 and α = 0.05. State your conclusion in terms of the problem
 t= 0.75; fail to reject the null hypothesis and conclude that the average scores on the abstract reasoning test are the same before and after course in logic
 t= 2.37; reject the null hypothesis and conclude that the course does improve the average score on the abstract reasoning test.
 t= 2.37; fail to reject the null hypothesis and conclude that the average scores on the abstract reasoning test are the same before and after the course in logic.
 t= 0.75; fail to reject the null hypothesis. There is no evidence to conclude that the course improves the average on the abstract reasoning test.
 t=2.37; fail to reject the null hypothesis. There is no enough evidence to conclude that the course improves the average score on the abstract reasoning test.
8.Provide an appropriate response.
You are interested in determining whether there is a difference in the mean calorie content of a serving of fries versus a serving of onion rings at fast food restaurants. Based on a sample of seventeen french fry choices from fast food restaurants, the mean caloric content is 543.35 with a standard deviation of 112.18. A sample of eight onion ring choices from fast food restaurants has a mean caloric content of 526.25 with a standard deviation of 142.32. Assuming both populations are normal with equal standard deviations, what is the test statistic for testing whether the mean caloric content is the same for french fry orders as for onion rings at fast food restaurants?
 0.14
 3.46
 0.33
 0.30
 None of these
9.Provide an appropriate response.
Do motivation levels between midlevel and upperlevel managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper Level 
Midlevel 
Sample size 
73 
109 
Mean score 
77.4 
79.71 
Standard Deviation 
10.6 
6.43 
Assuming equal population standard deviations, find the Pvalue for testing that the mean scores differ for upperlevel and midlevel managers. Interpret using a 5% significance level
a. Pvalue = 0.03; since the Pvalue < 0.05, we reject the null hypothesis
b. Pvalue = 0.08; since the Pvalue > 0.05. we fail to reject the null hypothesis
c. Pvalue =0.04; since the Pvalue < 0.05. we reject the null hypothesis
d. PValue =0.06; since the Pvalue > 0.05, we fail to reject the null hypothesis.
10..Construct the indicated confidence interval for the difference between the two population means. Assume that the assumptions and conditions for inference have been met.
The table below contains information pertaining to the gasoline mileage for random samples of trucks of two different types. Find a 95% confidence interval for the difference in the means μ_{X} – μ_{Y}.

Brand X 
Brand Y 
Number of trucks 
50 
50 
Mean mileage 
20.1 
24.3 
Standard deviation 
2.3 
1.8 
 (5.02, 3.38)
 (20.1,24.3)
 (3.7, 4.7)
 (4.7, 3.7)
 (3.38, 5.02)
11.Provide an appropriate response.
The weights before and after 9 randomly selected participants followed a particular diet were recorded. The mean of the before weights was 170.4444, the mean of the weights following the diet was 160.5556 and the standard error of the differences was 3.1333. Calculate the appropriate test statistic for testing that the average weight was lower following the diet and state your conclusion using a significance level of 0.01.
 t=3.16; reject the null hypothesis and conclude that the diet is effective for weight loss
 t= 9.47; fail to reject the null hypothesis; there is not enough information to conclude that the diet is effective for weight loss
 t=3.16; accept the null hypothesis and conclude that there is no difference in the average weight before and after the diet
 t=3.16; fail to reject the null hypothesis; there is not enough information to conclude that the diet is effective for weight loss.
 t= 9.47; reject the null hypothesis and conclude that the diet is effective fir weight loss.
12.Interpret the given confidence interval.
A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Subjects were randomly assigned to either a treatment group or a control group. The mean blood pressure was determined for each group, and a 95% confidence interval for the difference in the means for the treatment group versus the control group, UtUc was found to be (21, 6). ( t –stands for treatment group and C stands for control group)
 We are 95% confident that the average blood pressure of those who follow the diet is between 6 and 21 points higher than the average for those who do not follow the diet.
 The probability that the mean blood pressure for those on the diet is lower than for those not on the diet is 0.95.
 Since all of the values in the confidence interval are less than 0, we are unable to conclude that there is difference in blood pressure for those who follow the diet and those who do not.
 We are 95% confident that the average blood pressure of those who follow the diet is between 6 and 21 points lower than the average for those who do not follow the diet.
 The probability that the mean blood pressure for those on the diet is higher than for those not on the diet is 0.95.
13.Interpret the given confidence interval.
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 90% confidence interval for the difference, U1U2 between the mean weekly salary of all female employees and the mean weekly salary of all male employees was determined to be ($110,$10)
 90% of the time females at this company make less than males
 Since 0 is contained in the interval, the probability that the male employees at this company earn the same as females at this company is 0.9
 Based on these data, we are 90% confident that the male employees at this company average between $110 less and $10 more per week than the female employees.
