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	3.23E-06	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. Leave no cells blank - be certain to enter "0" wherever required.)

F
p-value

(Click to select)Do not rejectReject H₀: bottle design (Click to select)doesdoes 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.)

Point estimate Confidence interval

μB –μA: , [, ]

μC –μA: , [, ]

μC –μB: , [, ]

Bottle design (Click to select)ACB 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)

The F test statistics is

F = 63.14

The p-value of F test is 0.0000

The p-value is less than 0.05 so we reject the null hypothesis.

(b)

Here we have 3 groups and total number of observations are 15. So degree of freedom is

df= 15-3 = 12

Critical value for , df=12 and k=3 is

So Tukey's HSD will be

So confidence intervals are:

groups (i-j)	xbari	xbarj	ni	nj	HSD	xbari-xbarj	Lower limit	Upper limit	Significant(Yes/No)
mu1-mu2	15	32.4	5	5	4.13	-17.4	-21.53	-13.27	Yes
mu1-mu3	15	24.2	5	5	4.13	-9.2	-13.33	-5.07	Yes
mu2-mu3	32.4	24.2	5	5	4.13	8.2	4.07	12.33	Yes

(c)

For 95% confidence interval of individual means, degree of freedom of t is

The critical value of t for 95% confidence interval, using excel function "=TINV(0.05,12)", is 2.179.

Formula for confidence interval is

For treatment mean A:

For treatment mean B:

For treatment mean C: