 Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset. In addition to the computer’s calculations of miles per gallon, the driver also recorded the miles per gallon by dividing the miles driven by the number of gallons at each fillup. The following data are the differences between the computer’s and the driver’s calculations for that random sample of 20 records. The driver wants to determine if these calculations are different. Assume that the standard deviation of a difference is
σ = 3.0.
6.0  5.5  −0.6  1.6  3.7  4.5  7.0  2.2  4.7  3.0 
4.4  0.1  3.0  1.1  1.1  5.0  2.1  3.4  −0.6  −4.2 
(a) State the appropriate H_{0}and H_{a}to test this suspicion.
H_{0}: μ > 0 mpg; H_{a}: μ < 0 mpg
H_{0}: μ < 0 mpg; H_{a}: μ > 0 mpg
H_{0}: μ = 0 mpg; H_{a}: μ ≠ 0 mpg
H_{0}: μ > 3 mpg; H_{a}: μ < 3 mpg
H_{0}: μ = 3 mpg; H_{a}: μ ≠ 3 mpg
(b) Carry out the test. Give the Pvalue. (Round your answer to four decimal places.)
Interpret the result in plain language.
We conclude that μ = 3 mpg; that is, we have strong evidence that the computer’s reported fuel efficiency does not differ from the driver’s computed values.We conclude that μ ≠ 0 mpg; that is, we have strong evidence that the computer’s reported fuel efficiency does not differ from the driver’s computed values. We conclude that μ = 0 mpg; that is, we have strong evidence that the computer’s reported fuel efficiency differs from the driver’s computed values.We conclude that μ ≠ 3 mpg; that is, we have strong evidence that the computer’s reported fuel efficiency differs from the driver’s computed values.We conclude that μ ≠ 0 mpg; that is, we have strong evidence that the computer’s reported fuel efficiency differs from the driver’s computed values.
 Patients with chronic kidney failure may be treated by dialysis, in which a machine removes toxic wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood).
5.4  5.1  4.5  4.7  5.7  6.4 
The measurements are separated in time and can be considered an SRS of the patient’s blood phosphorus level. Assume that this level varies Normally with
σ = 0.8 mg/dl.
(Round your answers to three decimal places.)
(a) Give a 95% confidence interval for the mean blood phosphorus level.
,
(b) The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that this patient has a mean phosphorus level that exceeds 4.8?
Yes, there is strong evidence of a mean phosphorus level that exceeds 4.8.Yes, there is not strong evidence of a mean phosphorus level that exceeds 4.8. No, there is not strong evidence of a mean phosphorus level that exceeds 4.8.No, there is strong evidence of a mean phosphorus level that exceeds 4.8.
 An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has standard deviation σ= 8milligrams per gram (mg/g). A sample of 16 cuttings has mean cellulose content x = 144 mg/g.
(a) Give a 90% confidence interval for the mean cellulose content in the population. (Round your answers to two decimal places.)
,
(b) A previous study claimed that the mean cellulose content was μ = 140 mg/g, but the agronomist believes that the mean is higher than that figure. State H_{0} and H_{a}.
H_{0}: μ > 140 mg/g; H_{a}: μ = 140 mg/g
H_{0}: μ < 140 mg/g; H_{a}: μ = 140 mg/g
H_{0}: μ = 140 mg/g; H_{a}: μ ≠ 140 mg/g
H_{0}: μ = 140 mg/g; H_{a}: μ > 140 mg/g
H_{0}: μ = 140 mg/g; H_{a}: μ < 140 mg/g
 Carry out a significance test to see if the new data support this belief. (Use α= 0.05. Round your value for zto two decimal places and round your Pvalue to four decimal places.)
z  =  
Pvalue  = 
Do the data support this belief? State your conclusion.
Reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.Reject the null hypothesis, there not is significant evidence of a mean cellulose content greater than 140 mg/g. Fail to reject the null hypothesis, there is significant evidence of a mean cellulose content greater than 140 mg/g.Fail to reject the null hypothesis, there not is significant evidence of a mean cellulose content greater than 140 mg/g.
(c) The statistical procedures used in (a) and (b) are valid when several assumptions are met. What are these assumptions? (Select all that apply.)
Because our sample is not too large, the standard deviation of the population and sample must be less than 10.Because our sample is not too large, the population should be normally distributed, or at least not extremely nonnormal.We must assume that the 16 cuttings in our sample are an SRS.We must assume that the sample has an underlying distribution that is uniform.
 Many food products contain small quantities of substances that would give an undesireable taste or smell if they are present in large amounts. An exaple is the “offodors” caused by sulfur compounds in wine. Oenologists (wine experts) have determined the odor threshold, the lowest concentration of a compound that the human nose can detect. For example, the odor threshold for dimethyl sulfide (DMS) is given in the oenology literature as 25 micrograms per liter of wine (µg/l). Untrained noses may be less sensitive, however. Here are the DMS odor thresholds for 10 beginning students of oenology.
31  24  41  25  33  30  36  21  35  32 
Assume (this is not realistic) that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 µg/l. A normal quantile plot confirms that there are no systematic departures from normality.
