Answer to Question #157785 in Statistics and Probability for sfsd

Question

Mean life time of a sample of 400 of an product by a company is 1600 hours with a standard deviation of 150 hours. Test the hypothesis that the mean life of the product is more than 1570 hours at 1% level of significance

Answer 1

The provided sample mean is $\bar{x}=1600$ and the sample standard deviation is $s=150,$

and the sample size is $n=400.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq1570$

$H_1:\mu>1570$

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

Based on the information provided, the significance level is $\alpha=0.01,$ the number of degrees of freedom are $df=n-1=400-1=399,$ 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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1600-1570}{150/\sqrt{400}}=4$$

Since it is observed that $t=4>2.33573=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 1570, at the 1% significance level.

Using the P-value approach:

The p-value for $t=4, df=399, \alpha=0.01, right-tailed$ is $p=0.000038,$ and since $p=0.00003<0.01=\alpha,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 1570, at the 1% significance level.