Question

The manager of the service department of a local car dealership has noted that the service times of a sample of 24 new automobiles has a standard deviation of 5 minutes.
A.) Compute a 90% confidence interval estimate for the standard deviation of the service times (in minutes) for all their new automobiles.
B.) The manager of the service department of a local car dealership wants to test the claim that the variance of the service time does not exceed 27 minutes. Test this claim using a significance level of 0.1.

Answer 1

Solution :

A) The 90% confidence interval for population standard deviation is given as follows :

Where, n is sample size, s is sample standard deviation and the terms given in denominator are critical values of chi-square to construct 90% confidence interval for population standard deviation.

We have, n = 24, s = 5 minutes

Using chi-square table we get,

Hence, 90% confidence interval estimate for the standard deviation of the service times (in minutes) for all their new automobiles is,

The 90% confidence interval estimate for the standard deviation of the service times (in minutes) for all their new automobiles is (4.042 minutes, 6.628 minutes).

B) Null and alternative hypotheses :

The null and alternative hypotheses would be as follows :

Test statistic :

To test the hypothesis we shall use chi-square test for variance. The test statistic is given as follows :

Where, n is sample size, s is sample standard deviation and σ² is hypothesized value of population variance under H_{​​​​​​0}.

We have, n = 24, s = 5 and  σ² = 27

The value of the test statistic is 21.2963.

P-value :

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right-tailed p-value is given as follows :

P-value = P(χ² > value of the test statistic)

P-value = P(χ² > 21.2963)

P-value = 0.5630

The p-value is 0.5630.

Decision :

P-value = 0.5630

Significance level = 0.01

(0.5630 > 0.10)

Since, p-value is greater than the significance level of 0.10, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}) at 0.10 significance level.

Conclusion : At 0.10 significance level, there is not sufficient evidence to support the claim that the variance of the service time does not exceed 27 minutes.

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