Solutions to ECN627 Assignment 2

Solutions to ECN627 Assignment 2 1. In a survey of 400 likely voters, 215 responded that they would vote for the incumbent, and 185 responded that they would vote for the challenger. Let 𝑝𝑝 denote the fraction of all likely voters who preferred the incumbent at the time of the survey, and let 𝑝𝑝̂ be the fraction of survey respondents who preferred the incumbent. a. Use the survey results to estimate 𝑝𝑝 . b. Use the estimator of the variance of 𝑝𝑝̂ , 𝑝𝑝̂ (1 − 𝑝𝑝̂ )/n, to calculate the standard error of your estimator. c. What is the p-value for the test 𝐻𝐻 0 : 𝑝𝑝 = 0.5 𝑣𝑣𝑣𝑣 . 𝐻𝐻 1 : 𝑝𝑝 ≠ 0.5 ? d. What is the p-value for the test 𝐻𝐻 0 : 𝑝𝑝 = 0.5 𝑣𝑣𝑣𝑣 . 𝐻𝐻 1 : 𝑝𝑝 > 0.5 ? e. Why do the results from (c) and (d) differ? f. Did the survey contain statistically significant evidence that the incumbent was ahead of the challenger at the time of survey? Solution: Denote each voter’s preference by Y , with Y = 1 if the voter prefers the incumbent and Y = 0 if the voter prefers the challenger. Y is a Bernoulli random variable with probability Pr ( 1) Y p = = and Pr ( 0) 1 . Y p = = − From the solution to Exercise 3.2, Y has mean p and variance (1 ). p p −

(a) 215 400 ˆ 0 5375. p = = . (b) The estimated variance of ˆ p is The standard error is SE 1 2 ˆ ˆ ( ) (var( )) 0 0249. p p = = . (c) The computed t -statistic is 0 ˆ 0 5375 0 5 1 506 ˆ SE( ) 0 0249 p act p t p µ , − . − . = = = . . . Because of the large sample size ( 400), n = we can use Equation (3.14) in the text to compute the p -value for the test 0 0 5 H p : = . vs. 1 0 5: H p : ≠ . -value 2 ( | |) 2 ( 1 506) 2 0 066 0 132 act p t = Φ − = Φ − . = × . = . (d) Using Equation (3.17) in the text, the p -value for the test 0 0 5 H p : = . vs. 1 0 5 H p : > . is -value 1 ( ) 1 (1 506) 1 0 934 0 066 act p t = − Φ = − Φ . = − . = (e) Part (c) is a two-sided test and the p -value is the area in the tails of the standard normal distribution outside ± (calculated t -statistic). Part (d) is a one-sided test and the p -value is the area under the standard normal distribution to the right of the calculated t -statistic. (f) For the test H 0 : p = 0.5 versus H 1 : p > 0.5, we cannot reject the null hypothesis at the 5% significance level. The p -value 0.066 is larger than 0.05. Equivalently the calculated t - statistic 1.506 is less than the critical value 1.64 for a one-sided test with a 5% significance level. The test suggests that the survey did not contain statistically significant evidence that the incumbent was ahead of the challenger at the time of the survey.

3. To investigate possible gender discrimination in a firm, a sample of 100 men and 64 women with similar job descriptions are selected at random. A summary of the resulting monthly salaries follows: Average Salary ( 𝑌𝑌 � ) Standard Deviation ( 𝑣𝑣 𝑌𝑌 ) n Men $3100 $200 100 Women $2900 $320 64 What do these data suggest about wage differences in the firm? Do they represent statistically significant evidence that average wages of men and women are different? (To answer this question, first state the null and the alternative hypotheses; second, compute the relevant t-statistic; third, compute the 𝑝𝑝 -value associated with the t-statistic; and finally, use the 𝑝𝑝 -value to answer the question.) Solution: Sample size for men 1 100, n = sample average 1 3100, Y = sample standard deviation 1 200. s = Sample size for women 2 64, n = sample average 2 2900, Y = sample standard deviation 2 320. s = The standard error of 1 2 Y Y − is: The hypothesis test for the difference in mean monthly salaries is 0 1 2 1 1 2 0 vs 0 H H µ µ µ µ : − = . : − ≠ The t -statistic for testing the null hypothesis is 1 2 1 2 3100 2900 4 4722 SE( ) 44 721 act Y Y t Y Y − − = = = . . − . Use Equation (3.14) in the text to get the p -value: 6 6 -value 2 ( | |) 2 ( 4 4722) 2 (3 8744 10 ) 7 7488 10 act p t − − = Φ − = Φ − . = × . × = . × The extremely low level of p -value implies that the difference in the monthly salaries for men and women is statistically significant. We can reject the null hypothesis with a high degree of confidence.

4. Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 453 randomly selected students in Florida; in this sample, the mean is 1013, and the standard deviation (s) is 108. a. Construct a 95% confidence interval for the average test score for Florida students. b. Is there statistically significant evidence that Florida students perform differently than other students in the United States? c. Another 503 students are selected at random from Florida. They are given a 3-hour preparation course before the test is administered. Their average test score is 1019, with a standard deviation of 95. i. Construct a 95% confidence interval for the change in average test score associated with the prep course. ii. Is there statistically significant evidence that the prep course helped? d. The original 453 students are given the prep course and then are asked to take the test a second time. The average change in their test scores is 9 points, and standard deviation of the change is 60 points. i. Construct a 95% confidence interval for the change in average test scores. ii. Is there statistically significant evidence that students will perform better on their second attempt, after taking the prep course? iii. Students may have performed better in their second attempt because of the prep course or because they gained test-taking experience in their first attempt. Describe an experiment that would quantify these two effects. (a) The 95% confidence interval if 108 453 1.96 SE( ) or 1013 1.96 or 1013 9.95. Y Y ± ± × ± (b) The confidence interval in (a) does not include µ = 1000, so the null hypothesis that µ = 1000 (Florida students have the same average performance as students in the U.S.) can be rejected at the 5% level. (c) (i) The 95% confidence interval is 1.96 ( ) prep Non prep prep Non prep Y Y SE Y Y − − − ± − where 2 2 2 2 95 108 503 453 SE( ) 6.61; prep non prep prep non prep S S prep Non prep n n Y Y − − − − = + = + = the 95% confidence interval is (1019 1013) 12.96 6 12.96. or − ± ± (ii) No. The 95% confidence interval includes These data do not provide statistically significant evidence that the prep-course changes the average test score.