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
	18			31			28
	13			31			25
	14			29			22
	13			29			26
	15			29			20

  

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	73	14.6	4.3
Design B	5	149	29.8	1.2
Design C	5	121	24.2	10.2

ANOVA
Source of Variation	SS	df	MS	F	P-Value	F crit
Between Groups	590.9333	2	295.4667	56.46	3.23E-06	3.88529
Within Groups	62.8	12.0	5.2333
Total	653.7333	14
(b) Consider the pairwise differences ?B – ?A, ?C – ?A , and ?C – ?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 ?B –?A: , [, ] ?C –?A: , [, ] ?C –?B: , [, ] Bottle design (Click to select)BAC maximizes sales. (c) Find a 95 percent confidence interval for each of the treatment means ?A, ?B, and ?C. Interpret these intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) Confidence interval ?A: [, ] ?B: [, ] ?C: [, ]

Answer 1

(b)

The point estaimtes are:

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 df=n-1=4.

The critical value of t for 95% confidence interval is 2.776.

Formula for confidence interval is

Following table shows the confidence intervals:

	Mean	SD	Lower limit	Upper limit
A	14.6	2.074	12.03	17.17
B	29.8	1.095	28.44	31.16
C	24.2	3.194	20.23	28.17