Question

For the past few years, the number of customers of a drive-in bottle shop has averaged 20 per hour. This year, another bottle shop one kilometre away opened a drive-in window. The manager of the first shop believes that this will result in a decrease in customers. A random sample of 50 hours showed an average of 18.7 customers per hour with a standard deviation of 3.0. Can we conclude at the 5% level of significance that the manager’s belief is correct?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq20$

$H_1:\mu<20$

This corresponds to a left-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 $\alpha=0.05,$ and the critical value for a left-tailed test for $\alpha=0.05$ and $df=n-1=50-1=49$ degrees of freedom is $t_c=-1.676551.$ The rejection region for this left-tailed test is $R=\{t: t<-1.676551\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{18.7-20}{3/\sqrt{50}}\approx-3.064129$$

Since it is observed that $t=-3.064129< -1.676551=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 20, at the $\alpha=0.05$  significance level.

Using the P-value approach: The p-value for left-tailed, $t=-3.064129, df=49,$ $\alpha=0.05$ is $p=0.001772,$ and since $p=0.001772<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 20, at the $\alpha=0.05$  significance level.