Question

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. For a 0.01 level of significance, what is the critical value for the rejection region? a. 7.173 b. 2.33 c. 6.635 d. 7.106

Answer 1

Answer:

Critical value: b. 2.33

As the test statistic z=7.17 is greater than the critical value, it falls in the rejection region.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

The significance level is 0.01.

The sample 1 (women), of size n1=150 has a proportion of p1=0.62.

The sample 2 (men), of size n2=200 has a proportion of p2=0.24.

 

The difference between proportions is (p1-p2)=0.38.

$p_d=p_1-p_2=0.62-0.24=0.38$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{93+48}{150+200}=\dfrac{141}{350}=0.403$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.403*0.597}{150}+\dfrac{0.403*0.597}{200}}\\\\\\s_{p1-p2}=\sqrt{0.001604+0.001203}=\sqrt{0.002807}=0.053$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.38-0}{0.053}=\dfrac{0.38}{0.053}=7.17$

The critical value for a right-tailed test with a signficance level of 0.01 is zc=2.33 (see picture attached).

As the test statistic z=7.17 is greater than the critical value, it falls in the rejection region.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of women who think sexual harassment is a major problem in the American workplace is significantly higher than the proportion of men.