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 228 minutes and standard deviation 36 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?

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

Here in the example is given that ,

sample size = n = 85

sample mean =   = 228

standard deviation =

standard error =    =    = 3.9

using the 1 % level of significance the z value is z = 2.58

margin of error = ( z value ) * ( stadard error ) = (2.58)*(3.9) = 10.1

confidence interval formula for the 1% level of significance is ,

= ( sample mean - margin of error , sample mean + margin of error )

= ( 228 - 10.1 , 228 + 10.1 )

=(217.9 , 238.1 )

here the lower bound = 217.9 and the upper bound = 238.1

if the hypothesized value H0 lies within the interval ( not less than the lower bound or greater than the upper bound ) then fail to reject H0.

here the hypothesized value H0 : 237 lies within the interval hence we fail to reject Ho or we accept H0 Here.