Imagine a utility company shared population variance for all LEED and nonLEED certified buildings in their database. If we want to compare water usage between two samples of LEEDcertified and nonLEED 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 LEEDCertified Buildings (in thousands of gallons) 
Gallons of Water Used per Hour non LEEDCertified 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) onetail 
0.01 

z Critical onetail 
1.645 

P(Z<=z) twotail 
0.02 

z Critical twotail 
1.96 
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
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