In an effort to reduce energy costs a major university has i

In an effort to reduce energy costs, a major university has installed more efficient lights as well as automatic sensors that turn the lights off when no movement is present in a room. Historically, the cost of lighting an average classroom for 1 week has been $265. To determine whether the changes have signficantly reduced costs, the university takes a sample of 50 classrooms. They find that the average cost for 1 week is $247 with a standard deviation of $60. When testing the hypothesis (at the 5% level of significance) that the average energy use has decreased from the past, what is the p-value? (please round your answer to 4 decimal places)

Solution

So first we need to formulate null and alternate hypothesis

We always assume null hypothesis to be true and its is also that there is no change

So null hypothesis H₀ : average energy cost is $265

alternate hypothesis H₁ : average energy cost is less then $265

Now we need to calculate the Z score

mean of the population is $265

SD of sample is $60 and as we don\'t know population SD we can assume that sample SD represents the population well as sample size is 50(greater the 30).

SD for sampling distribution = 60/50^1/2 = 60/7.071 = 8.48

Z= (247 - 265)/ 8.48 = -18/8.48 =-2.122

At 5% level of significance

Z is - 1.645

As over calculated value is greater we can safely reject the null hypothesis

Which mean alternate hypothesis is accepted and energy cost have decreased

And p value is .0170