Reference no: EM132329481
Assignment - Need these math problems done.
1. A test of the null hypothesis H0: μ = μ0 gives test statistic z = -1.83. (Round your answers to four decimal places.)
(a) What is the P-value if the alternative is Ha: μ > μ0?
(b) What is the P-value if the alternative is Ha: μ < μ0?
(c) What is the P-value if the alternative is Ha: μ ≠ μ0?
2. The P-value for a two-sided test of the null hypothesis H0: μ = 35 is 0.037
(a) Yes, 35 is in the 95% confidence interval, because P = 0.037 means we reject H0 at α = 0.05.
Yes, 35 is in the 95% confidence interval, because P = 0.037 means we fail to reject H0 at α = 0.01.
Yes, 35 is in the 95% confidence interval, because P = 0.037 means we fail to reject H0 at α = 0.05
No, 35 is not in the 95% confidence interval, because P = 0.037 means we reject H0 at α = 0.05.
No, 35 is not in the 95% confidence interval, because P = 0.037 means we fail to reject H0 at α = 0.05.
Does the 95% confidence interval include the value 35? Why?
(b) Yes, 35 is in the 99% confidence interval, because P = 0.037 means we reject H0 at α = 0.01.
Yes, 35 is in the 99% confidence interval, because P = 0.037 means we fail to reject H0 at α = 0.01.
No, 35 is not in the 99% confidence interval, because P = 0.037 means we reject H0 at α = 0.01.
No, 35 is not in the 99% confidence interval, because P = 0.037 means we reject H0 at α = 0.10.
No, 35 is not in the 99% confidence interval, because P = 0.037 means we fail to reject H0 at α = 0.01.
3. In each of the following situations explain what is wrong and then either explain why it is wrong or change the wording of the statement to make it true.
(a) A researcher wants to test H0: x1 = x2 versus the two-sided alternative Ha: x1 ≠ x2.
The null hypothesis (but not the alternative hypothesis) should involve μ1 and μ2 (population means) rather than x1 and x2 (sample means).
The alternative hypothesis Ha should indicate that x1 ≥ x2.
The null hypothesis H0 should indicate that the two means are not equal.
Hypotheses should involve μ1 and μ2 (population means) rather than x1 and x2 (sample means).
This alternative hypothesis indicates a one-sided hypothesis instead of a two-sided hypothesis.
(b) A study recorded the IQ scores of 100 college freshmen. The scores of the 56 males in the study were compared with the scores of all 100 freshmen using the two-sample methods of this section.
The samples are too small to be used for hypothesis testing.
The samples are too large to be used for hypothesis testing.
A two-sample method is not appropriate in this situation.
The samples are not independent; we would need to compare the 56 males to the 44 females.
The sample sizes are too different to be used for hypothesis testing; we would need to have more males in the sample.
(c) A two-sample t statistic gave a P-value of 0.94. From this we can reject the null hypothesis with 90% confidence.
We need the P-value to be negative to reject H0.
We need the P-value to be small to reject H0.
We can reject the null hypothesis, but with more than 90% confidence.
We can reject the null hypothesis, but with less than 90% confidence.
A P-value of this size is impossible.
(d) A researcher is interested in testing the one-sided alternative Ha: μ1 < μ2.
The significance test gave t = 2.15.
Because the P-value for the two-sided alternative is 0.036, he concluded that his P-value was 0.018.
The alternative hypothesis should state that Ha: μ1 ≤ μ2.
At statistic of this size should have a much larger P-value associated with it.
A one-sided alternative should never be used. The alternative hypothesis should state that Ha: μ1 ≠ μ2.
Assuming the researcher computed the t statistic using x1 - x2, a positive value of t does not support Ha.
4. A recent study of food portion sizes reported that over a 17-year 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.)
t and degrees of freedom could be used to find the confidence interval. In this case we could compute SED and use degrees of freedom to find t*.
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.
Standard deviations and degrees of freedom could be used to find the confidence interval. In this case we could now find the P-value, which could be used to find SED
Degrees of freedom and a more accurate P-value could be used to find the confidence interval. In this case we could determine t, then calculate SED and t*.
Sample sizes and a more accurate P-value could be used to find the confidence interval. In this case we could determine standard deviations and the confidence interval in the usual way.
Do you think that a confidence interval would provide useful additional information? Explain why or why not.
No, the confidence interval could give us no more useful information because the P-value 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 variability between sample participants 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.
