Question

Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012). Suppose the survey sampled 200 men and 200 women. If 104 of the men replied yes and 74 of the women replied yes, are the results statistically significant in that you can conclude a greater proportion of men are expecting to get a raise or a promotion this year?
a. State the hypothesis test in terms of the population proportion of men and the population proportion of women?


H0: p1 - p2- Select your answer -greater than 0greater than or equal to 0less than 0less than or equal to 0equal to 0not equal to 0Item 1
Ha: p1 - p2- Select your answer -greater than 0greater than or equal to 0less than 0less than or equal to 0equal to 0not equal to 0Item 2


b. What is the sample proportion for men? Round your answer to two decimal places.


What is the sample proportion for women? Round your answer to two decimal places.


c. Use a .01 level of significance. What is the p-value? Round your answer to four decimal places.

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $p_1-p_2$ $\leq$ 0  or  $p_1$ $\leq$ $p_2$  

    Alternate Hypothesis, $H_a$ : $p_1-p_2$ > 0  or  $p_1$ > $p_2$

(b)  $\hat p_1$ = sample proportion of men = $\frac{104}{200}$ = 0.52

     $\hat p_2$ = sample proportion of women = $\frac{74}{200}$ = 0.37

(c) P-value is 0.0011.

Step-by-step explanation:

We are given that Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012).

Suppose the survey sampled 200 men and 200 women. If 104 of the men replied yes and 74 of the women replied yes.

Let $p_1$ = population proportion of men who replied yes

     $p_2$ = population proportion of women who replied yes

(a) Null Hypothesis, $H_0$ : $p_1-p_2$ $\leq$ 0  or  $p_1$ $\leq$ $p_2$   {means that the population proportion of men is less than or equal to the population proportion of women}

Alternate Hypothesis, $H_a$ : $p_1-p_2$ > 0  or  $p_1$ > $p_2$   {means that the population proportion of men is more than the population proportion of women}

(b)  $\hat p_1$ = sample proportion of men = $\frac{104}{200}$ = 0.52

     $\hat p_2$ = sample proportion of women = $\frac{74}{200}$ = 0.37

The test statistics that will be used here is Two-sample z proportion statistics;

               T.S.  =  $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }$  ~ N(0,1)

where, $\hat p_1$ = sample proportion of men who replied yes = 0.52

            $\hat p_2$ = sample proportion of women who replied yes = 0.37

            $n_1$ = sample of men = 200

            $n_2$ = sample of women = 200

So, test statistics  =  $\frac{(0.52-0.37)-(0)}{\sqrt{\frac{0.52(1-0.52)}{200} +\frac{0.37(1-0.37)}{200} } }$

                               =  3.05

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the population proportion of men is greater than the population proportion of women.

P-value is given by the following formula;

     P(Z > 3.05) = 1 - P(Z $\leq$ 3.05)

                        = 1 - 0.99886 = 0.0011

Hence, the p-value is 0.0011.