(a) Make a stemplot to verify that the distribution is roughly symmetric with no outliers. (A Normal quantile plot confirms that there are no systematic departures from Normality. Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)
2  
2  
3  
3  
4 
(b) Give a 95% confidence interval for the mean DMS odor threshold among all beginning oenology students. (Round your answers to three decimal places.)
,
(c) Are you convinced that the mean odor threshold for beginning students is higher than the published threshold, 25 µg/l? Carry out a significance test to justify your answer. (Use α = 0.05. Round your value for z to two decimal places and round your Pvalue to four decimal places.)
z  =  
Pvalue  = 
State your conclusion.
Reject the null hypothesis, there is significant evidence that the mean odor threshold for beginning students is higher than the published threshold.Reject the null hypothesis, there is not significant evidence that the mean odor threshold for beginning students is higher than the published threshold. Fail to reject the null hypothesis, there is significant evidence that the mean odor threshold for beginning students is higher than the published threshold.Fail to reject the null hypothesis, there is not significant evidence that the mean odor threshold for beginning students is higher than the published threshold.
 A study of the pay of corporate chief executive officers (CEOs) examined the increase in cash compensation of the CEOs of 96companies, adjusted for inflation, in a recent year. The mean increase in real compensation was x= 6.4%, and the standard deviation of the increases was s = 51%. Is this good evidence that the mean real compensation μ of all CEOs increased that year?
H_{o}:  μ = 0  (no increase) 
H_{a}:  μ > 0  (an increase) 
Because the sample size is large, the sample s is close to the population σ, so take σ = 51%.
(a) Sketch the normal curve for the sampling distribution of x when H_{o} is true. Shade the area that represents the Pvalue for the observed outcome x = 6.4%. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) Calculate the Pvalue. (Round your answer to four decimal places.)
(c) Is the result significant at the α = 0.05 level? Do you think the study gives strong evidence that the mean compensation of all CEOs went up?
Reject the null hypothesis, there is significant evidence that the mean compensation of all CEOs went up.Reject the null hypothesis, there is not significant evidence that the mean compensation of all CEOs went up. Fail to reject the null hypothesis, there is not significant evidence that the mean compensation of all CEOs went up.Fail to reject the null hypothesis, there is significant evidence that the mean compensation of all CEOs went up.
A study based on a sample of size 81 reported a mean of 93 with a margin of error of 11 for 95% confidence.
(a) Give the 95% confidence interval.
,
(b) If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 11? Explain your answer.
The confidence level should never be changed for the same study.The new margin of error would be equal to 11. The margin of error will not change when the confidence level is increased. The new margin of error would be greater than 11. A wider margin of error is needed to be more confident that the interval includes the true mean.The new margin of error would be less than 11. A smaller margin of error is needed to be more confident that the interval includes the true mean.It is impossible to determine if the margin of error will change using the information given.
 You want to rent an unfurnished onebedroom apartment in Dallas next year. The mean monthly rent for a random sample of 15apartments advertised in the local newspaper is $1050. Assume the monthly rents in Dallas follow a Normal distribution with a standard deviation of $280. Find a 95% confidence interval for the mean monthly rent for unfurnished onebedroom apartments available for rent in this community. (Round your answers to two decimal places.)
$ , $
 A test of the null hypothesis H_{0}: μ= μ_{0}gives test statistic z= 1.08. (Round your answers to four decimal places.)
(a) What is the Pvalue if the alternative is
H_{a}: μ > μ_{0}?
(b) What is the Pvalue if the alternative is
H_{a}: μ < μ_{0}?
(c) What is the Pvalue if the alternative is
H_{a}: μ ≠ μ_{0}?
 A test of the null hypothesis H_{0}: μ= μ_{0}gives test statistic z= −1.60.
(Round your answers to four decimal places.)
(a) What is the Pvalue if the alternative is
H_{a}: μ > μ_{0}?
(b) What is the Pvalue if the alternative is
H_{a}: μ < μ_{0}?
(c) What is the Pvalue if the alternative is
H_{a}: μ ≠ μ_{0}?
 The Pvalue for a twosided test of the null hypothesis
H_{0}: μ = 25 is 0.02.
(a) Does the 95% confidence interval include the value 25? Why?
Yes, 25 is in the 95% confidence interval, because P = 0.02 means we would reject
H_{0}
at α = 0.05.No, 25 is not in the 95% confidence interval, because P = 0.02 means we would reject
H_{0}
at α = 0.05. Yes, 25 is in the 95% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.05.Yes, 25 is in the 95% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.01.No, 25 is not in the 95% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.05.
(b) Does the 90% confidence interval include the value 25? Why?
Yes, 25 is in the 90% confidence interval, because P = 0.02 means we would reject
H_{0}
at α = 0.10.Yes, 25 is in the 90% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.01. No, 25 is not in the 90% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.10.Yes, 25 is in the 90% confidence interval, because P = 0.02 means we would not reject
H_{0}
at α = 0.10.No, 25 is not in the 90% confidence interval, because P = 0.02 means we would reject
H_{0}
at α = 0.10.
 In each of the following situations explain what is wrong and why.
