Question

1.   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

b.     Remember that the Null hypothesis will be rejected if . Give the value of

Answer 1

Male worker:

$\bar{X}_m=67952 \\ σ_m = 3202 \\ n_m = 24$

Female worker:

$\bar{X}_w=61295 \\ σ_w = 3758 \\ n_w = 27$

H₀: the mean salary of males is equal to that of females.

H₁: the mean salary of males is higher than that of females.

The test statistic:

$Z_c = \frac{\bar{X}_m -\bar{X}_w}{\sqrt{\frac{σ_m^2}{n_m}+\frac{σ_w^2}{n_w}}} \\ = \frac{67952 -61295}{\sqrt{\frac{3202^2}{24}+\frac{3758^2}{27}}} \\ = 6.4597$

Z_t = 1.645 for one tailed test

Z_c>Z_t

Reject the null hypothesis.