A consumer organization collected data on two types of automobile batteries, A and B. The summary statistics for 12 observations of each type are . X(bar)A=36.51, X(bar)B=34.21, SA=1.43, SB=.93. Assume that the data are normally distributed with OA=OB.

Is there evidence to support the claim that type A battery mean life exceeds that of type B? Test the hypothesis at alpha = 0.05

Construct a 99% Confidence interval (CI) for the difference in the mean battery life.

There evidence to support the claim that type A battery mean life exceeds that of type B.

The 99% Confidence interval (CI) for the difference in the mean battery life is (0.91, 3.69).

Step-by-step explanation:

The hypothesis can be defined as follows:

H₀:The mean life of battery A is not more than that of battery B, i.e. .

Hₐ:The mean life of battery A is more than that of battery B, i.e. .

The significance level of the test is, α = 0.05.

The test statistic is:

Compute the value of standard error as follows:

Compute the test statistic as follows:

Decision rule:

If the test statistic value is more than the critical value, , then the null hypothesis will be rejected. And vice-versa.

Compute the degrees of freedom (df) as follows:

The critical value is,

The test statistic, t = 4.67 > = 1.729.

Thus, the null hypothesis is rejected at 5% level of significance.

Conclusion:

Thus, there evidence to support the claim that type A battery mean life exceeds that of type B.

The critical value of t for a 99% confidence level and degrees of freedom 19 is:

The pooled standard deviation is:

Compute the 99% confidence interval for the difference in the mean battery life as follows:

Thus, the 99% Confidence interval (CI) for the difference in the mean battery life is (0.91, 3.69).

answer

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step-by-step explanation:

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