Consumers can purchase nonprescription medications at food stores, mass merchand

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 191 people chosen at random from the Indianapolis telephone directory. Here are the means and standard deviations of the performance scores for the sample.

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

(b) Give 98% confidence intervals for the mean performance for each type of store. (Round your answers to three decimal places.)

food stores ( ) , ( )

Mass merchandisers ( ), ( )

Explanation / Answer

a.
given that,
sample mean, x =18.49
standard deviation, s =25.09
sample size, n =191
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 190 d.f is 2.346
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 18.49 ± Z a/2 ( 25.09/ Sqrt ( 191) ]
= [ 18.49-(2.346 * 1.815) , 18.49+(2.346 * 1.815) ]
= [ 14.231 , 22.749 ]
b.
given that,
sample mean, x =32.4
standard deviation, s =33.26
sample size, n =191
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 190 d.f is 2.346
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 32.4 ± Z a/2 ( 33.26/ Sqrt ( 191) ]
= [ 32.4-(2.346 * 2.407) , 32.4+(2.346 * 2.407) ]
= [ 26.754 , 38.046 ]
c.
given that,
sample mean, x =48.49
standard deviation, s =35.48
sample size, n =191
level of significance, = 0.02
from standard normal table, two tailed value of |t /2| with n-1 = 190 d.f is 2.346
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 48.49 ± Z a/2 ( 35.48/ Sqrt ( 191) ]
= [ 48.49-(2.346 * 2.567) , 48.49+(2.346 * 2.567) ]
= [ 42.467 , 54.513 ]
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