Solution to A firm employs 300 women and 100 men. The mean number of days absent last … - Sikademy
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Archangel Macsika

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?

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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:



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:




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:



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.

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