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Results from a 2 Factor ANOVA produced the following FA(2, 48) = 2.25, p >.05, FB(1,48) = 12.07, p<.05 and FaxB(2, 48) = 7.48 I need help with this question please
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Answer #1

Solution:

We have:

dfA = r - 1 = 2,

thus r = 2+1=3.

dfB = c - 1 = 1,

thus c = 1+1=2.

dfAXB = (r-1)*(c-1) = 1

dferror = r*c*(n-1) = 48

thus using

r*c*(n-1) = 48

3*2*(n-1) =48

6*(n-1) = 48

n-1 = 48/6

n-1= 8

n = 8+ 1

n = 9

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