(a) A researcher wants to test H_{0}: x_{1} = x_{2} versus the twosided
alternative H_{a}: x_{1} ≠ x_{2}.
The null hypothesis (but not the alternative hypothesis) should involve μ_{1} and μ_{2} (population means) rather than
x_{1} and x_{2}
(sample means).The alternative hypothesis H_{a} should indicate that
x_{1} ≥ x_{2}
. This alternative hypothesis indicates a onesided hypothesis instead of a twosided hypothesis.The null hypothesis H_{0} should indicate that the two means are not equal.Hypotheses should involve μ_{1} and μ_{2} (population means) rather than
x_{1} and x_{2}
(sample means).
(b) A study recorded the IQ scores of 50 college freshmen. The scores of the 24 males in the study were compared with the scores of all 50 freshmen using the twosample methods of this section.
The samples are too small to be used for hypothesis testing.The sample sizes are too different to be used for hypothesis testing; we would need to have more males in the sample. The samples are not independent; we would need to compare the 24 males to the 26 females.The samples are too large to be used for hypothesis testing.A twosample method is not appropriate in this situation.
(c) A twosample t statistic gave a Pvalue of 0.93. From this we can reject the null hypothesis with 90% confidence.
We need the Pvalue to be negative to reject H_{0}.A Pvalue of this size is impossible. We can reject the null hypothesis, but with more than 90% confidence.We need the Pvalue to be small to reject H_{0}.We can reject the null hypothesis, but with less than 90% confidence.
(d) A researcher is interested in testing the onesided alternative
H_{a}: μ_{1} < μ_{2}.
The significance test gave
t = 2.25.
Since the Pvalue for the twosided alternative is 0.04, he concluded that his Pvalue was 0.02.
A onesided alternative should never be used.The alternative hypothesis should state that
H_{a}: μ_{1} ≤ μ_{2}.
The alternative hypothesis should state that
H_{a}: μ_{1} ≠ μ_{2}.
Assuming the researcher computed the t statistic using
x_{1} − x_{2},
a positive value of t does not support H_{a}.A t statistic of this size should have a much larger Pvalue associated with it.
 A recent study of food portion sizes reported that over a 17year period, the average size of a soft drink consumed by Americans aged 2 years and older increased from 13.1 ounces (oz) to 19.9 oz. The authors state that the difference is statistically significant with P< 0.01.†
Explain what additional information you would need to compute a confidence interval for the increase, and outline the procedure that you would use for the computations. (Select all that apply.)
Standard deviations and degrees of freedom could be used to find the confidence interval. In this case we could now find the Pvalue, which could be used to find SE_{D}.Sample sizes and standard deviations could be used to find the confidence interval. In this case we could find the interval in the usual way.Sample sizes and a more accurate Pvalue could be used to find the confidence interval. In this case we could determine standard deviations and the confidence interval in the usual way.t and degrees of freedom could be used to find the confidence interval. In this case we could compute SE_{D} and use degrees of freedom to find t*.Degrees of freedom and a more accurate Pvalue could be used to find the confidence interval. In this case we could determine t, then calculate SE_{D} and t*.
Do you think that a confidence interval would provide useful additional information? Explain why or why not.
Yes, the confidence interval could give us useful information about the variability between sample participants in the study.No, the confidence interval could give us no more useful information because the Pvalue already tells us that the interval does not contain 0. No, the confidence interval could give us no more useful information because it cannot tell us the sample size in the study.Yes, the confidence interval could give us useful information about the average size of soft drinks.Yes, the confidence interval could give us useful information about the magnitude of the difference.
 A friend has performed a significance test of the null hypothesis that two means are equal. His report states that the null hypothesis is rejected in favor of the alternative that the first mean is larger than the second. In a presentation on his work, he notes that the first sample mean was larger than the second mean and this is why he chose this particular onesided alternative.
(a) Explain what is wrong with your friend’s procedure and why.
The null hypothesis in this case should have been that the two means were not equal.The null hypothesis in this case should have been that the first mean is larger than the second. We should never choose a onesided alternative.The first mean can never be larger than the second mean; this indicates a mistake was made during statistical analysis.We should only choose a onesided alternative if we have some reason to expect a specific directional outcome before looking at the sample results.
(b) Suppose he reported t = 1.90 with a Pvalue of 0.04. What is the correct Pvalue that he should report?
 A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breastfed infants, while the children in another group were fed a standard baby formula without any iron supplements. Here are summary results on blood hemoglobin levels at 12 months of age.
Group  n  x  s 
Breastfed  24  13.2  1.5 
Formula  19  12.6  1.6 
(a) Is there significant evidence that the mean hemoglobin level is higher among breastfed babies? State H_{0} and H_{a}.
H_{0}: μ_{breastfed} = μ_{formula}; H_{a}: μ_{breastfed} > μ_{formula}H_{0}: μ_{breastfed} < μ_{formula}; H_{a}: μ_{breastfed} = μ_{formula} H_{0}: μ_{breastfed} > μ_{formula}; H_{a}: μ_{breastfed} = μ_{formula}H_{0}: μ_{breastfed} ≠ μ_{formula}; H_{a}: μ_{breastfed} < μ_{formula}
Carry out a t test. Give the Pvalue. (Use α = 0.01. Use μ_{breastfed} − μ_{formula}. Round your value for t to three decimal places, and round your Pvalue to four decimal places.)
t  =  
Pvalue  = 
What is your conclusion?
Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breastfed babies.Reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breastfed babies. Fail to reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breastfed babies.Reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breastfed babies.
(b) Give a 95% confidence interval for the mean difference in hemoglobin level between the two populations of infants. (Round your answers to three decimal places.)
,
(c) State the assumptions that your procedures in (a) and (b) require in order to be valid.
We need the data to be from a skewed distribution.We need two independent SRSs from normal populations. We need sample sizes greater than 40.We need two dependent SRSs from normal populations.
 On the morning of March 5, 1996, a train with 14 tankers of propane derailed near the center of the small Wisconsin town of Weyauwega. Six of the tankers were ruptured and burning when the 1700 residents were ordered to evacuate the town. Researchers study disasters like this so that effective relief efforts can be designed for future disasters. About half of the households with pets did not evacuate all of their pets. A study conducted after the derailment focused on problems associated with retrieval of the pets after the evacuation and characteristics of the pet owners. One of the scales measured “commitment to adult animals,” and the people who evacuated all or some of their pets were compared with those who did not evacuate any of their pets. Higher scores indicate that the pet owner is more likely to take actions that benefit the pet. Here are the data summaries.
Group  n  x  s 
Evacuated all or some pets  118  7.86  3.65 
Did not evacuate any pets  128  6.23  3.52 
Analyze the data and prepare a short report describing the results. (Use α = 0.01. Round your value for t to three decimal places and your Pvalue to four decimal places.)
t  =  
Pvalue  = 
State your conclusion.
Reject the null hypothesis. There is not significant evidence of a higher mean score for people who evacuated all or some pets.Fail to reject the null hypothesis. There is significant evidence of a higher mean score for people who evacuated all or some pets. Reject the null hypothesis. There is significant evidence of a higher mean score for people who evacuated all or some pets.Fail to reject the null hypothesis. There is not significant evidence of a higher mean score for people who evacuated all or some pets.
 Do various occupational groups differ in their diets? A British study of this question compared 85drivers and 66conductors of London doubledecker buses. The conductors’ jobs require more physical activity. The article reporting the study gives the data as “Mean daily consumption ± (se).” Some of the study results appear below.
Drivers  Conductors  
Total calories  2825 ± 40  2843 ± 45 
Alcohol (grams)  0.3 ± 0.09  0.45 ± 0.07 
What justifies the use of the pooled twosample t test?
The similarity of the sample means suggests that the population standard deviations are likely to be different.The similarity of the sample standard deviations suggests that the population standard deviations are likely to be different. The similarity of the sample standard deviations suggests that the population standard deviations are likely to be similar.The similarity of the sample means suggests that the population standard deviations are likely to be similar.

Is there significant evidence at the 5% level that conductors consume more calories per day than do drivers? Use the pooled twosample ttest to obtain the Pvalue. (Give answers to 3 decimal places.)
t =  
df =  
Pvalue = 
 Assume a sample size of n= 20. Draw a picture of the distribution of the tstatistic under the null hypothesis. Use the t distribution critical values table and your picture to illustrate the values of the test statistic that would lead to rejection of the null hypothesis at the 5% level for a twosided alternative.
What is/are the value(s) of the critical t in this case? (Enter your answer as a commaseparated list using three decimal places.)
The onesample t statistic for testing
H_{0}: μ = 10
H_{a}: μ > 10
from a sample of n = 23 observations has the value t = 2.26.
(a) What are the degrees of freedom for this statistic?
(b) Give the two critical values t* from the t distribution critical values table that bracket t.
< t <
(c) Between what two values does the Pvalue of the test fall?
0.005 < P < 0.010.01 < P < 0.02 0.02 < P < 0.0250.025 < P < 0.050.05 < P < 0.1
(d) Is the value t = 2.26 significant at the 5% level?
YesNo
Is it significant at the 1% level?
YesNo
(e) If you have software available, find the exact Pvalue. (Round your answer to four decimal places.)
 The onesample tstatistic for testing
H_{0}: μ = 60
H_{a}: μ ≠ 60
from a sample of n = 24 observations has the value t = 2.14.
(a) What are the degrees of freedom for t?
(b) Locate the two critical values t* from the t distribution critical values table that bracket t.
< t <
(c) Between what two values does the Pvalue of the test fall?
0.005 < P < 0.010.01 < P < 0.02 0.02 < P < 0.040.04 < P < 0.050.05 < P < 0.1
(d) Is the value t = 2.14 statistically significant at the 5% level?
YesNo
Is it significant at the 1% level?
YesNo
(e) If you have software available, find the exact Pvalue. (Round your answer to four decimal places.)
