Question

The average running time for Broadway shows is 2 hours and 12 minutes. A producer in another city claims that the length of time of productions in his city is the same. He samples 8 shows and finds the time to be 2 hours and 5 minutes with a standard deviation of 11 minutes. Using , is the producer correct?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=132\ min$

$H_1:\mu\not=132\ min$

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

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=8-1=7$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.364619.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.364619\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{125-132}{11/\sqrt{8}}=-1.7999$$

Since it is observed that$|t| = 1.7999 \le 2.364619=t_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for $df=7$ degrees of freedom, $t=-1.7999$ is $p=0.114901,$ and since $p=0.114901>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 2 hours and 12 minutes, at the $\alpha = 0.05$ significance level.