Question

Supermarkets have limited the number of customers who are in their stores at any one time. At one particular store, store managers decide that 100 customers at one time is necessary to ensure social distancing guidelines mandated by the government. One store manger suspects that the number of customers is more than 100 so he counts the number of customers for a week at different times of the day and on different days of the week. Here are the numbers: 105, 98, 111, 123, 88, 95, 109.

(3pts) Using StatKey, find the p-value:

Answer 1

The sample mean is and the sample standard deviation is s = 11.61075, and the sample size is n = 6.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 100

Ha: μ > 100

This corresponds to a right-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 right-tailed test is tc​=2.015.

The rejection region for this right-tailed test is R = t : t > 2.015

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = 0.874 ≤ tc ​= 2.015, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.211, and since p = 0.211 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is greater than 100, at the 0.05 significance level.

Therefore, there is not enough evidence to claim that the number of customers is more than 100.