Question

In a wage discrimination case involving male and female employees, it is assumed that male employees have a mean salary equal to that of female employees. To justify this, independent random samples of male and female employees were taken and the following result obtained.

Male Employees Female Employees

n1 = 100 n2 = 100

= Birr 20,600 = Birr 19,700

S1 = 3,000 S2 = Birr 2,500

Test the hypothesis with ( = 0.025. Does wage discrimination appear to exist in this case?

Answer 1

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

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{3000^2}{2500^2}=1.44$$

The critical values for $df_1=100-1=99=df_2, \alpha=0.025$ are $F_L = 0.6353,$ $F_U=1.574,$ and since $F = 1.44,$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

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.

Based on the information provided, the significance level is $\alpha = 0.025,$ and the degrees of freedom are $df=n_1-1+n_2-1 = 100-1+100-1=198,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test for $\alpha=0.025,df=198$ degrees of freedom is $t_c=2.258576.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.258576\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{20600-19700}{\sqrt{\dfrac{(100-1)3000^2+(100-1)2500^2}{100+100-2}(\dfrac{1}{100}+\dfrac{1}{100})}}$$

$$=2.304664$$

Since it is observed that $|t| = 2.304664 > 2.258576=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $df=198$ degrees of freedom, $t=2.304664$ is $p = 0.022222,$ and since $p=0.022222<0.025=\alpha,$ it is concluded that the null hypothes is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.025$ significance level.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq\mu_2$

$H_1:\mu_1>\mu_2$

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

Based on the information provided, the significance level is $\alpha = 0.025,$ and the degrees of freedom are $df=n_1-1+n_2-1 = 100-1+100-1=198,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this right-tailed test for $\alpha=0.025,df=198$ degrees of freedom is $t_c=1.972017.$

The rejection region for this two-tailed test is $R = \{t: t > 1.972017\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{20600-19700}{\sqrt{\dfrac{(100-1)3000^2+(100-1)2500^2}{100+100-2}(\dfrac{1}{100}+\dfrac{1}{100})}}$$

$$=2.304664$$

Since it is observed that $t = 2.304664 >1.972017=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $df=198$ degrees of freedom, $t=2.304664$ is $p = 0.011111,$ and since $p=0.011111<0.025=\alpha,$ it is concluded that the null hypothes is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha = 0.025$ significance level.