Question

In: Statistics and Probability

A consumer preference study compares the effects of three different bottle designs (A, B, and C)...

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.

Bottle Design Study Data
A B C
15 32 20
13 32 26
14 32 24
19 35 26
16 35 25

  

The Excel output of a one-way ANOVA of the Bottle Design Study Data is shown below.

SUMMARY
Groups Count Sum Average Variance
Design A 5 77 15.4 5.3
Design B 5 166 33.2 2.7
Design C 5 121 24.2 6.2
ANOVA
Source of Variation SS df MS F P-Value F crit
Between Groups 792.1333 2 396.0667 83.68 3.23E-06 3.88529
Within Groups 56.8 12.0 4.7333
Total 848.9333 14

(a) Test the null hypothesis that μA, μB, and μC are equal by setting α = .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.)

F
p-value

  (Click to select)   Do not reject   Reject   H0: bottle design   (Click to select)   does not   does  have an impact on sales.

(b) Consider the pairwise differences μBμA, μCμA , and μCμB. 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.)

Point estimate Confidence interval
μBμA:  , [   ,  ]
μCμA:  , [   ,  ]
μCμB:  , [   ,  ]

Bottle design   (Click to select)   B   A   C   maximizes sales.

(c) Find a 95 percent confidence interval for each of the treatment means μA, μB, and μC. Interpret these intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)

Confidence interval
μA: [   ,  ]
μB: [   ,  ]
μC: [   ,  ]

Solutions

Expert Solution

a)

Test statistic:

F = 83.68

p-value = F.DIST.RT(83.6761, 2, 12) = 0.0000

Reject H0: bottle design does have an impact on sales.

b)

At α = 0.05, k = 3, N-K = 12, Q value = 3.77

Critical Range, CV = Q*√(MSW/n) = 3.77*√(4.7333/5) = 3.67

Comparison Diff. = (xi - xj) Critical Range Confidence interval
(xi - xj) - CV (xi - xj) + CV
x̅B - x̅A 17.8 3.67 14.13 21.47
x̅C - x̅A 8.8 3.67 5.13 12.47
x̅C - x̅B -9 3.67 -12.67 -5.33

c)

df = n-p = 15-3 = 12

Critical value, t-crit = T.INV.2T(0.05, 12) = 2.179

95% confidence interval for A:

Lower Bound = x̅ - t-crit*√(MSE/n1) = 15.4 - 2.179 *√(4.7333/5) = 13.28

Upper Bound = x̅ + t-crit*√(MSE/n1) = 15.4 + 2.179 *√(4.7333/5) = 17.52

13.28 < µ < 17.52

95% confidence interval for B:

Lower Bound = x̅ - t-crit*√(MSE/n1) = 33.2 - 2.179 *√(4.7333/5) = 31.08

Upper Bound = x̅ + t-crit*√(MSE/n1) = 33.2 + 2.179 *√(4.7333/5) = 35.32

31.08 < µ < 35.32

95% confidence interval for C:

Lower Bound = x̅ - t-crit*√(MSE/n1) = 24.2 - 2.179 *√(4.7333/5) = 22.08

Upper Bound = x̅ + t-crit*√(MSE/n1) = 24.2 + 2.179 *√(4.7333/5) = 26.32

22.08 < µ < 26.32


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