5. 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 one-sided alternative.
(a) Explain what is wrong with your friend's procedure and why.
We should only choose a one-sided alternative if we have some reason to expect a specific directional outcome before looking at the sample results.
The first mean can never be larger than the second mean; this indicates a mistake was made during statistical analysis.
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 one-sided alternative.
(b) Suppose he reported t = 1.80 with a P-value of 0.05. What is the correct P-value that he should report?
6. A study of iron deficiency among infants compared samples of infants following different feeding regimens. One group contained breast-fed infants, while the infants 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
|
|
Breast-fed
|
22
|
13.3
|
1.6
|
|
Formula
|
19
|
12.4
|
1.7
|
(a) Is there significant evidence that the mean hemoglobin level is higher among breast-fed babies? State H0 and Ha.
H0: μbreast-fed < μformula; Ha: μbreast-fed = μformula
H0: μbreast-fed > μformula; Ha: μbreast-fed = μformula
H0: μbreast-fed ≠ μformula; Ha: μbreast-fed < μformula
H0: μbreast-fed = μformula; Ha: μbreast-fed > μformula
Carry out a t test. Give the P-value. (Use α = 0.01. Use μbreast-fed - μformula. Round your value for t to three decimal places, and round your P-value to four decimal places.)
t =
P-value =
What is your conclusion?
Reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed babies.
Fail to reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies.
Reject the null hypothesis. There is significant evidence that the mean hemoglobin level is higher among breast-fed babies.
Fail to reject the null hypothesis. There is not significant evidence that the mean hemoglobin level is higher among breast-fed 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 two dependent SRSs from normal populations.
We need sample sizes greater than 40.
7. 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
|
115
|
7.83
|
3.67
|
|
Did not evacuate any pets
|
129
|
6.26
|
3.55
|
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 P-value to four decimal places.)
t =
P-value =
State your conclusion.
Fail to 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.
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 not significant evidence of a higher mean score for people who evacuated all or some pets.
8. Do various occupational groups differ in their diets? A British study of this question compared 86 drivers and 60 conductors of London double-decker 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
|
2816 ± 45
|
2847 ± 47
|
|
Alcohol (grams)
|
0.28 ± 0.09
|
0.41 ± 0.05
|
What justifies the use of the pooled two-sample t test?
The similarity of the sample standard deviations suggests that the population standard deviations are likely to be different.
The similarity of the sample means suggests that the population standard deviations are likely to be different.
The similarity of the sample means suggests that the population standard deviations are likely to be similar.
The similarity of the sample standard deviations 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 two-sample t test to obtain the P-value. (Give answers to 3 decimal places.)
t =
df =
P-value =
9. Assume a sample size of n = 16. Draw a picture of the distribution of the t statistic under the null hypothesis. Use Table D 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 two-sided alternative.
What is/are the value(s) of the critical t in this case? (Enter your answer as a comma-separated list using three decimal places.)
10. The one-sample t statistic for testing
H0: μ = 10
Ha: μ > 10
from a sample of n = 16 observations has the value t = 1.79.
(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.
(c) Between what two values does the P-value of the test fall?
0.005 < P < 0.01
0.01 < P < 0.02
0.02 < P < 0.025
0.025 < P < 0.05
0.05 < P < 0.1
(d) Is the value t = 1.79 significant at the 5% level?
Yes
No
Is it significant at the 1% level?
Yes
No
(e) If you have software available, find the exact P-value. (Round your answer to four decimal places.)
11. The one-sample t statistic for testing
H0: μ = 40
Ha: μ ≠ 40
from a sample of n = 12 observations has the value t = 2.77.
(a) What are the degrees of freedom for t?
(b) Locate the two critical values t* from the Table D that bracket t.
(c) Between what two values does the P-value of the test fall?
0.005 < P < 0.01
0.01 < P < 0.02
0.02 < P < 0.04
0.04 < P < 0.05
0.05 < P < 0.1
(d) Is the value t = 2.77 statistically significant at the 5% level?
Yes
No
Is it significant at the 1% level?
Yes
No
(e) If you have software available, find the exact P-value. (Round your answer to four decimal places.)
12. Dual-energy X-ray 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
|
1.327
|
1.338
|
1.079
|
1.226
|
0.938
|
1.006
|
1.182
|
1.287
|
|
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 sample is too small to make judgments about skewness or symmetry.
