Answer to Question #154724 in Statistics and Probability for nathan

Question

The mean annual wage of male workers in a particular company was Ksh. 67952 and that of female workers was Ksh. 61295 each year. The standard deviation for the two samples is Ksh. 3202 and Kshs.3758, respectively. Assuming the measurements are from random samples of 24 males and 27 we want to investigate whether the mean salary of males is higher than that of females at α=0.05.

Answer 1

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$Ho: \mu_1 = \mu_2 ​$

$Ha: \mu_1 > \mu_2$

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.

$df= \frac{ \left(\dfrac{s_{1}^{2}}{n_1}+\dfrac{s_{2}^{2}}{n_2}\right)^{\!2} }{ \dfrac{(s_{1}^{2}/n^{}_1)^2}{n_1-1}+ \dfrac{(s_{2}^{2}/n^{}_2)^2}{n_2-1} }$ , substituting the values we get df= 48.923

Hence, it is found that the critical value using t table for this right-tailed test is

 $\\ tc ​=1.677, for \\ α=0.05 ,df = 48.923$

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

$t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}}$

=$\frac{{67952 }-{61295}}{\sqrt{{\frac{3202^2}{24}+\frac{3758^2}{27}}}}$ =6.829

Since it is observed that $t=6.829>tc ​=1.677,$ it is then concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is greater than μ2​, at the 0.05 significance level.