The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women.The article claimed that women perceived the problem to be much more prevalent than did men.One question asked to both men and women was: "Do you think sexual harassment is a major problem in the American workplace?" Some 24% of the men compared to 62% of the women responded "Yes." Suppose that 150 women and 200 men were interviewed.What is the value of the test statistic?
A)7.173
B)7.106
C)6.635
D)2.33

Question

Quizplus · Accepted Answer

The test statistic for the difference in proportions is calculated using the formula for a two-sample z-test. Given the proportions of men and women who responded "Yes" (24% of men and 62% of women), and the sample sizes (150 women and 200 men), the test statistic is calculated as follows:$$ z = \frac{p_1 - p_2}{\sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2})}} $$Where:- $ p_1 = 0.62 $ (proportion of women who said "Yes")- $ p_2 = 0.24 $ (proportion of men who said "Yes")- $ n_1 = 150 $ (number of women)- $ n_2 = 200 $ (number of men)- $ p = \frac{x_1 + x_2}{n_1 + n_2} $ (combined proportion)$$ p = \frac{(0.62 \times 150) + (0.24 \times 200)}{150 + 200} $$$$ p = \frac{93 + 48}{350} = \frac{141}{350} = 0.402857 $$$$ z = \frac{0.62 - 0.24}{\sqrt{0.402857(1-0.402857)(\frac{1}{150} + \frac{1}{200})}} $$$$ z = \frac{0.38}{\sqrt{0.402857 \times 0.597143 \times (0.006667 + 0.005)}} $$$$ z = \frac{0.38}{\sqrt{0.402857 \times 0.597143 \times 0.011667}} $$$$ z = \frac{0.38}{\sqrt{0.002813}} $$$$ z = \frac{0.38}{0.053} \approx 7.173 $$Therefore, the value of the test statistic is approximately 7.173, which corresponds to choice A.

The Wall Street Journal Recently Ran an Article Indicating Differences