Question

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

Answer 1

Answer: The null and the alternative hypotheses are given below:

$H_{0}:\mu=18.6$

$H_{a}:\mu<18.6$

Under null hypothesis, the test statistic is:

$z=\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

        $=\frac{18.4730-18.6}{\frac{3.9045}{\sqrt{65}} }$

        $=-0.26$

$\therefore z=-0.26$

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

The critical value is:

$z_{critical}=-2.33$

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.