Consider the e-billing case. the mean and the standard deviation of the sample of n = 65 payment times are xbar= 18.4730 and s = 3.9045. test h0: μ = 18.6 versus ha: μ < 18.6 by setting α equal to .01 and using a critical value rule and assume normality of the population

The null and the alternative hypotheses are given below:

Under null hypothesis, the test statistic is:

Now we have to find the left tailed critical value at 0.01 significance level using the standard normal table.

The critical value is:

Since the test statistic is -0.26 lies in acceptance region because the critical region is beyond -2.33, therefore, we fail to reject the null hypothesis.

here

step-by-step explanation: