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NOT EQUAL VARIANCES! A bank is concerned about the time it takes customers to be served during lunch. It loo
branches (one downtown and one not) and takes random samples from each branch to
(in mins) a customer is in the bank during the lunch hour. Construct a 93% confidence in the average service times during lunch between the downtown branch and the othe
Interpret your answer. Two‐Sample T‐Test and CI: downtown, other Two‐sample T for downtown vs other N Mean StDev SE Mean
downtown 17 4.06 1.07 0.26
other 21 4.10 1.59 0.35 Difference = mu (downtown) ‐ mu (other)
Estimate for difference: ‐0.036
93% CI for difference: (‐0.849, 0.776)
T‐Test of difference = 0 (vs not =): T‐Value = ‐0.08 P‐Value = 0.934 DF = 34 e served during lunch. It looks at its two
samples from each branch to record the amount a time Construct a 93% confidence interval for the difference wntown branch and the other branch . = 0.934 DF = 34 We are 93% confident in the method used the adifference in the avg
service times during lunch between the downtown branch and the other branch falls between ‐0.849 and 0.776 minutes. ...
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 Spring '13
 Leffakis
 branch, Downtown, SE Mean downtown

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