Five-step hypothesis testing

[debbiebdunn]

1. The new director of special programs in XYZ Corporation felt the customers were waiting too long to receive and complete forms needed to enroll in special programs. After collecting some data, Ms. Jones determined the mean wait time was 28 minutes. She felt this time period was excessive and she instituted new procedures to streamline the process. One month later, a sample of 127 customers was selected. The mean wait time recorded was 26.9 minutes and the standard deviation of the sampling was 8 minutes. Using the .02 level of significance, conduct a five-step hypothesis testing procedure to determine if the new processes significantly reduced the wait time.

What I came up. I'm not sure if it's correct.

Null Hypothesis : = 28
H : not equal to 28
Reject H if Z < - 2.05
Computed Z = -1.55, found by:
Z = 26.9-28 = -1.55
8/sq. root 127
Fail to reject H . The change did not reduce the wait time significantly.

 

[galactus]

Since they think the claimed mean is excessive, I think a left tail may be more appropriate.

\(\displaystyle H_{0}:{\mu}\geq 28\)
\(\displaystyle H_{a}:{\mu}<28, \;\ \text{claim}\)

The test stat is -1.5495 and the critical value is -2.0537, p value = .0619

Since the test stat is not in the rejection region, we DO NOT reject the null hypothesis.

Also, since the p value is > alpha, we do not reject the null hyp.

Which means that their claim of 28 minutes holds water.

There is not enough evidence at the .02 alpha level to support the claim that the mean time is less than 28 minutes.

You are correct. We have a do not reject with a two tail as well. So, that option is OK too.