Answer to Question #154456 in Statistics and Probability for Alexander

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. Find:

a. The test statistic 𝑍𝑐

b. Remember that the Null hypothesis will be rejected if 𝑍𝑐 > 𝑍𝑡 . Give the value of 𝑍�

Answer 1

$Z_C= \frac{x_1-x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}= \frac{67952-61295}{974.81}=6.829\\ Z_t =Z_{1-\alpha}= Z_{0.95}=1.6449\\ Z_C>Z_t$