Question

10.6 A company claims that its batteries have a mean life of 100 hours. You try to verify this for a sample of size 21 with mean 97 hours and variance 9 hours.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=100$

$H_1:\mu\not=100$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ $df=n-1=21-1=20$ degrees of freedom and the critical value for a two-tailed test is $t_c=2.086.$

The rejection region for this two-tailed test is $R=\{t: |t|>2.086\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{97-100}{9/\sqrt{21}}=-1.5275$$

Since it is observed that $|t|=1.5275<2.086=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for $\alpha=0.05, df=20$ degrees of freedom, two-tailed, $t=-1.5275$ is $p=0.142298,$ and since $p=0.142298>0.05=\alpha,$ 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 $100,$ at the $\alpha=0.05$ significance level.