A 2005 study found that 295 women held board seats out of a total of 1,109 seats

Question

A 2005 study found that 295 women held board seats out of a total of 1,109 seats in the Fortune 100 companies. A 2003 study found that 774 women held board seats out of a total of 5,772 seats in the Fortune 500 companies.

(a-2)

     Reject the null hypothesis if zcalc > .

(b)

(c-1)

Find the test statistic zcalc and p-value. (Do not round the intermediate calculations but round the proportions to 4 decimal places. Round your answers to 4 decimal places.)

(a-2)

     Reject the null hypothesis if zcalc > .

(b)

(c-1)

Find the test statistic zcalc and p-value. (Do not round the intermediate calculations but round the proportions to 4 decimal places. Round your answers to 4 decimal places.)

Explanation / Answer

Solution:-

Proportion of Women board members in Fortune 100 companies = 295/1109 = 0.266

Proportion of Women board members in Fortune 500 companies = 774/5772 = 0.1341

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.1554

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.01188

z = (p1 - p2) / SE

z = 11.10

zcritical = + 1.96

Reject the null hypothesis if zcalc > 1.96

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 11.10 or greater than 11.10.

Thus, the P-value = less than 0.00001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.