F Test: $F_{4, 81}(5\%) \approx F_{4,60}(5\%) - (81 - 60)(F_{4, 60}(5\%) - F_{4,120}(5\%))/60$. Where does this come from?

Question

I had to compute the F-test in my ANOVA question and use interpolation if necassary.

The first time, I had to work out the values of $F_{4,90}(5\%)$ which I said was approximatley $F_{4, 60}(5 \%)$ and then did my test and got the right answer.

In the second bit, in my 3-way ANOVA table, when I'm checking to see if two of the factors interact, I need to get the values for $F_{4, 81}(5 \%)$ and so I thought I'd do the same thing and so its roughly $F_{4, 60}(5 \%)$ and then do it like this. However it turns out you get something like:

$$ F_{4, 81}(5\%) \approx F_{4,60}(5\%) - (81 - 60)(F_{4, 60}(5\%) - F_{4,120}(5\%))/60$$

Where does this come from and why is it like this?

For higher accuracy, you should have. I suspect that the size of the error made (from not interpolating) was small enough that it did not matter. Jaime's answer explains this nicely. — whuber, Jan 12 '13 at 17:32

score 4 · Answer 1 · answered Jan 12 '13 at 00:37

4

This graph shows $F_{4,n}(5\%)$ for $n$ in the range $60$ to $120$, plus the two approximations you have used with your ANOVAs.

enter image description here

The maximum error you are going to make by using $F_{4,60}$ for every value in the range is below $3\%$, but if you go for linear interpolation, you can bring that down to $0.6\%$. Why it would be necessary for one part, but not the other, I don't have a clue.

answered Jan 12 '13 at 00:37

Jaime

256

1

Incidentally, it's common to interpolate in the reciprocal of the degrees of freedom. This tends to improve the accuracy. – Glen_b Jul 16 '13 at 03:13

F Test: $F_{4, 81}(5\%) \approx F_{4,60}(5\%) - (81 - 60)(F_{4, 60}(5\%) - F_{4,120}(5\%))/60$. Where does this come from?

1 Answers1

Linked