Answer to Question #337534 in Statistics and Probability for cia

Question

A supermarket boasts that checkout times for customers are never more than 15 minutes. A random sample of 36 costumers reveals a mean checkout time of 17 minutes with a standard deviation of 3 minutes. What can you conclude about the supermarket’s boast at the 0.05 level? 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le15$

$H_a:\mu>15$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=35$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.689572.$

The rejection region for this right-tailed test is $R = \{t: t > 1.689572\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{17-15}{3/\sqrt{36}}=4$$

Since it is observed that $t = 4 >1.689572= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $df=35$ degrees of freedom, $t=4$ is $p = 0.000156,$ and since $p= 0.000156<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 15, at the $\alpha = 0.05$ significance level.