Question

Suppose you are working for a regional residential natural gas utility. For a sample of 85 customer visits, the staff time per reported gas leak has a mean of 231 minutes and standard deviation 35 minutes. The VP of network maintenance hypothesizes that the average staff time devoted to reported gas leaks is 237 minutes. At a 1 percent level of significance, what is the lower bound of the interval for determining whether to accept or reject the VP's hypothesis? At a 1 percent level of significance, what is the upper bound of the interval for determining whether to accept or reject the VP's hypothesis? Note that the correct answer will be evaluated based on the z-values in the summary table in the Teaching Materials section. Please round your answer to the nearest tenth. Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.

Answer 1

Ideally we would have used a t test since the population standard deviation is not known, but since it is specially mentioned that we have to evaluate using the z value, hence we are going to be using the z test for one population mean instead.

Based on the information provided, the significance level is α=0.01, and the critical value for a two-tailed test is

The rejection region for this two-tailed test is

Lower bound = -2.58

Upper bound = 2.58

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