The Coca-Cola Company introduced New Coke in 1986. Within three months of this i
ID: 3246602 • Letter: T
Question
The Coca-Cola Company introduced New Coke in 1986. Within three months of this introduction, negative consumer reaction forced Coca-Cola to reintroduce the ordinal formula of Coke as Coca-Cola Classic. Suppose that two years later, in 1987, marketing research firm in Chicago compared tie sales of Coca-Cola Classic. New Coke, and Pepsi in public butting 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: Two-way ANOVA: Cans versus Drink. Building Calculate the value of the test statistic and p-value. At the 0.05 significance level, what is the conclusion? (b) What is the Tukey simultaneous 95 percent confidence interval for the following? Calculate the value of the test statistic and p-value. At the 0.05 significance level, what is the conclusion? Reject H_0 Do not reject H_0 (b) What is the Turkey simultaneous 95 percent confidence interval for the following?Explanation / Answer
I am really sorry to say that, the data you pasted here is not the same as you copied in Minitab. As per as I see the data, the mean value you got from the Minitab is not correct and hence the whole anova table. We will make it correct. but as I don't have Minitab in my computer, I will solve the data in R and this way you will be able to know the R also.
Here is the code to input the data.
Install package Rmisc from tools.
library(Rmisc)
drink<-rep(c("Coke Classic","New Coke","Pepsi"),each=10)
building<-rep(1:10,3)
building
cans<-c(46,143,136,42,151,31,79,220,115,88,
10,106,62,62,42,20,32,140,68,137,
31,82,104,37,52,47,70,124,68,107)
data<-data.frame(drink,building,cans)
data
Output:
> drink<-rep(c("Coke Classic","New Coke","Pepsi"),each=10)
> building<-rep(1:10,3)
> building
[1] 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
> cans<-c(46,143,136,42,151,31,79,220,115,88,
+ 10,106,62,62,42,20,32,140,68,137,
+ 31,82,104,37,52,47,70,124,68,107)
> data<-data.frame(drink,building,cans)
> data
drink building cans
1 Coke Classic 1 46
2 Coke Classic 2 143
3 Coke Classic 3 136
4 Coke Classic 4 42
5 Coke Classic 5 151
6 Coke Classic 6 31
7 Coke Classic 7 79
8 Coke Classic 8 220
9 Coke Classic 9 115
10 Coke Classic 10 88
11 New Coke 1 10
12 New Coke 2 106
13 New Coke 3 62
14 New Coke 4 62
15 New Coke 5 42
16 New Coke 6 20
17 New Coke 7 32
18 New Coke 8 140
19 New Coke 9 68
20 New Coke 10 137
21 Pepsi 1 31
22 Pepsi 2 82
23 Pepsi 3 104
24 Pepsi 4 37
25 Pepsi 5 52
26 Pepsi 6 47
27 Pepsi 7 70
28 Pepsi 8 124
29 Pepsi 9 68
30 Pepsi 10 107
>
I think this is the data as per you posted.
Now, from this data, if you calculate the mean value across different value of drink, you will get the mean as :
Code:
sum<-summarySE(data,measurevar ="cans",groupvars ="drink")
sum
Output:
drink N cans sd se ci
1 Coke Classic 10 105.1 59.38284 18.77850 42.47993
2 New Coke 10 67.9 46.06867 14.56819 32.95554
3 Pepsi 10 72.2 31.62910 10.00200 22.62610
>
See the mean value are not same as you posted. Note: None of the softwares works incorrectly. So, it is the manual error leading to different results. If I have pasted any values incorrectly then change it in the value of cans vector where the values are given as input. And the results will follow.
Now, do the anova. here is the code.
data$building<-as.character(data$building)
anova<-aov(cans~drink+building,data)
summary(anova)
Output:
> data$building<-as.character(data$building)
> anova<-aov(cans~drink+building,data)
> summary(anova)
Df Sum Sq Mean Sq F value Pr(>F)
drink 2 8282 4141 5.241 0.016081 *
building 9 45619 5069 6.415 0.000424 ***
Residuals 18 14223 790
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Now, our variable of interest for which we are looking for the effects is Drinks. Now, look at the output of the ANOVA, We see that F-value for Drink as 5.24. and the value Pr(>F) is 0.0161. So,
a-1) Test- Statistic = 5.24, P-value = 0.0161
a-2) as the p-value is less than 0.05, the significance level, we reject H0 .
Answer: Reject H0.
b) For Tukey test write the following code:
TukeyHSD(anova, "drink",conf.level = 0.95, ordered=TRUE)
Output:
> TukeyHSD(anova, "drink",conf.level = 0.95, ordered=TRUE)
Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = cans ~ drink + building, data = data)
$drink
diff lwr upr p adj
Pepsi-New Coke 4.3 -27.7834268 36.38343 0.9377622
Coke Classic-New Coke 37.2 5.1165732 69.28343 0.0217850
Coke Classic-Pepsi 32.9 0.8165732 64.98343 0.0439268
So, from the output we get the confidence interval as:
Coke-Claissic - New Coke = (5.117,69.283)
Coke-Claissic - Pepsi = (0.817,64.983)
New Coke - Pepsi = (-36.383,27.783) (Just the change of sign of the confidence interval of Pepsi - New Coke in the table)