A consumer preference study compares the effects of three different bottle desig
ID: 3262156 • Letter: A
Question
A consumer preference study compares the effects of three different bottle designs (A, B, and C) on sales of a popular fabric softener. A completely randomized design is employed. Specifically, 15 supermarkets of equal sales potential are selected, and 5 of these supermarkets are randomly assigned to each bottle design. The number of bottles sold in 24 hours at each supermarket is recorded. The data obtained are displayed in the following table.
The Excel output of a one-way ANOVA of the Bottle Design Study Data is shown below.
(a) Test the null hypothesis that A, B, and C are equal by setting = .05. Based on this test, can we conclude that bottle designs A, B, and C have different effects on mean daily sales? (Round your answers to 2 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)
(Click to select)RejectDo not reject H0: bottle design (Click to select)doesdoes not have an impact on sales.
(b) Consider the pairwise differences B – A, C – A , and C – B. Find a point estimate of and a Tukey simultaneous 95 percent confidence interval for each pairwise difference. Interpret the results in practical terms. Which bottle design maximizes mean daily sales? (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
Bottle design (Click to select)BCA maximizes sales.
(c) Find a 95 percent confidence interval for each of the treatment means A, B, and C. Interpret these intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
Bottle Design Study Data A B C 19 34 26 18 33 24 14 35 23 17 30 21 14 31 27
Explanation / Answer
Solution
Part (a)
To test H0: µA = µB = µC , test statistics is F = 68.65 [refer ANOVA Table]
p-value of F2, 12 for a value greater than 68.65 = 3.23E-06 [refer ANOVA Table]
Since p-value < 0.05 [i.e., 5% level of significance stipulated in the question], H0 is rejected.
=> bottle design does have impact on sales. All ANSWERS in bold.
Part (b)
Back-up Theory
Given n observations on each of two variables, X and Y, the point estimate for (µX - µY) is:
(Xbar - Ybar)/{s(2/n)} and 100(1 - )% confidence interval for (µX - µY) is:
(Xbar - Ybar) ± t /2{s(2/n)}
t /2 is the upper /2 percent point of t-distribution with degrees of freedom = (2n - 2).
From the Fcrit value given under ANOVA, F2, 12. = 3.88269 => level of significance, = 0.05 (i.e., 5%). So, t8, 0.025 = 2.306 and (2/n) = 0.4 = 0.6325
All computations are given in the table below:
CI for
Xbar
Ybar
s12
s22
s2
s
Point Estimate
CI Lower Bound
CI Upper Bound
µB - µA
32.6
16.4
4.3
5.3
4.8
2.1909
15.8
13.00
19.40
µC - µA
24.2
16.4
5.7
5.3
5.5
2.3452
7.8
4.38
11.22
µC - µB
24.2
32.6
5.7
4.3
5.0
2.2361
- 8.4
- 11.66
5.14
Part (c)
100(1 - )% confidence interval for µX is: Xbar ± t /2{sX(1/n)}
t /2 is the upper /2 percent point of t-distribution with degrees of freedom = (n - 1) = 4.
CI for
Xbar
sX2
s
CI Lower Bound
CI Upper Bound
µA
16.4
5.3
2.1909
13.54
19.26
µB
32.6
4.3
2.3452
30.03
35.17
µC
24.2
5.7
2.2361
21.24
27.16
DONE
CI for
Xbar
Ybar
s12
s22
s2
s
Point Estimate
CI Lower Bound
CI Upper Bound
µB - µA
32.6
16.4
4.3
5.3
4.8
2.1909
15.8
13.00
19.40
µC - µA
24.2
16.4
5.7
5.3
5.5
2.3452
7.8
4.38
11.22
µC - µB
24.2
32.6
5.7
4.3
5.0
2.2361
- 8.4
- 11.66
5.14
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