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.

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 μ_A, μ_B, and μ_C 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 μ_B – μ_A, μ_C – μ_A , and μ_C – μ_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.)

Point estimate Confidence interval

μB –μA:  , [   ,  ]

μC –μA:  , [   ,  ]

μC –μB:  , [   ,  ]

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

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

Confidence interval

μA: [   ,  ]

μB: [   ,  ]

μC: [   ,  ]

Answer 1

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