 Dualenergy Xray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body bone mineral content (TBBMC). A highly skilled operator is required to take the measurements. Recently, a new DXA machine was purchased by a research lab and two operators were trained to take the measurements. TBBMC for eight subjects was measured by both operators. The units are grams (g). A comparison of the means for the two operators provides a check on the training they received and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Here are the data:
Subject  
Operator  1  2  3  4  5  6  7  8 
1  1.325  1.340  1.074  1.229  0.936  1.004  1.178  1.288 
2  1.323  1.322  1.073  1.233  0.934  1.019  1.184  1.304 
(a) Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. (Use Operator 1 minus Operator 2. Round your answers to four decimal places.)
x  = 
s  = 
Describe the distribution of these differences using words.
The distribution is left skewed.The distribution is right skewed. The distribution is uniform.The distribution is Normal.The sample is too small to make judgments about skewness or symmetry.
(b) Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)
t =
Give the degrees of freedom.
Give the Pvalue. (Round your answer to four decimal places.)
Give your conclusion.
We cannot reject H_{0} based on this sample.We can reject H_{0} based on this sample.
(c) The sample here is rather small, so we may not have much power to detect differences of interest. Use a 95% confidence interval to provide a range of differences that are compatible with these data. (Round your answers to four decimal places.)
,
(d) The eight subjects used for this comparison were not a random sample. In fact, they were friends of the researchers whose ages and weights were similar to the types of people who would be measured with this DXA. Comment on the appropriateness of this procedure for selecting a sample, and discuss any consequences regarding the interpretation of the significance testing and confidence interval results.
The subjects from this sample, test results, and confidence interval are representative of future subjects.The subjects from this sample may be representative of future subjects, but the test results and confidence interval are suspect because this is not a random sample.
 In a study of children with a particular disorder, parents were asked to rate their child on a variety of items related to how well their child performs different tasks. One item was “Has difficulty organizing work,” rated on a fivepoint scale of 0 to 4 with 0 corresponding to “not at all” and 4 corresponding to “very much.” The mean rating for 276boys with the disorder was reported as 2.39with a standard deviation of 1.17. (Round your answers to four decimal places.)
Compute the 90% confidence interval.
,
Compute the 95% confidence interval.
,
Compute the 99% confidence interval.
,
Explain the effect of the confidence level on the width of the interval.
We see that the width of the interval decreases with confidence level.We see that the width of the interval increases with confidence level. We see that the width of the interval does not change with confidence level.
 Selfefficacy is a general concept that measures how well we think we can control different situations. A multimedia program designed to improve dietary behavior among lowincome women was evaluated by comparing women who were randomly assigned to intervention and control groups. Participants were asked, “How sure are you that you can eat foods low in fat over the next month?” The response was measured on a fivepoint scale with 1 corresponding to “not sure at all” and 5 corresponding to “very sure.” Here is a summary of the selfefficacy scores obtained about 2 months after the intervention:
Group  n  x  s 
Intervention  163  4.11  1.19 
Control  213  3.65  1.12 
(a) Do you think that these data are Normally distributed? Explain why or why not.
The distribution is Normal because the sample sizes are large.The distribution is not Normal because all scores are integers. The distribution is not Normal because the sample included only women.The distribution is Normal because the sample was randomly assigned.The distribution is Normal because the standard deviation is smaller than the mean.
(b) Is it appropriate to use the twosample t procedures that we studied in this section to analyze these data? Give reasons for your answer.
The t procedures should be appropriate because we have two large samples with no outliers.The t procedures should not be appropriate because the sample sizes are not large enough. The t procedures should not be appropriate because we do not have Normally distributed data.The t procedures should not be appropriate because the two groups are different sizes.The t procedures should be appropriate because we have Normally distributed data.
(c) Describe appropriate null and alternative hypotheses.
H_{0}: μ_{Intervention} ≠ μ_{2}; H_{a}: μ_{Intervention} > μ_{Control} (or μ_{Intervention} = μ_{Control})
H_{0}: μ_{Intervention} ≠ μ_{2}; H_{a}: μ_{Intervention} < μ_{Control} (or μ_{Intervention} = μ_{Control})
H_{0}: μ_{Intervention} = μ_{2}; H_{a}: μ_{Intervention} > μ_{Control} (or μ_{Intervention} = μ_{Control})
H_{0}: μ_{Intervention} = μ_{2}; H_{a}: μ_{Intervention} > μ_{Control} (or μ_{Intervention} < μ_{Control})
H_{0}: μ_{Intervention} = μ_{2}; H_{a}: μ_{Intervention} < μ_{Control} (or μ_{Intervention} ≠ μ_{Control})
Some people would prefer a twosided alternative in this situation while others would use a onesided significance test. Give reasons for each point of view.
The onesided alternative reflects the researchers’ (presumed) belief that the intervention would increase scores on the test. The twosided alternative allows for the possibility that the intervention might have had a negative effect.The twosided alternative reflects the researchers’ (presumed) belief that the intervention would decrease scores on the test. The onesided alternative allows for the possibility that the intervention might have had a positive effect. The twosided alternative reflects the researchers’ (presumed) belief that the intervention would increase scores on the test. The onesided alternative allows for the possibility that the intervention might have had a negative effect.The onesided alternative reflects the researchers’ (presumed) belief that the intervention would decrease scores on the test. The twosided alternative allows for the possibility that the intervention might have had a negative effect.The onesided alternative reflects the researchers’ (presumed) belief that the intervention would decrease scores on the test. The twosided alternative allows for the possibility that the intervention might have had a positive effect.