The distribution is Normal.
The distribution is right skewed.
The distribution is left skewed.
The distribution is uniform.
(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 P-value. (Round your answer to four decimal places.)
Give your conclusion. (Use the significance level of 5%.)
We cannot reject H0 based on this sample.
We can reject H0 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 machine. 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 may be representative of future subjects, but the test results and confidence interval are suspect because this is not a random sample.
The subjects from this sample, test results, and confidence interval are representative of future subjects.
13. 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 five-point scale of 0 to 4 with 0 corresponding to "not at all" and 4 corresponding to "very much." The mean rating for 289 boys with the disorder was reported as 2.22 with a standard deviation of 1.06. (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 increases with confidence level.
We see that the width of the interval decreases with confidence level.
We see that the width of the interval does not change with confidence level.
14. Self-efficacy is a general concept that measures how well we think we can control different situations. A multimedia program designed to improve dietary behavior among low-income 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 five-point scale with 1 corresponding to "not sure at all" and 5 corresponding to "very sure." Here is a summary of the self-efficacy scores obtained about 2 months after the intervention:
|
Group
|
n
|
x
|
s
|
|
Intervention
|
167
|
4.12
|
1.19
|
|
Control
|
217
|
3.61
|
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 Normal because the standard deviation is smaller than the mean.
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.
(b) Is it appropriate to use the two-sample t procedures that we studied in this section to analyze these data? Give reasons for your answer.
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 not be appropriate because the sample sizes are not large enough.
The t procedures should be appropriate because we have two large samples with no outliers.
The t procedures should be appropriate because we have Normally distributed data.
(c) Describe appropriate null and alternative hypotheses.
H0: μIntervention = μControl; Ha: μIntervention < μControl (or μIntervention ≠ μControl)
H0: μIntervention ≠ μControl; Ha: μIntervention > μControl (or μIntervention = μControl)
H0: μIntervention = μControl; Ha: μIntervention > μControl (or μIntervention ≠ μControl)
H0: μIntervention ≠ μControl; Ha: μIntervention < μControl (or μIntervention = μControl)
H0: μIntervention = μControl; Ha: μIntervention > μControl (or μIntervention < μControl)
Some people would prefer a two-sided alternative in this situation while others would use a one-sided significance test. Give reasons for each point of view.
The one-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a positive effect.
The one-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a negative effect.
The one-sided alternative reflects the researchers' (presumed) belief that the intervention would increase scores on the test. The two-sided alternative allows for the possibility that the intervention might have had a negative effect.
The two-sided alternative reflects the researchers' (presumed) belief that the intervention would increase scores on the test. The one-sided alternative allows for the possibility that the intervention might have had a negative effect.
The two-sided alternative reflects the researchers' (presumed) belief that the intervention would decrease scores on the test. The one-sided alternative allows for the possibility that the intervention might have had a positive effect.
(d) Carry out the significance test using a one-sided alternative. Report the test statistic with the degrees of freedom and the P-value. (Use μIntervention - μControl. Round your test statistic to three decimal places, your degrees of freedom to the nearest whole number, and your P-value to four decimal places.)
t =
df =
P-value =
Write a short summary of your conclusion.
We reject H0 and conclude that the intervention increased test scores.
We do not reject H0 and conclude that the intervention had no significant effect on 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.
15. 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 juvenile and 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
|
62
|
48
|
|
Male
|
132
|
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.)
p^F =
SEF =
p^M =
SEM =
(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 p^F - p^M. Round your value for z to two decimal places and round your P-value to four decimal places.)
z =
P-value =
State your conclusion.
There is sufficient evidence to conclude that the two proportions are different.
There is not sufficient evidence to conclude that the two proportions are different.
16. Castaneda v. Partida is 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 z statistic and correspond to P-values 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 180,425 persons eligible for jury duty, of whom 143,300 were Mexican Americans. Of the 883 people selected for jury duty, 335 were Mexican Americans.
(a) What proportion of eligible voters were Mexican Americans? Let this value be po. (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 Ho: p = po. Find the value of p^ for this problem, compute the z-statistic, and find the P-value. 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 P-value to four decimal places.)
z =
P-value =
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 two-sample 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 p1 be the probability that a randomly selected juror is a Mexican American, and let p2 be the probability that a randomly selected nonjuror is a Mexican American. Find the z statistic and its P-value. (Use α = 0.01. Round your test statistic to two decimal places and your P-value to four decimal places.)
z
P-value
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 different
very similar
none of the above
17. 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 1250 for 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% said they felt burdened.
