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?

  • 3
    $\begingroup$ This is called linear interpolation. $\endgroup$
    – whuber
    Commented Jan 11, 2013 at 21:05
  • $\begingroup$ @whuber Why is it that I don't do this for $F_{4,80}$ then? $\endgroup$
    – Kaish
    Commented Jan 11, 2013 at 22:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Jan 12, 2013 at 17:32

1 Answer 1


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.

  • 1
    $\begingroup$ Incidentally, it's common to interpolate in the reciprocal of the degrees of freedom. This tends to improve the accuracy. $\endgroup$
    – Glen_b
    Commented Jul 16, 2013 at 3:13

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