Final answer:
Since the p-value is less than 0.05 (α = 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that women are more satisfied with their jobs than men. Therefore, the correct option is a. P-value < 0.01.
Step-by-step explanation:
In this hypothesis test, we are comparing the proportions of job satisfaction between women (PW) and men (PM). The null hypothesis (H₀) states that the difference in proportions is zero, while the alternative hypothesis (Hₐ) states that the difference is greater than zero.
Using the formula for a z-test of two proportions, we calculate the test statistic z as:
z = (Pw - PM) / sqrt[(Pw(1 - Pw) / nw) + (Pm(1 - Pm) / nm)]
where Pw is the proportion of satisfied women, PM is the proportion of satisfied men, nw is the number of women in the sample, and nm is the number of men in the sample.
Using the given data, we find that Pw = 46/100 = 0.46, PM = 42/120 = 0.35, nw = 100, and nm = 120.
Plugging these values into the formula, we get z = (0.46 - 0.35) / √[(0.46(1 - 0.46) / 100) + (0.35(1 - 0.35) / 120)]
= 1.98.
Next, we find the p-value associated with this test statistic. Since the alternative hypothesis is one-tailed (PW-PM > 0), we look up the p-value in the z-table for a one-tailed test at 1.98, which is approximately 0.025.
Since the p-value is less than 0.05 (α = 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that women are more satisfied with their jobs than men. Therefore, the correct option is a. P-value < 0.01.