Question

In: Statistics and Probability

A consumer preference study compares the effects of three different bottle designs (A, B, and C)...

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 μBμA, μCμA , and μCμB. 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
μBμA: 16.2, [?, ?]
μCμA: 8.2, [?, ?]
μCμB: -8, [?,? ]

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

Confidence interval
μA: [?,? ]
μB: [?, ?]
μC: [?, ?]

Solutions

Expert Solution

(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:


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