Question

Suppose you are working for a regional residential natural gas utility. For a sample of 95 customer visits, the staff time per reported gas leak has a mean of 219 minutes and standard deviation 34 minutes. The VP of network maintenance hypothesizes that the average staff time devoted to reported gas leaks is 226 minutes. At a 5 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.

Answer 1

Solution :

Given that,

= 219

s = 34

n = 95

Degrees of freedom = df = n - 1 = 95- 1 = 94

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,94 = 1.99

Margin of error = E = t_/2,df * (s /n)

= 1.99 * ( 34/ 95)

= 6.94

The 95% confidence interval estimate of the population mean is,

- E < < + E

219 - 6.94 < < 219 - 6.94

212.06 < < 225.94

212.1 < < 225.9

upper bound =225.9

Reject VP's hypothesis because  is out of upper bound . Value of is not between confidence interval