Answer to Question #209171 in Statistics and Probability for rex

Question

samples of 12 foremen in one division and 10 foremen in another division of a factory were selected at random and the following data were obtained.

Sample Size Average monthly salary (Rs.) S.D. of salary (Rs.)

Division I 12 6000 150

Division II 10 5500 165

Can you conclude that the foremen in division I get more salary than foremen in division II at 5% level ?

Answer 1

Assume that the data come from approximately normally distributed populations. Since variance are not relatively equal, unequal variances assumption is assumed. Therefore, two independent samples t-test with unequal variances assumed is appropriate for this test.

$x_1=6000$

$s_1=150$

$n_1=12$

$x_2=5500$

$s_2=165$

$n_2=10$

$H0:\mu_1=\mu_2$

$Ha:\mu_1>\mu_2$

$t = \frac{\overline{x}_{1} - \overline{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}$

$t = \frac{6000 - 5500}{\sqrt{\frac{150^{2}}{12} + \frac{165^{2}}{10}}}=7.37$

$df = \frac{ \left ( \frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}} \right ) ^{2} }{ \frac{1}{n_{1}-1} \left ( \frac{s_{1}^2}{n_{1}} \right ) ^{2} + \frac{1}{n_{2}-1} \left ( \frac{s_{2}^2}{n_{2}} \right ) ^{2}}$

$df = \frac{ \left ( \frac{150^2}{12} + \frac{165^2}{10} \right ) ^{2} }{ \frac{1}{11} \left ( \frac{150^2}{12} \right ) ^{2} + \frac{1}{9} \left ( \frac{165^2}{10} \right ) ^{2}}=18.49$ which is approximately 18

$cv=t_{18,0.05}=1.734$ (Right tailed)

Since the test statistic 7.37 is greater than the critical value 1.734, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that foremen in division I get more salary than foremen in division II.