(d) Carry out the significance test using a onesided alternative. Report the test statistic with the degrees of freedom and the Pvalue. (Round your test statistic to three decimal places, your degrees of freedom to the nearest whole number, and your Pvalue to four decimal places.)
t  = 
df  = 
Pvalue  = 
Write a short summary of your conclusion.
We do not reject H_{0} and conclude that the intervention had no significant effect on test scores.We reject H_{0} and conclude that the intervention increased test scores.
(e) Find a 95% confidence interval for the difference between the two means. Compare the information given by the interval with the information given by the significance test.
,
(f) The women in this study were all residents of Durham, North Carolina. To what extent do you think the results can be generalized to other populations?
The results for this sample may not generalize well to other areas of the country.The results for this sample will generalize well to all other areas of the country.
 To what extent do syntax textbooks, which analyze the structure of sentences, illustrate gender bias? A study of this question sampled sentences from 10 texts. One part of the study examined the use of the words “girl,” “boy,” “man,” and “woman.” We will call the first two words juvenileand the last two adult. Is the proportion of female references that are juvenile (girl) equal to the proportion of male references that are juvenile (boy)? Here are data from one of the texts:
Gender  n  X(juvenile) 
Female  59  49 
Male  131  53 
(a) Find the proportion of juvenile references for females and its standard error. Do the same for the males. (Round your answers to three decimal places.)
p̂_{F}  = 
SE_{F}  = 
p̂_{M}  = 
SE_{M}  = 
(b) Give a 90% confidence interval for the difference. (Do not use rounded values. Round your final answers to three decimal places.)
,
(c) Use a test of significance to examine whether the two proportions are equal. (Use p̂_{F} − p̂_{M}. Round your value for z to two decimal places and round your Pvalue to four decimal places.)
z  =  
Pvalue  = 
State your conclusion.
There is not sufficient evidence to conclude that the two proportions are different. There is sufficient evidence to conclude that the two proportions are different.
 Castaneda v. Partidais an important court case in which statistical methods were used as part of a legal argument. When reviewing this case, the Supreme Court used the phrase “two or three standard deviations” as a criterion for statistical significance. This Supreme Court review has served as the basis for many subsequent applications of statistical methods in legal settings. (The two or three standard deviations referred to by the Court are values of the zstatistic and correspond to Pvalues of approximately 0.05 and 0.0026.) In Castaneda the plaintiffs alleged that the method for selecting juries in a county in Texas was biased against Mexican Americans. For the period of time at issue, there were 181,675 persons eligible for jury duty, of whom 143,150 were Mexican Americans. Of the 862 people selected for jury duty, 341 were Mexican Americans.
(a) What proportion of eligible voters were Mexican Americans? Let this value be p_{o}. (Round your answer to four decimal places.)
(b) Let p be the probability that a randomly selected juror is a Mexican American. The null hypothesis to be tested is H_{o}: p = p_{o}. Find the value of p̂ for this problem, compute the z statistic, and find the Pvalue. What do you conclude? (A finding of statistical significance in this circumstance does not constitute a proof of discrimination. It can be used, however, to establish a prima facie case. The burden of proof then shifts to the defense.) (Use α = 0.01. Round your test statistic to two decimal places and your Pvalue to four decimal places.)
z  
Pvalue 
Conclusion
Reject the null hypothesis, there is significant evidence that Mexican Americans are underrepresented on juries.Reject the null hypothesis, there is not significant evidence that Mexican Americans are underrepresented on juries. Fail to reject the null hypothesis, there is not significant evidence that Mexican Americans are underrepresented on juries.Fail to reject the null hypothesis, there is significant evidence that Mexican Americans are underrepresented on juries.
(c) We can reformulate this exercise as a twosample problem. Here we wish to compare the proportion of Mexican Americans among those selected as jurors with the proportion of Mexican Americans among those not selected as jurors. Let p_{1} be the probability that a randomly selected juror is a Mexican American, and let p_{2} be the probability that a randomly selected nonjuror is a Mexican American. Find the z statistic and its Pvalue. (Use α = 0.01. Round your test statistic to two decimal places and your Pvalue to four decimal places.)
z  
Pvalue 
Conclusion
Reject the null hypothesis, there is significant evidence of a difference in proportions.Reject the null hypothesis, there is not significant evidence of a difference in proportions. Fail to reject the null hypothesis, there is not significant evidence of a difference in proportions.Fail to reject the null hypothesis, there is significant evidence of a difference in proportions.
How do your answers compare with your results in (b)?
very differentvery similar none of the above
 The National Student Loan Survey asked the student loan borrowers in their sample about attitudes toward debt. Below are some of the questions they asked, with the percent who responded in a particular way. Assume that the sample size is 1285for all these questions. Compute a 95% confidence interval for each of the questions, and write a short report about what student loan borrowers think about their debt. (Round your answers to three decimal places.)
(a) “To what extent do you feel burdened by your student loan payments?” 57.9% said they felt burdened.
