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 100 shoppers who recently purchased a consumer electronics item online after making a visit to a retail store yielded a mean savings of $58 and a standard deviation of $55.
a. Construct a 95% confidence interval estimate for the mean savings for all showroomers who purchased a consumer electronics item.

b. Suppose the owners of a consumer electronics retailer wants to estimate the total value of lost sales attributed to the next 1,000 showroomers that enter their retail store. How are the results in (a) useful in assisting the consumer electronics retailer in their estimation?

Answer 1

a)

Data provided:

Sample size n = 100

Mean savings = $58

Standard deviation s = $55

Our sample statistic is mean.

Standard error of the mean

Thus, SE

= 5.5

Critical z-value for 95% interval is 1.96

Margin of error M.E. = critical value x standard error

= 1.96 * 5.5

= 10.78

Thus, the 95% confidence interval is:

b) Now this data of confidence intervals can be used for various purposes. One such example is to estimate the total value of lost sales attirbuted to the next 1000 showroomers.

Given the 95% confidence interval, each showroomer will make a saving of minimum $(58 - 10.78), to maximum of $(58 + 10.78)

That is, minimum saving of $47.22 and maximum saving of $68.78

Now suppose all 1000 showroomers visited the showroom, but bought the item online.

This means, the showroom lost sales of minimum $47220 to maximum $68780. These are big numbers, and the showroom must try and encourage the customers to buy from them, rather than online.