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
	19			29			24
	18			33			23
	16			32			26
	17			33			23
	18			32			24

SUMMARY
Groups	Count	Sum	Average	Variance
Design A	5	88	17.6	1.3
Design B	5	159	31.8	2.7
Design C	5	120	24.0	1.5

ANOVA
Source of Variation	SS	df	MS	F	P-Value	F crit
Between Groups	505.7333	2	252.8667	137.93	5.25E-09	3.88529
Within Groups	22.0	12.0	1.8330
Total	527.7333	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.)

F=______ ; P-Value = ______; (reject/do not reject) H0: bottle design (Does/Does not) 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.)

Ub-Ua:	(confidence interval)	[	]
Uc-Ua		[	]
Uc-Ub		[	]

(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.)

Ua:[
Ub:[
Uc:[

Answer 1

a)
F =137.93

p value =0.000

reject Ho, bottle design Does have an impact on sales.

b)

pooled standard deviation        =Sp =√MSE      =				1.354
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)         =						2.28

			Confidence interval
			Lower bound	Upper bound
	(xi-xj )		(xi-xj)-ME	(xi-xj)+ME
μB-μA	14.20		11.92	16.48
μC-μA	6.40		4.12	8.68
μB-μA	-7.80		-10.08	-5.52

c)

	lower	Upper
	bound	bound
A	16.28	18.92
B	30.48	33.12
C	22.68	25.32