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 given below. Let μA, μB, and μC represent the mean daily sales using bottle designs A, B, and C, respectively.

a. State the null and alternative hypotheses

2. What is the test statistic?
3. Using a .05 significance level, what is the decision rule?
4. Show the test statistic and essential calculations.
5. Interpret you results
6. Should you conduct a Post Hoc Tukey on this data? Why.
7. Based on your answer for part f conduct a Post Hoc Tukey

Bottle Design Study Data
A	B	C
16	33	23
18	31	27
19	37	21
17	29	28
13	34	25

Answer 1

	A	B	C	Total
Sum	83	164	124	371
Count	5	5	5	15
Mean, Sum/n	16.6	32.8	24.8
Sum of square, Ʃ(xᵢ-x̅)²	21.2	36.8	32.8

Number of treatment, k = 3

Total sample Size, N = 15

df(between) = k-1 = 2

df(within) = N-k = 12

df(total) = N-1 = 14

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 656.1333

SS(within) = SS1 + SS2 + SS3 = 90.8

SS(total) = SS(between) + SS(within) = 746.9333

MS(between) = SS(between)/df(between) = 328.0667

MS(within) = SS(within)/df(within) = 7.566667

F = MS(between)/MS(within) = 43.3568

p-value = F.DIST.RT(43.3568, 2, 12) = 0.0000

a)

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: Not all means are equal.

b)

We will use One way ANOVA test

Test statistic:

F = 43.36

c)

Critical value Fc = F.INV.RT(0.05, 2, 12) = 3.885

Rejection region: Reject Ho if F > 3.885

d) Essential calculation are shown above.

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	656.1333	2	328.0667	43.3568	0.0000
Within Groups	90.8000	12	7.5667
Total	746.9333	14

e)

As F > Fc, we reject the null hypothesis.

There is enough evidence to conclude that the not all means are equal.

f)

Yes, a post-Hoc tukey test should be performed as the null hypothesis is rejected.

g)

At α = 0.05, k = 3, N-K = 12, Q value = 3.78

Critical Range, CV = Q*√(MSW/n) = 3.78*√(7.5667/5) = 4.65

Comparison	Diff. = (xi - xj)	Critical Range	Results
x̅1-x̅2	-16.2	4.65	Means are different
x̅1-x̅3	-8.2	4.65	Means are different
x̅2-x̅3	8	4.65	Means are different