(b) "If you could begin again, taking into account your current experience, what would you borrow?" 54.2% 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." 34.6% disagreed.
(d) "Making loan payments is unpleasant but I know that the benefits of education loans are worth it." 59.9% agreed.
(e) "I am satisfied that the education I invested in with my student loan(s) was worth the investment for career opportunities." 58.4% agreed.
(f) "I am satisfied that the education I invested in with my student loan(s) was worth the investment for personal growth." 69.4% agreed.
Conclusion
While a minority feel that loans are a burden and wish they had borrowed more, a majority are satisfied with their education.
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.
18. A matched pairs experiment compares the taste of instant with fresh-brewed 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 80 subjects who participate in the study, 32 prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers fresh-brewed coffee to instant coffee. (In practical terms, p is the proportion of the population who prefer fresh-brewed coffee.)
(a) Test the claim that a majority of people prefer the taste of fresh-brewed coffee. Report the large-sample z statistic. (Round your answer to two decimal places.) Report its P-value. (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 P-value.
(c) Is your result significant at the 5% level?
Yes
No
What is your practical conclusion?
The result is significant at the 5% level, so we reject H0 and conclude that a majority of people prefer instant coffee.
The result is not significant at the 5% level, so we reject H0 and conclude that a majority of people prefer instant coffee.
The result is not significant at the 5% level, so we do not reject H0 and conclude that a majority of people prefer instant coffee.
The result is not significant at the 5% level, so we do not reject H0 and conclude that a majority of people prefer fresh-brewed coffee.
The result is significant at the 5% level, so we reject H0 and conclude that a majority of people prefer fresh-brewed coffee.
19. In a study of the relationship between pet ownership and physical activity in older adults, 602 subjects reported that they owned a pet, while 1943 reported that they did not. Give a 95%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
20. 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 95% margin of error to be 0.1 or less? (Round your answer up to the next whole number.)
21. 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, it believes that this proportion will be about 0.29. Find the sample size needed if the margin of error of the confidence interval is to be no more than 0.035. (Round your answer up to the next whole number.)
22. In the Health ABC Study, 543 subjects owned a pet and 1980 subjects did not. Among the pet owners, there were 296 women; 976 of the non-pet owners were women. Find the proportion of pet owners who were women. Do the same for the non-pet 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.)
p^1 =
p^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.)
23. A survey of Internet users reported that 18% downloaded music onto their computers. The A survey of Ifiling of lawsuits by the recording industry may be a reason why this percent has decreased from the estimate of 26% from a survey taken two years before. Assume that the sample sizes are both 1451. 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 P-value to four decimal places.)
z =
P-value =
Summarize your conclusion.
We conclude that the proportions are not different.
We conclude that the means are different.
We conclude that the proportions are different.
We cannot draw any conclusions using a significance test for this data.
We conclude that the means are not different.
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 significance test does not indicate the direction of change, but the interval shows that the music downloads decreased.
The interval does not provide any more information than the significance test would tell us.
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.
24. A survey of Internet users reported that 17% 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 17% 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 (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.
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.
25. 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, 374 Cubs 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 "die-hard fans" or "less loyal fans." Of the 134 die-hard fans, 91.0% reported that they had watched or listened to Cubs games when they were children. Among the 240 less loyal fans, 66.3% said that they watched or listened as children. (Let D = pdie-hard - pless loyal.)
(a) Find the numbers of die-hard 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.)
die-hard fans
less loyal fans
(b) Use a one sided significance test to compare the die-hard 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 z-value to two decimal places and your P-value to four decimal places.)
z =
P-value =
Conclusion
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
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.
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.
(c) Express the results with a 95% confidence interval for the difference in proportions. (Round your answers to three decimal places.)
26. 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, 393 Cubs 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 "die-hard fans" or "less loyal fans." The study found that 93 of the 132 die-hard fans attended Cubs games at least once a month, but only 43 of the 261 less loyal fans attended this often. Analyze these data using a significance test for the difference in proportions. (Let D = pdie-hard - pless loyal. Use α = 0.05. Round your value for z to two decimal places. Round your P-value to four decimal places.)
z =
P-value =
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.
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.
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.
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.
Attachment:- Assignment File.rar