A marketing researcher wants to estimate the mean savings ($) realized by shoppe

Question

A marketing researcher wants to estimate the mean savings ($) realized by shoppers who showroom. Showrooming is the practice of inspecting products in retail stores and then purchasing the products online at a lower price. A random sample of 81 shoppers who recently purchased a consumer electronics item online after making a visit to a retail store yielded a mean savings of $55 and a standard deviation of $48. Complete parts (a) and (b). a. Construct a 99% confidence interval estimate for the mean savings for all showroomers who purchased a consumer electronics item. $ lessthanorequalto mu lessthanorequalto $ (Type integers or decimals. Round to two decimal places as needed.)

Explanation / Answer

a) Margin of error = (2.638691*48)/Sqrt(81) = 14.07302

99% Interval

(55-14.07302, 55+14.07302)

= (40.93, 69.07)

So, Interval is (40.93, 69.07)

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