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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:
[tex]H_0: \mu_B - mu_A \leq 2[/tex]
At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:
[tex]H_1: \mu_B - \mu_A > 2[/tex]
- The sample means are: [tex]\mu_A = 8.65, \mu_B = 11.23[/tex]
- The standard deviations for the samples are [tex]s_A = s_B = 0.67[/tex]
Hence, the standard errors are:
[tex]s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934[/tex]
The distribution of the difference has mean and standard deviation given by:
[tex]\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58[/tex]
[tex]s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 2[/tex] is the value tested at the hypothesis.
Hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{2.58 - 2}{0.2735}[/tex]
[tex]t = 2.12[/tex]
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 [tex]t^{\ast} = 1.71[/tex]
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 https://brainly.com/question/13873630
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