Question

Consumers can purchase nonprescription medications at food stores, mass merchandise stores such as Target and Wal-Mart, or pharmacies. About 45% of consumers make such purchases at pharmacies. What accounts for the popularity of pharmacies, which often charge higher prices?

A study examined consumers' perceptions of overall performance of the three types of stores, using a long questionnaire that asked about such things as "neat and attractive store," "knowledgeable staff," and "assistance in choosing among various types of nonprescription medication." A performance score was based on 27 such questions. The subjects were 199 people chosen at random from the Indianapolis telephone directory. Here are the means and standard deviations of the performance scores for the sample.

Store type	x	s
Food stores	18.72	24.51
Mass merchandisers	32.55	33.49
Pharmacies	48.79	35.41

We do not know the population standard deviations, but a sample standard deviation s from so large a sample is usually close to σ. Use s in place of the unknown σ in this exercise.

Give 90% confidence intervals for the mean performance for each type of store. (Round your answers to three decimal places.)

Food stores ( __, ___ )

Mass Merchandisers ( __ , __)

Pharmacies ( __ , __ )

Answer 1

Solution:

(a) For food stores - 90% confidence interval will be calculated as-

$\dpi{100} \overline{X}\pm Z_{.05}*\frac{s }{\sqrt{n}}$

where $\overline{X}$ = 18.72 and n = 199 and s (sample standard deviation) = 24.51 and from Z-tables we have Z_.05 = 1.6449

Thus required confidence interval is

( 15.315 , 22.125 )

(b) For mass merchandisers - 90% confidence interval will be calculated as-

$\dpi{100} \overline{X}\pm Z_{.05}*\frac{s }{\sqrt{n}}$

where $\overline{X}$ = 32.55 and n = 199 and s (sample standard deviation) = 33.49 and from Z-tables we have Z_.05 = 1.6449

Thus required confidence interval is

( 27.897 , 37.203 )

(c) For pharmacies - 90% confidence interval will be calculated as-

$\dpi{100} \overline{X}\pm Z_{.05}*\frac{s }{\sqrt{n}}$

where $\overline{X}$ = 48.79 and n = 199 and s (sample standard deviation) = 35.41 and from Z-tables we have Z_.05 = 1.6449

Thus required confidence interval is

( 43.870, 53.709 )

TY!