1. Here are several situations in which there is an incorrect application of the ideas in statistical inference. Write a sentence or two in each case explaining what is wrong and why it is wrong.

(a) A significance test rejected the null hypothesis that the sample mean is equal to 500.

(b) A test preparation company wants to test that the average score of their students on the ACT is better than the national average score of 21.2. They state their null hypothesis as H0: µ > 21.2.

(c) A study summary says that the results are statistically significant and the p-value = 0.98.

(d) The computed value for the test statistic is Z0 = 0.018. Because this is less than 0.05, the null hypothesis is rejected.

2. I read about the following experiment in a recent NYTimes article. Researchers were interested in the effect of cellphone use on an individual’s willingness to assist others. On a street in Manhattan a man (apparently) using crutches dropped his newspaper and had obvious difficulty picking it up. An observer recorded whether a person passing the man was (a) on a cellphone, and (b) whether or not the person retrieved the man’s newspaper for him. Over the course of a week, this scenario was repeated many times. Of the 48 people who were on a cellphone, 19 retrieved the newspaper. Of the 205 people who were not on a cellphone, 134 retrieved the newspaper.

(a) Write a null and a two-sided alternative hypothesis in this case. Define the two parameters of interest as best you can.

(b) What are the explanatory and the response variables in this case?

(c) Construct a contingency table for these data with the rows being the categories for the response variable.

(d) Compute the appropriate sample percentages and write a brief summary of your results. Does the difference in the sample percentages seem to be of practical importance? Explain.

(e) Compute the p-value in this case.

(f) Can you reject the null hypothesis at the 5% level of significance? Explain. Can you reject the null hypothesis at the 1% level of significance? Explain.

(g) Write a conclusion in the context of the data.

3. Experiments on learning in animals sometimes measure how long it takes a mouse to find its way through a maze. Long experience indicates that the mean time is 20 seconds for one particular maze. A researcher thinks that playing loud rap music will, on average, cause mice to take longer to complete the maze.

(a) State the null and the alternative hypothesis in this case. Define (in words) the parameter of interest in this case.

A total of 12 mice attempted the maze while loud rap music was being played. The mean and standard deviation of the times to complete the maze are given below.

= 22.40 seconds S = 3.58 seconds

(b) Compute the value for the test statistic (t0) in this case and use the t table to find bounds on the corresponding p-value.

(c) Can you reject the null hypothesis at the 5% level of significance? Explain.

(d) Write a conclusion in the context of the problem.

(e) Find a 90% confidence interval for the mean time for mice to complete the maze with the loud rap music.

4. An alternative to the experiment in Q.3 to investigate whether loud music tends to increase the time it takes mice to complete a maze would involve randomly assigning say, 50 mice to two groups. The 25 mice in Group A (control) will navigate the maze in silence; those in Group B (rap) will complete the maze with loud rap music.

(a) State the null and the alternative hypothesis in this case. Define (in words) the parameters of interest in this case.

The results of such an experiment are shown in the output below.

Two-sample T for Time

Group N Mean StDev SE Mean

control 25 20.45 2.39 0.48

rap 25 21.10 2.78 0.56

Difference = mu (control) – mu (rap)

Estimate for difference: -0.650

95% CI for difference: (-2.126, 0.826)

T-Test of difference = 0 (vs not =): T-Value = -0.89 P-Value = 0.380 DF = 46

(b) Use the sample summary statistics in the output to verify the value for t in this case.

(c) What is the p-value corresponding to your alternative hypothesis?

(d) State a conclusion in the context of the data.

5. Starting in 1998, a series of credit card usage studies have been performed by Sallie Mae, a major provider of educational loans and savings programs. In a recent pilot study, a random pool of 122 loan applicants attending four-year colleges had their credit card data pulled for analysis. The sample mean credit card balance was $3173. The standard deviation was $3216.

(a) Compute a 95% confidence interval for the mean credit card balance among all undergraduate loan applicants (studying at four-year colleges).

(b) Provide the ‘correct’ interpretation of your interval.

(c) Sally Mae would like to repeat this survey but with a sample size large enough to achieve a margin of error of $100. How many credit card records should they analyze in order to achieve this margin of error with 95% confidence? You should use the results of the pilot survey in part (a) in answering this question.

(d) Researchers at Sally Mae were interested not only in the actual balances in the pilot survey described above. They were also interested in using the 122 records to estimate the percentage of all credit card balances that exceeded $10,000. In fact, 23 of the 122 balances in the pilot survey exceeded $10,000. Use this result to find an 80% confidence interval for the percentage of all credit card balances that exceeded $10,000.

(e) To the extent possible, check the validity of the conditions associated with your confidence interval in part (d).

(f) Explain carefully how you would use your confidence interval in part (d) to perform a two-sided test of whether a quarter of all balances exceed $10,000. What decision is appropriate? What level of significance is implied in this test?

6. This question is a bonus. I will assign three points for the correct answer to each of the four parts and an additional three points if you get each of the four parts correct. Please circle the correct answer to the following four questions.

(a) When performing a test of significance for a null hypothesis, H0 against an alternative hypothesis HA, the p-value is:

(i) The probability that H0 is true given the sample data

(ii) The probability that HA is true given the sample data

(iii) The probability of observing a sample result at least as extreme as that observed if H0 is true

(iv) The probability of observing a sample result at least as extreme as that observed if HA is true

(b) A 95% confidence interval for the (unknown) population proportion is 0.62 to 0.68. This means that:

(i) Ninety-five percent of the population lie between 0.62 and 0.68.

(ii) If we repeat the study over and over again, 95% of the resulting confidence intervals will contain the population proportion.

(iii) If we repeat the study over and over again, 95% of the times the population proportion will lie between 0.62 and 0.68.

(iv) If we repeat the study over and over again, 95% of the resulting confidence intervals will contain the sample proportion.

(c) A researcher wishes to obtain 600 patients requiring an operation for acute appendicitis. Three hundred (300) of them are to receive the standard anesthetic (A) and the remainder, an experimental anesthetic (B), which, it is hoped, will reduce the recovery time. She knows that randomization can help in greatly reducing the effects of confounding variables in this kind of experiment.

Which of the following statement is the closest to the truth?

(i) Selecting a random sample of such patients will substantially equalize the distribution of confounding variables over the two treatment groups.

(ii) Matching the two groups on several of the likely confounding variable will substantially equalize the distribution of confounding variables over the two treatment groups.

(iii) Only randomly selecting the patients and then randomly assigning them to the two treatments will substantially equalize the distribution of confounding variables over the two treatment groups.

(iv) Randomly assigning the 600 patients to the two treatment groups will substantially equalize the distribution of confounding variables over the two treatment groups.

(d) A recent editorial in the New York Times reported on a clinical trial in which two different drugs for treating breast cancer in older women were compared. The editorial contained the phrase “The difference fell just shy of statistical significance, so it remains possible that it occurred by chance, …” Which of the following possible p-values is the most consistent with this phrase?

(i) p-value = 0.46

(ii) p-value = 0.046

(iii) p-value = 0.064

(iv) p-value = 0.64