Question

2. Look at the result reported and indicate what was Grand N: F(3, 17)=3.44, p<.05 A 17 B. 21 C.3 D. 4
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Answer #1

Notice that the df of F distribution is 3 and 17. The first df is the degree of freedom of k treatment, which is k-1 and second df denotes the degree of freedom of error which is N-k. Thus

k-1=3 so k=4 and N-k=17 so N=17+4 = 21.

Thus Grand N is 21. So option B is correct.

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