Question

Imagine a utility company shared population variance for all LEED and non-LEED certified buildings in their database. If we want to compare water usage between two samples of LEED-certified and non-LEED certified buildings for which we have data, what type of test should we use and why? What is the main benefit of using this type of test? Interpret the findings.

	Gallons of Water Used per Hour LEED-Certified Buildings (in thousands of gallons)		Gallons of Water Used per Hour non LEED-Certified Buildings (in thousands of gallons)
Mean		49.22	61.80
Known Variance		182.25	144
Observations		40	50
Hypothesized Mean Difference		0
z		-2.461
P(Z<=z) one-tail		0.01
z Critical one-tail		1.645
P(Z<=z) two-tail		0.02
z Critical two-tail		1.96

Answer 1

WE SHOULD USE HERE LARGE SAMPLE TESTS FOR 2 VARIABLES ( DIFFERENCE BETWEEN 2 MEANS).

AS WE KNOW THE POULATION VARIANCES, WE CAN USE LARGE SAMPLE TESTS.

AS WE KNOW POULATION VARIANCES, WE CAN ASSUME THAT DATA ARE NORMALLY DISTRIBUTED

IF BOTH POPULATIONS HAVE A NORMAL DISTRIBUTION, THE CENTRAL LIMIT THEOREM ENABLES US TO CONCLUDE THAT THE SAMPLING DISTRIBUTIONS OF SAMPLE 1 AND 2 CAN BE APPROXIMATED BY A NORMAL DISTRIBUTION.

SO BENEFIT IS INSTEAD OF MAKING CALCULATION FOR 2 MEANS SEPARATELY, WE CAN HAVE CONCLUSION BY COMPARING BOTH THE SAMPLES AT A TIME

NOW FOR FINDINGS, SEE IMAGE