The marketing manager of a company producing a new cereal aimed for children wan
ID: 3361915 • Letter: T
Question
The marketing manager of a company producing a new cereal aimed for children wants to examine the effect of the color and shape of the box's logo on the approval rating of the cereal. He combined 4colors and 3 shapes to produce a total of 12 designs. Each logo was presented to 2 different groups (a total of 24 groups) and the approval rating for each was recorded and is shown below. The manager analyzed these data using the = 0.01 level of significance for all inferences, and a partially completed 2-way ANOVA table is provided.
Analysis of Variance
Source df SS MS F
Colors
Shapes 579.00
Interaction 150.33
Error 150.00
Total 3590.50
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1) What is the sum of squares due to factor Colors?(10 points)
2) What are the degrees of freedom for the factor Shapes and the Error ? (10 points)
3) How many treatment combinations are there in the experiment and how many replicates for each treatment combination? (10 points)
4) Perform the F- test on 'Colors'. Is the factor 'Colors' significant ( = 0.01)?(10 points)
5) Perform the F- test on 'Interaction'. Is the factor 'Interaction' significant (=0.01)?(10 points)
COLORS SHAPES Red Green Blue Yellow Circle 54 67 36 45 44 61 44 41 Square 34 56 36 21 36 58 30 25 Diamond 46 60 34 31 48 60 38 33Explanation / Answer
a) What is the sum of squares due to factor Colors?
Total - Shapes - Interaction - Error
= 3590.5-579-150.33-150
=2711.17
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b) What are the degrees of freedom for the factor Shapes and the Error ?
There are three shapes. Therefore:
df(shapes)=3-1=2
There are four colors. Therefore:
df(colors)=4-1=3
df(interaction)=df(shapes)*df(colors)=2*3=6
df(Total) =n-1= 24-1=23
The degree of freedom(error) = 23-3-2-6 =12
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c) How many treatment combinations are there in the experiment and how many replicates for each treatment combination?
4*3= 12 (because He combined 4 colors and 3 shapes to produce a total of 12 designs which is given by the question.)
Each treatment is measured twice, therefore:
replicates for each treatment combination = 2
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d) Perform the F- test on 'Colors'. Is the factor 'Colors' significant ( = 0.01)?
MS(Colors) = 2711.17/3 = 903.72
MSE =150/12 = 12.5
The test hyothesis:
Ho:the factor 'Colors' is not significant (i.e. null hypothesis)
Ha: the factor 'Colors' is significant (i.e. alternative hypothesis)
The test statistic is
F test= MS/ MSE =903.72 /12.5 =72.30
Given a=0.01, the critical value is F(0.99, df1=3, df2=12) =5.95 (from F table)
Since F=72.30 is larger than 5.95, we reject HO.
So we can conclude that the factor 'Colors' is significant
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e) Perform the F-test on 'Interaction.' Is the factor 'Interaction' significant?
MS=150.33/6 =25.055
MSE=150/12 = 12.5
The test hypothesis:
Ho: the factor 'Interaction' is not significant (i.e. null hypothesis)
Ha: the factor 'Interaction' is significant (i.e. alternative hypothesis)
The test statistic is
F= MS/MSE = 25.055/12.5 =2.0044
Given a=0.01, the critical value is F(0.99, df1=6, df2=12) =4.82 (from F table)
Since F=2.0044 is less than 4.82, we do not reject HO.
So we can not conclude that the factor 'Interaction' is significant
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