Factorial Anova Practice Example Answers PDF

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Factorial ANOVA Practice Example A silly researcher is interested in whether liking for pizza varies as a function of type of brand and gender. To test this he asks both men and women to taste the exact same pizza and rate their liking of the pizza on a 1 (This is disgusting) to 10 (Nom! NOM! NOM!!!) scale. Again, everyone eats the exact same pizza, but one group thinks they are eating a low-fat pizza and the other thinks they are eating a regular pizza. Both men and women were randomly assigned to either low-fat or regular pizza branding. Part A. Analyze the following data using a factorial ANOVA. I have decided that Type of Branding is Factor A and Gender is factor B. I have not presented it as a contingency table as I did in the notes so be careful that you are calculating the correct values.

Low-Fat Females Males

Factor A Factor B

M SD  x 2 x 

2

9

7

4

4

6

9

1

2

6

7

5

2

9

10

6

5

9

10

3

4

7

10

5

3.17 7.67 1.33 1.51 19 46 69 364 Factor A Low-Fat Regular (all)

SD  x

T grand total

142 1

T # grand totals A Σx for each

8.83 1.47 53 479

(all)

5.42 2.71 65

M

Regular Females Males

6.42 2.97 77

4.00 1.79 Grand Totals 24 142 112 1024 Factor B Females Males (all)

(all)

6.00 3.25 72

M SD  x

[T] = T2/abs =

5.83 2.48 70

1422 / 24 =

840.17

[A] =ΣA2/bs = 2 2

65 +77 / 12

846.17

[B] = ΣB /as = 722+702/ 12

840.33

2

65, 77

level of factor A

[AB] = ΣAB2/s a # levels F A B Σx for each

2 72, 70 2

level of factor B

b # levels F B AB Σx for each indv condition

ab # of conds ABS s # ps per cond abs # ps o-all

19, 46,53,24 4 individual scores 6 24

= 192+462+532+242 /6

[ABS] = 2 ΣABS

977.00 1024.00

Source A (Brand)

SS

[A] – [T] = 846.17-840.17

df

MS

sig.

6.00

a-1 = 2-1= 1

6.00

2.55

ns

0.16

b-1= 2-1= 1

0.16

0.07

ns

(a-1)(b-1)= 1*1 = 1 130.67

55.60

B (Gender) [B] –[T] = 840.33-840.17

F

AxB [AB]-[A]-[B]+[T]= =977-846.17840.33+840.17

130.67

Error [ABS]-[AB] = =1024-977

47.00

abs-ab= 24-4 = 20

183.83

abs-1= 24-1 = 23

Total [ABS]-[T] = =1024-840.17

< .01

2.35

For (1, 20) degrees of freedom

4.35 8.10

Fcrit(.05) Fcrit(.01)

Part B. Write up the results of this ANOVA (using a Fisher’s LSD if necessary). YOU DO NOT NEED TO MAKE A GRAPH ON THE EXAM! However, I may provide you one so you can get a better idea of what is going on with the data and how to write up the results. Remember, you only perform LSD tests for any effects that: 1) are significant and 2) there are more than 2 groups in the effect. So, even if the main effects for Brand and Gender were significant you wouldn’t calculate the LSD for them. Here the interaction is significant, so you need to follow up with LSD tests. Based on the results in the graph I just did two mean comparisons because they can explain the significant interaction. LSD FOR AN INTERACTION Error df for AxB interaction

20

tcrit (Error df) MS error s

2.09 2.35 6

LSD =

 tcrit (erro r_ df )     2.09 

1.85

M(Low-Fat/Females)-M(Regular/Females)= M(Low-Fat/Males) - M(Regular/Males)=

|3.17-8.83| =5.66 |7.67-4| =3.67

2MS erro r     s 

22.35  6

   

Write-up: A 2 X 2 Factorial ANOVA was conducted with brand (low-fat, regular) and gender (female, male) as the between-subjects factors. There was insufficient evidence to conclude that brand had an effect on taste of ratings, F(1,20) = 2.55, ns. Participants who ate the low-fat brand (M = 5.42, SD = 2.71) did not differ in taste ratings than those who ate the regular brand (M = 6.42, SD = 2.97). There was insufficient evidence to conclude that gender had an effect on taste ratings, F(1, 20) = 0.07, ns. Women (M = 6.00, SD = 3.25) did not differ in taste ratings from men (M = 5.83, SD = 2.48). The lack of main effects was qualified by a brand by gender interaction, F(1, 20) = 55.60, p < .01. A Fisher's LSD of 1.85 indicated that women liked the regular brand (M = 8.83, SD = 1.47) better than the low-fat brand (M = 3.17, SD = 1.33). However, men liked the low-fat brand (M = 7.67, SD = 1.51) better than the regular brand (M = 4.00, SD = 1.79). The means are presented in Figure 1....