Answer to Question #347060 in Statistics and Probability for giftseee

Question

A manufacturer claims that the average lifetime of his lightbulb is

3 years or 36 months. The standard deviation is 8 months. Fifty bulbs are selected, and the average life expectancy is found to be 32 months. Should the manufacturer statement be rejected at level of significance 0.01?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=36$

$H_1:\mu\not=36$

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 $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z:|z|>2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{32-36}{8/\sqrt{50}}\approx-3.5355$$

Since it is observed that $|z|=3.5355>2.5758=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z<-3.5355)= 0.000407,$ and since $p= 0.000407<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 36, at the $\alpha = 0.01$ significance level.