Question

The Coca-Cola Company introduced New Coke in 1985. Within three months of this introduction, negative consumer...

The Coca-Cola Company introduced New Coke in 1985. Within three months of this introduction, negative consumer reaction forced Coca-Cola to reintroduce the original formula of Coke as Coca-Cola Classic. Suppose that two years later, in 1987, a marketing research firm in Chicago compared the sales of Coca-Cola Classic, New Coke, and Pepsi in public building vending machines. To do this, the marketing research firm randomly selected 10 public buildings in Chicago having both a Coke machine (selling Coke Classic and New Coke) and a Pepsi machine.

The Coca-Cola Data and a MINITAB Output of a Randomized Block ANOVA of the Data:

Building
1 2 3 4 5 6 7 8 9 10
Coke Classic 44 126 144 53 145 30 77 232 107 84
New Coke 5 106 63 63 43 11 35 146 69 139
Pepsi 27 94 102 41 58 47 69 130 68 104

Two-way ANOVA: Cans versus Drink, Building

Source DF SS MS F P
Drink 2 9,206.6 4,603.30 6.57 .007
Building 9 54,523.6 6,058.18 8.64 .000
Error 18 12,614.1 700.78
Total 29 76,344.3
Descriptive Statistics: Cans
Variable Drink Mean
Cans Coke Classic 104.2
New Coke 62.7
Pepsi 74.0

(a-1) Calculate the value of the test statistic and p-value. (Round "test statistic" value to 2 decimal places and "p-value" to 3 decimal places.)

Test statistic
p-value

  

(a-2) At the 0.05 significance level, what is the conclusion?

  • Reject H0

  • Do not reject H0

(b) What is the Tukey simultaneous 95 percent confidence interval for the following? (Negative amounts should be indicated by a minus sign. Round your answers to 2 decimal places.)

Confidence interval
Coke Classic - New Coke [   ,  ]
Coke Classic – Pepsi [   ,  ]
New Coke – Pepsi [   ,  ]

Homework Answers

Answer #1

a-1)

from above given output:

test statistic F =6.57

p value =0.007

a-2)since p value <0.05

Reject H0

b)

MSE= 700.7800
df(error)= 18
number of treatments = 3
pooled standard deviation=Sp =√MSE= 26.472
critical q with 0.05 level and k=3, N-k=18 df= 3.61
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj)         = 30.220

Confidence interval =sample mean difference -/+ HSD

Lower bound Upper bound
(xi-xj)-ME (xi-xj)+ME
μ1-μ2 11.28 71.72
μ1-μ3 -0.02 60.42
μ2-μ3 -41.52 18.92
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