Answer to Question #228182 in Statistics and Probability for Karl

Question

an electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hours and the standard deviation of 40 hours. test the hypothesis at .05 level of significance that mean= 800 hours against the alternative mean=/ 800 hours if the random sample of 30 bulbs has an average life of 788 hours

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu=800$

$H_1:\mu\not=800$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{x-\mu}{\sigma/\sqrt{n}}=\dfrac{788-800}{40/\sqrt{30}}\approx-1.643$$

Since it is observed that$|z|=1.643<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=2P(Z<-1.643)=0.1003,$ and since $p=0.103>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 800, at the $\alpha=0.05$ significance level.