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 pham ades what accounts or the popularity of pharmacies, w h ch o en 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 amang various types af nonprescription medication. A performance score was based on 27 such quesions. The subjects were 184 pcople chosen at random from the ind anapolis telephone directory, Here are the means and standard deviations of the performance scores for the sample. Store type 18.39 25.07 Mass merchandisers 32.62 33.02 49.79 35.71 Food stores we do not know the population standard deviados, but a sample standard deviation s from so large a sample is usually dose to . use s in place of the unknown in this exercise. (a) What population do you think the authors of the study want to draw cond usions about? o the American public sick people citizens of Indianapolis What population are you certain they can draw conclusions about? o citizens of Indianapolis sick people the American public (b) Give 95% confidence intervals for the mean performance for each type of store. (Round your answers to three decimal places.) Food stores Hass merchand isers (c) Based on these confidence intervals, are you convinced that consumers think that pharmades offer higher perfarmance than the other types of stores? (in Chapter 12, we will study a statistical method for comparing means of several groups.) res, the pharmacy interval is well above the others. Yes, the mass merchandiser interval is well above the others. Yes, the food store interval is well below the others. No, there is no clear evidence of a sign ficant difference

Explanation / Answer

a.
i. consumers
ii. pharmacies
b
i.
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 ]

ii.
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 ]

iii.
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 ]

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

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