Question

.. A Cornell University study of average wage differentials between men and women reported that one of the reasons average wage for men is higher than wages for women is that men tend to have more years of work experience (exp.) than women. Assume the following random sample summaries were obtained for each group. Men Women Sample size 30 45 Sample Average 14.9 years of exp. 10.3 years of exp. 3.8 years of exp. Sample St. Deviation 5.2 years of exp. Formulate the null and alternative hypotheses. At a = 0.05, what are the test statistic and the p-value? Conduct the test and explain your statistical and plain English conclusions.

Answer 1

Since s₁/s₂ = 5.2 / 3.8 = 1.37 (it lies between 0.5 and 2) we used the pooled variance.

The degrees of freedom used is n1 + n2 - 2 = 30 + 45 -2 = 73 (since pooled variance is used)

$S_p^2 = \frac{(n1-1)*s1^2+(n2-1)*s2^2}{n1+n2-2} = \frac{29*5.2^2+44*3.8^2}{30+45-2} = 19.45$

The Hypothesis:

H0: $\mu_1$ = $\mu_2$

Ha: $\mu_1$ > $\mu_2$

This is a Right tailed test.

The Test Statistic:

$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{S_p^2}(\frac{1}{n1}+\frac{1}{n2})} = \frac{14.9-10.3}{\sqrt{19.45*(\frac{1}{30}+\frac{1}{45})}} = 4.16$

The p Value:    The p value (Right Tail) for t = 4.16,df = 73, is; p value = 0.0000

The Decision Rule: If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since P value (0.0000) is < $\alpha$ (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the average work experience for men is greater than the average work experience for women.