Question

The table below contains data on the enzyme activity of mannose-6-phosphate isomerase (MPI) for three different genotypes (SS, FS, FF) in the amphipod crustacean Platorchestia platensis. Because the effects of sex are not known, specimens were classified by sex. Conduct a two-way ANOVA on this dataset. Test for the effect of genotype, the effect of sex, and whether there is an interaction effect between two factors. Show all calculations and state your conclusions.

	SS	FS	FF
Females	3.65	3.57	3.99
	2.89	4.11	4.26
	3.22	3.34	4.02
	2.74	3.25	3.91
	3.16	3.51	3.87
Males	2.89	3.60	4.93
	3.41	3.47	3.98
	3.21	3.78	4.18
	3.57	3.69	4.01
	3.99	3.72	4.15

Answer 1

a = number of levels for Factor A (Gender) = 2
b = number of levels for Factor B (Genotypes ) = 3
n = number of subjects in each group = 5
abn = 30
For Factor A (Gender):    d.f.N. = a - 1 = 2 - 1 = 1
For Factor B (Genotypes ):    d.f.N. = b - 1 = 3 - 1 = 1
For Interaction (A×B):    d.f.N. = (a - 1)(b - 1) = 2
For Within (error):    d.f.D. = ab(n - 1) = 24

Compute SS_T = sum of squares total:
All data values: 3.65;2.89;3.22;2.74;3.16;3.57;4.11;3.34;3.25;3.51;3.99;4.26;4.02;3.91;3.87;2.89;3.41;3.21;3.57;3.99;3.60;3.47;3.78;3.69;3.72;4.93;3.98;4.18;4.01;4.15
N = Number of Values = 30
?X = Sum of all data values = 110.07
X_GM = Grand Mean = 110.07/30 = 3.669
[?X]²/abn = 110.07²/30 = 403.84683
SS_T = Sum of Squares Total = (3.65)²+ (2.89)²+ (3.22)²+ (2.74)²+ (3.16)²+ (3.57)²+ (4.11)²+ (3.34)²+ (3.25)²+ (3.51)²+ (3.99)²+ (4.26)²+ (4.02)²+ (3.91)²+ (3.87)²+ (2.89)²+ (3.41)²+ (3.21)²+ (3.57)²+ (3.99)²+ (3.60)²+ (3.47)²+ (3.78)²+ (3.69)²+ (3.72)²+ (4.93)²+ (3.98)²+ (4.18)²+ (4.01)²+ (4.15)² - 403.84683
SS_T = Sum of Squares Total = 6.47107

Compute SS_A = sum of squares for Factor A:
From Row 1:
Data: 3.65;2.89;3.22;2.74;3.16;3.57;4.11;3.34;3.25;3.51;3.99;4.26;4.02;3.91;3.87
n = Number of Values = 15
Sum of data values = 53.49
Mean = 53.49/15 = 3.566

From Row 2:
Data: 2.89;3.41;3.21;3.57;3.99;3.60;3.47;3.78;3.69;3.72;4.93;3.98;4.18;4.01;4.15
n = Number of Values = 15
Sum of data values = 56.58
Mean = 56.58/15 = 3.772

SS_A = sum of squares for Factor A = 15(3.566 - 3.669)²+ 15(3.772 - 3.669)²
SS_A = sum of squares for Factor A = 0.31827
a - 1 = 2 - 1 = 1
MS_A = SS_A/(a-1) = 0.31827

Compute SS_B = sum of squares for Factor B:
From Column 1
Data:3.65;2.89;3.22;2.74;3.16;2.89;3.41;3.21;3.57;3.99
n = Number of Values = 10
Sum of data values = 32.73
Mean = 32.73/10 = 3.273


From Column 2
Data:3.57;4.11;3.34;3.25;3.51;3.60;3.47;3.78;3.69;3.72
n = Number of Values = 10
Sum of data values = 36.04
Mean = 36.04/10 = 3.604

From Column 3
Data:3.99;4.26;4.02;3.91;3.87;4.93;3.98;4.18;4.01;4.15
n = Number of Values = 10
Sum of data values = 41.3
Mean = 41.3/10 = 4.13

SS_B = sum of squares for Factor B = 10(3.273 - 3.669)²+ 10(3.604 - 3.669)²+ 10(4.13 - 3.669)²
SS_B = sum of squares for Factor B = 3.73562
b - 1 = 3 - 1 = 2
MS_B = SS_B/(b-1) = 1.86781

Compute SS_AXB = sum of squares for Interaction:
X_GM = Grand Mean = 3.669
n = number of data values in each cell = 5

Sample Mean for cell (1 , 1) = 3.132
Sample Mean for Row 1 = 3.566
Sample Mean for Column 1 = 3.273

Sample Mean for cell (1 , 2) = 3.556
Sample Mean for Row 1 = 3.566
Sample Mean for Column 2 = 3.604

Sample Mean for cell (1 , 3) = 4.01
Sample Mean for Row 1 = 3.566
Sample Mean for Column 3 = 4.13

Sample Mean for cell (2 , 1) = 3.414
Sample Mean for Row 2 = 3.772
Sample Mean for Column 1 = 3.273

Sample Mean for cell (2 , 2) = 3.652
Sample Mean for Row 2 = 3.772
Sample Mean for Column 2 = 3.604

Sample Mean for cell (2 , 3) = 4.25
Sample Mean for Row 2 = 3.772
Sample Mean for Column 3 = 4.13

SS_A×B = sum of squares for Interaction (A×B) = ??n[Cell Mean(i, j) - Row Mean(i) - Column Mean(j) + Grand Mean]²
SS_A×B = sum of squares for Interaction (A×B) = 5(3.132 - 3.566 - 3.273 + 3.669)²+ 5(3.556 - 3.566 - 3.604 + 3.669)²+ 5(4.01 - 3.566 - 4.13 + 3.669)²+ 5(3.414 - 3.772 - 3.273 + 3.669)²+ 5(3.652 - 3.772 - 3.604 + 3.669)²+ 5(4.25 - 3.772 - 4.13 + 3.669)²
SS_A×B = sum of squares for Interaction (A×B) = 0.04758
(a - 1)(b - 1) = (2 - 1 )(3 - 1 ) = 2

MS_A×B = SS_A×B/[(a - 1)(b - 1)] = 0.02379

Compute SS_E = sum of squares for error term (within-group):
SS_E = SS_T - SS_A - SS_B - SS_A×B
SS_E = 6.47107 - 0.31827 - 3.73562 - 0.04758
SS_E = 2.3696
a = 2
b = 3
n = 5
ab(n - 1) = (2)(3)(5- 1) = 24

MS_E = SS_E/ ab(n - 1) = 0.0987333333

Based on above calculation we obtained following ANNOVA summary table:

Inference:

There is no significant effect of sex on enzyme activity, since the calculated value of F_A = 3.2235313977 is smaller than the Critical Value for F_A test = 4.259677214. Therefore the null hypothesis H0 cannot be rejected.

There is significant effect of genotypes on enzyme activity, since the calculated value of F_B = 18.9177245105 is greater than the Critical Value for F_B test = 3.402826105. Therefore the null hypothesis H0 is rejected.

There is no significant interaction between Gender*Genotypes, since the calculated value of F_AXB = 0.2409520594 is smaller than the Critical Value for F_AXB test = 3.402826105.