PLEASE EXPLAIN Consumers can purchase nonprescription medications at food stores

Question

PLEASE EXPLAIN

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 205 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 90% confidence intervals for the mean performance for each type of store. (Round your answers to three decimal places.)

(c) Based on these confidence intervals, are you convinced that consumers think that pharmacies offer higher performance than the other types of stores? (In Chapter 12, we will study a statistical method for comparing means of several groups.)

a.Yes, the pharmacy interval is well above the others.

b. Yes, the mass merchandiser interval is well above the others.    

c. Yes, the food store interval is well below the others.

d. No, there is no clear evidence of a significant difference.

Store type x s Food stores 18.4 25.22 Mass merchandisers 32.24 33.25 Pharmacies 48.68 35.60

Explanation / Answer

a.
given that,
sample mean, x =18.4
standard deviation, s =25.22
sample size, n =205
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 204 d.f is 1.652
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 = [ 18.4 ± Z a/2 ( 25.22/ Sqrt ( 205) ]
= [ 18.4-(1.652 * 1.761) , 18.4+(1.652 * 1.761) ]
= [ 15.49 , 21.31 ]

b.
given that,
sample mean, x =32.24
standard deviation, s =33.25
sample size, n =205
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 204 d.f is 1.652
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 = [ 32.24 ± Z a/2 ( 33.25/ Sqrt ( 205) ]
= [ 32.24-(1.652 * 2.322) , 32.24+(1.652 * 2.322) ]
= [ 28.404 , 36.076 ]

c.
given that,
sample mean, x =48.68
standard deviation, s =35.6
sample size, n =205
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 204 d.f is 1.652
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 = [ 48.68 ± Z a/2 ( 35.6/ Sqrt ( 205) ]
= [ 48.68-(1.652 * 2.486) , 48.68+(1.652 * 2.486) ]
= [ 44.572 , 52.788 ]

d. a.Yes, the pharmacy interval is well above the others.

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