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
	17			34			21
	16			30			21
	13			34			28
	13			30			26
	16			34			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	75	15.0	3.5
Design B	5	162	32.4	4.8
Design C	5	121	24.2	9.7

ANOVA
Source of Variation	SS	df	MS	F	P-Value	F crit
Between Groups	757.7333	2	378.8667	63.14	4.27E-07	3.88529
Within Groups	72.0	12.0	6.0000
Total	829.7333	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.)

(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.)

(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.)

Answer 1

a)

since test statistic F =63.14 is in critical region of p value< 0.05

reject null and conclude that bottle designs A, B, and C have different effects on mean daily sales

b)

pooled standard deviation        =Sp =√MSE      =				2.449
critical q with 0.05 level and at k=3 and N-k=12 degree of freedom=						3.77

Tukey's (HSD) for group i and j =                  (q/√2)*(sp*√(1/ni+1/nj)         =

4.13

	point estimate:		Lower bound	Upper bound	differ
	(xi-xj )		(xi-xj)-ME	(xi-xj)+ME
μB-μA	17.40		13.27	21.53	significant difference
μC-μA	9.20		5.07	13.33	significant difference
μC-μB	-8.20		-12.33	-4.07	significant difference

c)

	lower	Upper
Confidence interval for	bound	bound
A	12.61	17.39
B	30.01	34.79
C	21.81	26.59