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(b) “If you could begin again, taking into account your current experience, what would you borrow?” 54.4% said they would borrow less.
,
(c) “Since leaving school, my education loans have not caused me more financial hardship than I had anticipated at the time I took out the loans.” 34.7% disagreed.
,
(d) “Making loan payments is unpleasant but I know that the benefits of education loans are worth it.” 57.6% agreed.
,
(e) “I am satisfied that the education I invested in with my student loan(s) was worth the investment for career opportunities.” 59.8% agreed.
,
(f) “I am satisfied that the education I invested in with my student loan(s) was worth the investment for personal growth.” 71.5% agreed.
,
Conclusion
While many feel that loans are a burden and wish they had borrowed less, a majority are satisfied with their education.While many feel that loans are a burden and wish they had borrowed less, a minority are satisfied with their education. While a minority feel that loans are a burden and wish they had borrowed more, a minority are satisfied with their education.While a minority feel that loans are a burden and wish they had borrowed more, a majority are satisfied with their education.
 A matched pairs experiment compares the taste of instant versus freshbrewed coffee. Each subject tastes two unmarked cups of coffee, one of each type, in random order and states which he or she prefers. Of the 50subjects who participate in the study, 16prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers freshbrewed coffee to instant coffee. (In practical terms, p is the proportion of the population who prefer freshbrewed coffee.)
(a) Test the claim that a majority of people prefer the taste of freshbrewed coffee. Report the largesample z statistic. (Round your answer to two decimal places.)
Report its Pvalue. (Round your answer to four decimal places.)
(b) Draw a sketch of a standard Normal curve and mark the location of your z statistic. Shade the appropriate area that corresponds to the Pvalue.
(c) Is your result significant at the 5% level?
YesNo
What is your practical conclusion?
The result is significant at the 5% level, so we reject H_{0} and conclude that a majority of people prefer freshbrewed coffee.The result is significant at the 5% level, so we reject H_{0} and conclude that a majority of people prefer instant coffee. The result is not significant at the 5% level, so we do not reject H_{0} and conclude that a majority of people prefer freshbrewed coffee.The result is not significant at the 5% level, so we do not reject H_{0} and conclude that a majority of people prefer instant coffee.The result is not significant at the 5% level, so we reject H_{0} and conclude that a majority of people prefer instant coffee.
In a study of the relationship between pet ownership and physical activity in older adults, 608 subjects reported that they owned a pet, while 1905 reported that they did not. Give a 99% confidence interval for the proportion of older adults in this population who are pet owners. (Round your answers to three decimal places.)
Lower limit  
Upper limit 
 One of your employees has suggested that your company develop a new product. You decide to take a random sample of your customers and ask whether or not there is interest in the new product. The response is on a 1 to 5 scale with 1 indicating “definitely would not purchase”; 2, “probably would not purchase”; 3, “not sure”; 4, “probably would purchase”; and 5, “definitely would purchase.” For an initial analysis, you will record the responses 1, 2, and 3 as “No” and 4 and 5 as “Yes.” What sample size would you use if you wanted the 99%margin of error to be 0.1or less? (Round your answer up to the next whole number.)
participants
 An automobile manufacturer would like to know what proportion of its customers are dissatisfied with the service received from their local dealer. The customer relations department will survey a random sample of customers and compute a 95% confidence interval for the proportion that are dissatisfied. From past studies, they believe that this proportion will be about 0.27. Find the sample size needed if the margin of error of the confidence interval is to be no more than 0.02. (Round your answer up to the next whole number.)
customers
 In this exercise we examine the effect of the sample size on the significance test for comparing two proportions. In each case suppose that p̂_{1}= 0.5 and p̂_{2}= 0.4, and take n to be the common value of n_{1} and n_{2}. Use the z statistic to test H_{0}: p_{1} = p_{2} versus the alternative H_{a}: p_{1} ≠ p_{2}. Compute the statistic and the associated Pvalue for the following values of n: 50, 60, 90, 110, 410, 510, and 1010. Summarize the results in a table. (Test the difference p_{1} − p_{2}. Round your values for z to two decimal places and round your Pvalues to four decimal places.)
n  z  Pvalue 
50  
60  
90  
110  
410  
510  
1010  
Explain what you observe about the effect of the sample size on statistical significance when the sample proportions p̂_{1} and p̂_{2} are unchanged.
As sample size increases, the test becomes less significant.As sample size increases, the test becomes more significant. As sample size increases, there is no effect on significance.There is not enough information.
 We have studied the effect of the sample size on the margin of error of the confidence interval for a single proportion. In this exercise we perform some calculations to observe this effect for the twosample problem. Suppose that p̂_{1}= 0.6 and p̂_{2}= 0.5, and n represents the common value of n_{1} and n_{2}. Compute the 95% margins of error for the difference in the two proportions for n = 50, 60, 90, 110, 410, 510, and 1010. Present the results in a table. (Give the largesample margins of error. Round your answers to three decimal places.)
n  m  
50  
60  
90  
110  
410  
510  
1010  
Write a short summary of your findings.
As sample size increases, margin of error remains constant.As sample size increases, margin of error increases. As sample size increases, margin of error decreases.There is not enough information.
