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
	16			33			23
	18			31			27
	19			37			21
	17			29			28
	13			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	83	16.6	5.3
Design B	5	164	32.8	9.2
Design C	5	124	24.8	8.2

ANOVA
Source of Variation	SS	df	MS	F	P-Value	F crit
Between Groups	656.1333	2	328.0667	43.35683	3.23E-06	3.88529
Within Groups	90.8	12	7.566667
Total	746.9333	14

F= 43.36

p-value= .000

(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: 16.2, [?, ?]

μC –μA: 8.2, [?, ?]

μC –μB: -8, [?,? ]

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

Confidence interval

μA: [?,? ]

μB: [?, ?]

μC: [?, ?]

Answer 1

(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:

For :

Since confidence interval does not contain zero so we can conclude that populaiton means differ.

For :

Since confidence interval does not contain zero so we can conclude that populaiton means differ.

For :

Since confidence interval does not contain zero so we can conclude that populaiton means differ.

(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: