Answer to Question #300429 in Statistics and Probability for stat

Question

 the mean working hours of 400 college teachers was found to be 15.7 hours with standard deviation of 1.5 hours. test the hypothesis that the mean working hours is 16 hours against the hypothesis that it is greater than 16hrs at 1% level of significance

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=16$

$H_1:\mu>16$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01, df=n-1=400-1=399$ degrees of freedom, and the critical value for a right-tailed test is $t_c =2.33573.$

The rejection region for this right-tailed test is $R = \{t: t > 2.33573\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{15.7-16}{1.5/\sqrt{400}}=-4$$

Since it is observed that $t = -4 \le 2.33573=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $t=-4, df=399$ degrees of freedom is $p=0.999962,$ and since $p=0.999962>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 16, at the $\alpha = 0.01$ significance level.