 In the Health ABC Study, 538subjects owned a pet and 1965subjects did not. Among the pet owners, there were 307 women; 951 of the nonpet owners were women. Find the proportion of pet owners who were women. Do the same for the nonpet owners. (Be sure to let Population 1 correspond to the group with the higher proportion so that the difference will be positive. Round your answers to three decimal places.)
p̂_{1}  = 
p̂_{2}  = 
Give a 95% confidence interval for the difference in the two proportions. (Do not use rounded values. Round your final answers to three decimal places.)
,
 A survey of Internet users reported that 18%downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 25%from a survey taken two years before. Assume that the sample sizes are both 1401. Using a significance test, evaluate whether or not there has been a change in the percent of Internet users who download music. Provide all details for the test. (Round your value for z to two decimal places. Round your Pvalue to four decimal places.)
z  =  
Pvalue  = 
Summarize your conclusion.
We conclude that the proportions are not different.We conclude that the means are not different. We conclude that the proportions are different.We conclude that the means are different.We cannot draw any conclusions using a significance test for this data.
Also report a 95% confidence interval for the difference in proportions. (Round your answers to four decimal places.)
,
Explain what information is provided in the interval that is not in the significance test results.
The interval shows no significant change in music downloads.The interval gives us an idea of how large the difference is between the first survey and the second survey. The interval tells us there was a significant change in music downloads, but the test statistic is inconclusive.The interval does not provide any more information than the significance test would tell us.The significance test does not indicate the direction of change, but the interval shows that the music downloads decreased.
 A survey of Internet users reported that 19%downloaded music onto their computers. The filing of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 29%from a survey taken two years before. Suppose we are not exactly sure about the sizes of the samples. Perform the calculations for the significance tests and 95% confidence intervals under each of the following assumptions. (Use previous − recent. Round your test statistics to two decimal places and your confidence intervals to four decimal places.)
(i) Both sample sizes are 1000.
z =  
95% C.I.

,

(ii) Both sample sizes are 1600.
z =  
95% C.I.

,

(iii) The sample size for the survey reporting 29% is 1000 and the sample size for the survey reporting 19% is 1600.
z =  
95% C.I.

,

Summarize the effects of the sample sizes on the results.
We see in (i) and (ii) that smaller samples result in larger z (weaker evidence) and smaller intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated. We see in (i) and (ii) that smaller samples result in smaller z (weaker evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.We see in (i) and (ii) that smaller samples result in larger z (stronger evidence) and narrower intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.We see in (i) and (ii) that smaller samples result in smaller z (stronger evidence) and wider intervals, while larger samples have the reverse effect. The results of (iii) show that the effect of varying unequal sample sizes is more complicated.
 According to literature on brand loyalty, consumers who are loyal to a brand are likely to consistently select the same product. This type of consistency could come from a positive childhood association. To examine brand loyalty among fans of the Chicago Cubs, 366Cubs fans among patrons of a restaurant located in Wrigleyville were surveyed prior to a game at Wrigley Field, the Cubs’ home field. The respondents were classified as “diehard fans” or “less loyal fans.” Of the 131diehard fans, 90.8% reported that they had watched or listened to Cubs games when they were children. Among the 235 less loyal fans, 68.1% said that they watched or listened as children. (Let D = p_{diehard} − p_{less loyal}.)
(a) Find the numbers of diehard Cubs fans who watched or listened to games when they were children. Do the same for the less loyal fans. (Round your answers to the nearest whole number.)
diehard fans 
less loyal fans 
(b) Use a one sided significance test to compare the diehard fans with the less loyal fans with respect to their childhood experiences relative to the team. (Use your rounded values from part (a). Use α = 0.01. Round your zvalue to two decimal places and your Pvalue to four decimal places.)
z  = 
Pvalue  = 
Conclusion
Reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children.Fail to reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children. Fail to reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children.Reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans watched or listened to Cubs games as children.
(c) Express the results with a 95% confidence interval for the difference in proportions. (Round your answers to three decimal places.)
,
 According to literature on brand loyalty, consumers who are loyal to a brand are likely to consistently select the same product. This type of consistency could come from a positive childhood association. To examine brand loyalty among fans of the Chicago Cubs, 394Cubs fans among patrons of a restaurant located in Wrigleyville were surveyed prior to a game at Wrigley Field, the Cubs’ home field. The respondents were classified as “diehard fans” or “less loyal fans.” The study found that 69.3%of the 137 diehard fans attended Cubs games at least once a month, but only 17.1% of the 257 less loyal fans attended this often. Analyze these data using a significance test for the difference in proportions. (Let D = p_{diehard} − p_{less loyal}. Use α = 0.05. Round your value for z to two decimal places. Round your Pvalue to four decimal places.)
z  =  
Pvalue  = 
Analyze these data using a 95% confidence interval for the difference in proportions. (Round your answers to three decimal places.)
,
Write a short summary of your findings.
Fail to reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.Reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month. Fail to reject the null hypothesis, there is not significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.Reject the null hypothesis, there is significant evidence that a higher proportion of die hard Cubs fans attend games at least once a month.