Question

The PACE project at the University of Wisconsin in Madison deals with problems associated with high-risk drinking on college campuses. Based on random samples, the study states that the percentage of UW students who reported bingeing at least three times within the past two weeks was 42.2% in 1999 (n = 334) and 21.2% in 2009 (n = 843). Test that the proportion of students reporting bingeing in 1999 is different from the proportion of students reporting bingeing in 2009 at the 10% significance level.

(A.1) Construct a 90% confidence interval around the difference-in-proportions estimate. Enter the lower bound of the interval you calculated in the box below. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

(A.2) Construct a 90% confidence interval around the difference-in-proportions estimate. Enter the upper bound of the interval you calculated in the box below. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

(B.1) How would you interpret the confidence interval?

(B.2) What connections can you draw between the confidence interval and the hypothesis test?(Because zero falls outside/inside the confidence interval, we reject/fail to reject the null hypothesis.)

Answer 1

H₀: P1 = P2

H₁: P1 P2

The pooled sample proportion(P) = ( * n1 + * n2)/(n1 + n2)

                                                = (0.422 * 334 + 0.212 * 843)/(334 + 843)

                                                = 0.272

The standard error = sqrt(P(1 - P)(1/n1 + 1/n2))

                           = sqrt(0.272 * (1 - 0.272) * (1/334 + 1/843))

                           = 0.0288

The test statistic is

  

  

At = 0.10, the critical value is +/- z_0.05 = +/- 1.645

Since the test statistic value is greater than the positive critical value(7.29 > 1.645), so we should reject H₀.

At 90% confidence level, the critical value is z_0.05 = 1.645

The 90% confidence interval is

() +/- z_0.05 * SE

= (0.422 - 0.212) +/- 1.645 * 0.0288

= 0.21 +/- 0.047

= 0.163, 0.257

A.1) Lower bound = 0.163

A.2) Upper bound = 0.257

B.1) We are 90% confident that the true difference in proportions of students bingeing in 1999 and the students bingeing in 2009 lies between the confidence bounds 0.163 and 0.257.

B.2) Because zero falls outside the confidence interval, we reject the null hypothesis.