Question

The average lifetime of 120 Brand X Alkaline AA batteries and 120 Brand Y alkaline AA batteries were found to be 9.1 hours and 9.6 hours respectively. Suppose the standard deviations of lifetimes are 1.9 for Brand X batteries and 2.1 hours for Brand Y batteries, test the hypothesis using α = 0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\ge\mu_2$

$H_a:\mu_1<\mu_2$

This corresponds to a left-tailed test, and a z-test for two means, with known population standard deviations 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:

$$z=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$=\dfrac{9.1-9.6}{\sqrt{1.9^2/120+2.1^2/120}}=-1.934$$

Since it is observed that $z = -1.934<-1.6449= z_c,$

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.934)=0.026557,$ and since $p=0.026557<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is less than $\mu_2,$ at the $\alpha = 0.05$ significance level.