Answer to Question #349354 in Statistics and Probability for Cecel

Question

A company which produces batteries claims that the life expectancy of their batteries is 90 hours in order to test the claim a consumer interest group tested a random sample of 40 batteries. The test resulted to a mean life expectancy of 87 hours. Using a 0.05 level of significance, can it be concluded that the life expectancy of their batteries is less than 90 hours? Assume that the population standard deviation is known to be 10 hours.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge90$

$H_1:\mu<90$

This corresponds to a left-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 left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{87-90}{10/\sqrt{40}}=-1.8974$$

Since it is observed that $z=-1.8974<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z<-1.8974)= 0.028888,$ and since $p= 0.028888<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 90, at the $\alpha = 0.05$ significance level.