Answer to Question #339673 in Statistics and Probability for pgi akcadaaff

Question

A company that produces batteries claims that the life expectancy of their batteries is 90 hours with a standard deviation of 10 hours. A consumer interest group believes that the actual battery life expectancy of these batteries is shorter than the claimed battery life. In order to prove this, they test a random sample of 20 batteries which resulted a mean of 87 hours. Conduct a hypothesis test with a significance level of 0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=90$

$H_a:\mu<90$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{87-90}{10/\sqrt{20}}\approx-1.34164$$

Since it is observed that $z = -1.34164>-1.6449= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.34164)=0.089856,$ and since $p=0.089856>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

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