Answer to Question #347801 in Statistics and Probability for Maekey

Question

2. A sample of 40 sales receipts from a grocery store has sample mean of 137 and population sd= 30.2. Use these values to test whether or not the mean sales the grocery store is less than 150. Use a 95% confidence level

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge150$

$H_1:\mu<150$

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 $\alpha = 0.05,$ $df=n-1=39$ and the critical value for a left-tailed test is $t_c =-1.684875.$

The rejection region for this left-tailed test is $R = \{t:t<-1.684875\}.$

The t-statistic is computed as follows:

$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{137-150}{30.2/\sqrt{40}}=-2.7225$

Since it is observed that $t=-2.7225<-1.684875=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=39$ degrees of freedom, $t=-2.7225$ is $p=0.004819,$ and since $p=0.004819<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 150, at the $\alpha = 0.05$ significance level.