Question

A survey was conducted to estimate the difference in salaries of university professors in the private sector and government sector of Rawalpindi. A random survey of 100 professors in private universities showed an average 9 month salary of 26000$ with standard deviation of 1300$. A random sample of 200 professors in government universities showed an average salary of 26,900$ with a standard deviation of 1400$. Test the hypothesis that the average salary for professors teaching in government universities does not exceed the average salary for professors teaching in private universities by no more than 500$. Use a 0.05 level of significance

Answer 1

The following null and alternative hypotheses need to be tested:

$\mu_1-\mu_2\leq500$

$\mu_1-\mu_2>500$

This corresponds to a right-tailed test, for which a t-test will be used.

The degrees of freedom (DF) is

$$df=\dfrac{\dfrac{s_1^2}{n_1^2}+\dfrac{s_2^2}{n_2^2}}{\dfrac{\dfrac{s_1^2}{n_1^2}}{n_1-1}+\dfrac{\dfrac{s_2^2}{n_2^2}}{n_2-1}}$$

$$=\dfrac{\dfrac{1300^2}{100^2}+\dfrac{1400^2}{200^2}}{\dfrac{\dfrac{1300^2}{100^2}}{100-1}+\dfrac{\dfrac{1400^2}{200^2}}{200-1}}=112$$

It is found that the critical value for this right-tailed test is $t_c=1.658573,$ for $\alpha=0.05$ and $df=112.$

The rejection region for this right-tailed test is $R=\{t: t>1.658573\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2-(\mu_1-\mu_2)}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

$$=\dfrac{26900-26000-500}{\sqrt{\dfrac{1300^2}{100}+\dfrac{1400^2}{200}}}\approx2.447960$$

Since it is observed that $t=2.447960>1.658573=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $df=112, \alpha=0.05,$

$t=2.447960$ is $p=0.007959,$ and since $p=0.007959<0.05=\alpha,$ it is concluded that the null hypothes is rejected.

Therefore, there is enough evidence to claim that the difference between two means is greater than $\$500,$ at the $\alpha=0.05$ significance level.