Question

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Battery A: . Battery B: . Assume equal variances and that we calculated . Perform a .05 level test to determine if Battery B outlasts Battery A by more than 2 hours.

Answer 1

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

• The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

• The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630