Answer to Question #302215 in Statistics and Probability for Wells

Question

A firm employs 300 women and 100 men. The mean number of days absent last year for the women was 5.3 with a s.d. of 2.2 and for the men the corresponding figures were 6.2 and 2.9. Is the difference between the means significant at 5% level of significance?

Answer 1

A F-test is used to test for the equality of variances. The following F-ratio is obtained: $F=\dfrac{s_1^2}{s_2^2}=\dfrac{2.2^2}{2.9^2}\approx0.5755$

The critical values for $df_1=300-1=299$ degrees of freedom, $df_2=100-1=99$ degrees of freedom, are $F_L = 0.7337$ and $F_U = 1.3995,$ and since $F = 0.5755,$ then the null hypothesis of equal variances is 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 $\alpha = 0.05,$ and the degrees of freedom are computed as follows, assuming that the population variances are unequal:

$df_{total}=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}$

$=\dfrac{(\dfrac{2.2^2}{300}+\dfrac{2.9^2}{100})^2}{\dfrac{(2.2^2/300)^2}{300-1}+\dfrac{(2.9^2/100)^2}{100-1}}$

$=138.9337$

Hence, it is found that the critical value for this two-tailed test is $t_c = 1.9772,$ for $\alpha = 0.05$  and $df=138.9337.$

The rejection region for this two-tailed test is $R = \{t: |t| > 1.9772\}.$

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

$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$

$=\dfrac{5.2-6.3}{\sqrt{\dfrac{2.2^2}{300}+\dfrac{2.9^2}{100}}}=-2.8427$

Since it is observed that $|t| = 2.8427 > t_c = 1.9772,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $df=138.9337, t=-2.8427,$ two-tailed is $p=0.005148,$ and since $p=0.005148<0.05=\alpha,$ 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 $\alpha = 0.05$ significance level.