Question

You are the manager of a restaurant that delivers pizza to college dormitory rooms. You have just changed your delivery process in an effort to reduce the mean time between the order and completion of delivery from the current 25 minutes. A sample of 36 orders using the new delivery process yields a sample mean of 22.4 minutes and a sample standard deviation of 6 minutes. a. Using the six-step critical value approach, at the 0.05 level of significance, is there evidence that the population mean delivery time has been reduced below the previous population mean value of 25 minutes? b. At the 0.05 level of significance, use the five-step p-value approach. c. Interpret the meaning of the p-value in (b). d. Compare your conclusions in (a) and (b)

Answer 1

a.

The sample mean is and the sample standard deviation is s = 6, and the sample size is n = 36.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 25

Ha: μ < 25

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

The significance level is α=0.05, and the critical value for a left-tailed test is tc​=−1.69.

The rejection region for this left-tailed test is R={t : t < − 1.69}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = −2.6 < tc ​= −1.69, it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 25, at the 0.05 significance level.

There is evidence that the population mean delivery time has been reduced below the previous population mean value of 25 minutes at the 0.05 level of significance.

b.

The sample mean is and the sample standard deviation is s = 6, and the sample size is n = 36.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 25

Ha: μ < 25

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

The significance level is α=0.05, and the critical value for a left-tailed test is tc​=−1.69.

The rejection region for this left-tailed test is R={t : t < − 1.69}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Using the P-value approach: The p-value is p = 0.0068, and since p = 0.0068 < 0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 25, at the 0.05 significance level.

There is evidence that the population mean delivery time has been reduced below the previous population mean value of 25 minutes at the 0.05 level of significance.

c.

The p-value is the probability of extremum of the value of the statistic obtained under the null hypothesis is true.

d.

By both the approach, we are rejecting the null hypothesis and concluding that population mean delivery time has been reduced below the previous population mean value of 25 minutes at the 0.05 level of significance.