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 3 colors and 2 shapes to produce a total of 6 designs. Each logo was presented to 3 different groups (a total of 18 groups) and the approval rating for each was recorded and is shown below. The manager analyzed these data using the α = 0.05 level of significance for all inferences, and a partially completed 2-way ANOVA table is also provided below. COLORS Red Green Blue 54 67 36 Circle 44 61 40 SHAPES 49 64 44 34 56 36 Square 36 57 33 35 58 30 Analysis of Variance Source df SS MS F Shapes 392.00 Colors 1897.00 Interaction 49.00 Error Total 2460.00 -------------------------------------------------------------------------------- A.) What is the sum of squares due to Error ? B.) What are the degrees of freedom for the factor Colors and the Error ? C.) How many treatment combinations are there in the experiment and how many replicates for each treatment combination? D.) Perform the F- test on 'Shapes'. Is the factor 'Shapes' significant (α = 0.05)? Please show all the relevant calculations. E.) Perform the F- test on 'Interaction'. Is 'Interaction' significant (α=0.05)? Please show all the relevant calculations.

Answer 1

Answer:

There are two factors which are color and shape.

Color have three levels red, green and yellow and shapes have two levels circle and square.

THis is the problem of two way anova.

The formulaes for completing ANOVA table is,

a) Sum of squares for shapes (SSshapes) = SStotal - sum of (SSshapes + SScolors + SSinteraction)

= 2460.00 - (1897.00 + 49.00 + 122.00) = 392.00

b) Degrees of freedoms :

dfshapes = no. of levels for shapes - 1 = 2-1 = 1

dfcolors = no. of levels of colors - 1 = 3-1 = 2

dfinteraction = 2*1= 2

dftotal = 18-1 = 17

dferror = 17 - 2-2-1 = 12

Now to find Mean sum of squares.

MSshapes = SSshapes / dfshapes = 392.00 / 1 = 392.00

MScolors = SScolors / dfcolors = 1897.00 / 2 = 948.50

MSerror = SSerror / dferror = 122.00 / 12 = 10.17

d) F-test for shapes :

Hypothesis :

H0 : There is no difference in population mean for rows.

H1 : There is difference in population mean for rows.

Assume alpha = level of significane = 0.05  

Fshapes = MSshapes / MSerror = 392.00 / 10.17 = 38.56

Now we have to find p-value for F-test.

P-value we can find in excel.

syntax :

=FDIST(x, deg_freedoms1, deg_freedoms2)

x = 38.56

deg_freedoms1 = 1

deg_freedoms2 = 12

P-value = 0.0000

P-value < alpha

Reject H0 at 5% level of significance.

COnclusion : Atleast one of the population mean for rows is differ than 0.

e).

Hypothesis :

H0 : There is no difference in population mean for columns.

H1 : There is difference in population mean for columns.

Assume alpha = level of significane = 0.05  

Fcolors = MScolors / MSerror = 948.50 / 10.17 = 93.30

Now we have to find p-value for F-test.

P-value we can find in excel.

syntax :

=FDIST(x, deg_freedoms1, deg_freedoms2)

x = 93.30

deg_freedoms1 = 1

deg_freedoms2 = 12

P-value = 0.0000

P-value < alpha

Reject H0 at 5% level of significance.

COnclusion : Atleast one of the population mean for columns is differ than 0.

The complete ANOVA table will be,

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
shapes 1 392.00 392.00 38.56 0.000
colors 2 1897.00 948.50 93.30 0.000
shapes*colors 2 49.00 24.50 2.41 0.132
Error 12 122.00 10.17
Total 17 2460.00

NOTE:: I HOPE YOUR HAPPY WITH MY ANSWER.....

**PLEASE SUPPORT ME WITH YOUR RATING

....THANK YOU.....