Question

A prospective MBA student was made to estimate the difference in the salaries of professors in private and state universities. An independent study of simple random samples of the most recent MBA graduates of both universities revealed the following statistics: Conduct a test using 0.01 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_0:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{2.4^2}{2.1^2}=1.306$$

Based on the information provided, the significance level is $\alpha = 0.01,$ and the degrees of freedom are $df_1=n_1-1=48$ degrees of freedom, $df_2=n_2-1=48$ degrees of freedom and the critical value for a two-tailed test are $F_L=0.4695$ and $F_U=2.13,$ and since $F=1.306,$ then the null hypothesis of equal variances is not rejected.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n_1+n_2-2=96$ degrees of freedom, and the critical value for a two-tailed test is $t_c =2.628016.$

The rejection region for this two-tailed test is $R = \{t:|t|> 2.628016\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-1}(1/n_1+2/n_2)}}$$ $$=\dfrac{52.285−50.188}{\sqrt{\dfrac{(49-1)2.4^2+(49-1)2.1^2}{49+49-1}(1/49+2/49)}}$$ $$=4.603$$

Since it is observed that $|t|= 4.603>2.628016=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed $df=96$ degrees of freedom, $t=4.603$ is $p=0.000013,$ and since $p=0.000013<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$is different than $\mu_2,$ at the $\alpha = 0.01$ significance level.