 The probability that a randomly selected female employee at this company makes between $110 less and $10 more per week than a randomly selected male employee is 0.9.
 Based on these data, we are 90% confident that the female employees at this company average between $ 110 less and $10 more per week than the male employees.
14. In a positive association between two variables
 A decrease in the value of one variable is associated with a decrease in the value of a second variable.
 An increase in the value of one variable is associated with a decrease in the value of a second variable.
 An increase in the value of one variable is associated with an increase in the value of a second variable.
 Both A and C
15.Comparison of means can be used when
 A researcher wants to compare responses to an ordinal variable by the categories of a nominal variable.
 A researcher wants to compare responses to a numerical variable by the categories of a nominal or ordinal variable
 A researcher wants to compare responses to a nominal variable by the categories of an ordinal variable
 A researcher wants to compare responses to a numerical variable by categories of an ordinal variable
16. The independent samples t test assesses differences in means between
 An independent nominal variable and dependent ordinal variable
 A dependent numerical variable and an independent categorical variable
 A dependent nominal variable and an independent categorical variable
 A dependent categorical variable and an independent numerical variable
17. In a negative association between two variables,
 A decrease in the value of one variable is associated with a decrease in the value of a second variable
 An increase in the value of one variable is associated with a decrease in the value of a second variable
 A decrease in the value of one variable is associated with an increase in the value of a second variable
 Both B and C
18.For the independent samples t test, if equality of variances cannot be assumed then you
 Have to use a formula for the t statistic that is different from the one you would use if equality of variances can be assumed
 Have to use a formula for degrees of freedom that is different from the one you would use if equality of variances can be assumed
 Cannot reject the null hypothesis , no matter the value of t
 Both A and B
19. Provide an appropriate response.
A 95% confidence interval for the difference in means for a collection of paired sample data is (0, 3.4) Based on the same sample, a traditional significance test fails to support the claim of μd > 0. What can you conclude about the significance level α (α = 1 – .95) of the hypothesis test?
α > 0.05
α < 0.05
α = 0.01
α = 0.05
α = 0.95
20.From the sample statistics, find the value of P1capP2cap, the point estimate of the difference of proportions. Unless otherwise indicated, round to the nearest thousandth when necessary.
n1 = 100 n2 = 100
x1 = 34 x2 = 30
0.04
none of these
0.02
0.02
0.04
20. The statistic that answers the question, how likely is it that the difference between the means for two categories of a variable that we observe in a sample is merely a chance occurrence, is the
independent samples ttest
t statistic for Pearson’s r
onesample t test
oneway analysis of variance
21.Interpret the given confidence interval.
A high school coach uses a new technique in training middle distance runners. He records the times for 4 different athletes to run 800 meters before and after this training. A 90% confidence interval for the difference of the means before and after the training, μB – μA, was determined to be(2.7, 4.2)
The probability that the average time for the 800meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds shorter after the training is 0.9.
We are 90% confident that a randomly selected middle distance runner at this high school will have a time for the 800meter run that is between 2.7 and 4.2 seconds shorter after the training than before the training.
Based on this sample, we are 90% confident that the average time for the 800meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds longer after the new training.
Based on this sample, we are 90% confident that the average time for the 800meter run for middle distance runners at this high school is between 2.7 and 4.2 seconds shorter after the new training.
We know that 90% of the middle distance runners shortened their times between 2.7 and 4.2 seconds after the training.
22. Select the most appropriate answer.
The central limit theorem predicts that the sampling distribution of X1bar – X2bar is approximately normal
when the total number sampled is greater than or equal to 30.
when both of the sample sizes are greater than or equal to 30.
when either one of the sample sizes is greater than or equal to 30.
when at least one of the sample sizes is greater than or equal to 30.
regardless of both of the sample sizes.
From the sample statistics, find the value of P1capP2cap , the point estimate of the difference of proportions. Unless otherwise indicated, round to the nearest thousandth when necessary.
A survey asked respondents whether marijuana should be made legal. A 95% confidence interval for PAPB, is given by (0.08,0.14) where PA is the proportion of respondents who answered “legal” in state A and PB is the proportion of respondents who responded “legal” in state B. Based on the 95% confidence interval, what can we conclude about the percentage of respondents who favor legalization in state B versus state A?
Since all of the values in the confidence interval are less than 1, we can conclude that there is a significant difference between the percentage in favor of legalization in state B and the percentage in favor of legalization in state A.
Since all of the values in the confidence interval are less than 1, we are unable to conclude that there is a significant difference between the percentage in favor of legalization in state B and the percentage in favor of legalization in state A.
Since all of the values in the confidence interval are greater than 0, we can conclude that the percentage in favor of legalization was greater in state A than it was in state B.
Since all of the values in the confidence interval are greater than 0, we can conclude that the percentage in favor of legalization was greater in state B than it was in state A.