Answer to Question #273600 in Statistics and Probability for Serine

Question

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a mean of 800 hours and a standard deviation 40 hours. Test the null hypothesis that =800 hours against the alternative ≠800 hours if a random sample of 20 bulbs have an average of 700 hours. Use alpha = 0.05 level of significance.

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 $p=2P(Z<-1.643)=0.100383,$ and since $p=0.100383>0.05=\alpha,$ 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.