# What do you do if your degrees of freedom goes past the end of your tables?

The degrees of freedom in my F table don't go up high enough for my big sample.

For example, if I have an F with 5 and 6744 degrees of freedom, how do I find the 5% critical value for an ANOVA?

What if I was doing a chi-square test with big degrees of freedom?

[A question like this was posted a while ago but the OP made an error and actually had a smaller d.f., reducing it to a duplicate -- but the original large d.f. question should have an answer somewhere on site]

• Get a larger table? – Federico Poloni May 8 '17 at 17:56

## 1 Answer

F tables:

1. The easiest way of all -- if you can -- is to use a statistics package or other program to give you the critical value. So for example, in R, we can do this:

 qf(.95,5,6744)
 2.215425


(but you can as easily calculate an exact p-value for your F).

2. Usually F tables come with an "infinity" degrees of freedom at the end of the table, but a few don't. If you have a really large d.f. (for example, 6744 is really large), you can use the infinity ($\infty$) entry in its place.

So you might have tables for $\nu_1=5$ that give 120 df and $\infty$ df:

      ...    5      ...
⁞
120        2.2899
∞         2.2141


The $\infty$ d.f. row there will work for any really large $\nu_2$ (denominator d.f.). If we use that we have 2.2141 instead of the exact 2.2154 but that's not too bad.

3. If you don't have an infinity degrees of freedom entry, you can work one out from a chi-square table, using the critical value for the numerator d.f. divided by those d.f.

So for example, for a $F_{5,\infty}$ critical value, take a $\chi^2_5$ critical value and divide by $5$. The 5% critical value for a $\chi^2_5$ is $11.0705$. If we divide by $5$ that's $2.2141$ which is the $\infty$ row from the table above.

4. If your degrees of freedom may be a bit too small to use the "infinity" entry (but still a lot bigger than 120 or whatever your table goes up to) you can use inverse interpolation between the highest finite d.f. and the infinity entry. Let's say we want to calculate a critical value for $F_{5, 674}$ d.f.

   F       df     120/df
------   ----    -------
2.2899    120      1
C       674    0.17804
2.2141     ∞       0


Then we compute the unknown critical value, $C$ as

$C \approx 2.2141 + (2.2899-2.2141) \times (0.17804-0)/(1-0) \approx 2.2276$

(The exact value is $2.2274$, so that works pretty well.)

More details on interpolation and inverse interpolation are given at that linked post.

Chi-squared tables:

If your chi-squared d.f. are really large you can use normal tables to get an approximation.

For large d.f. $\nu$ the chi-squared distribution is approximately normal with mean $\nu$ and variance $2\nu$. To get the upper 5% value, take the one-tailed 5% critical value for a standard normal ($1.645$) and multiply by $\sqrt{2\nu}$ and add $\nu$.

For example, imagine we needed an upper 5% critical value for a $\chi^2_{6744}$.

We would calculate $1.645 \times \sqrt{2 \times 6744} + 6744 \approx 6935$. The exact answer (to $5$ significant figures) is $6936.2$.

If the degrees of freedom are smaller, we can use the fact that if $X$ is $\chi^2_\nu$ then $\sqrt{2X}\dot\sim N(\sqrt{2\nu-1},1)$.

So for example, if we had $674$ d.f. we might use this approximation. The exact upper 5% critical value for a chi-square with 674 d.f. is (to 5 figures) $735.51$. With this approximation, we would calculate as follows:

Take the upper (one tailed) 5% critical value for a standard normal (1.645), add $\sqrt{2\nu-1}$, square the total and divide by 2. In this case:

$(1.645+\sqrt{2\times 674-1})^2/2 \approx 735.2$.

As we see, this is quite close.

For considerably smaller degrees of freedom, the Wilson-Hilferty transformation could be used -- it works well down to only a few degrees of freedom -- but the tables should cover that. This approximation is that $(\frac{X}{\nu})^{\frac13}\dot\sim N(1-\frac{2}{9\nu},\frac{2}{9\nu})$.

• +1 The $\chi^2$ idea can be improved. Use the fact that $F$ limits to a rational function of a $\chi^2$ distribution as its second parameter grows large. In R, for instance, you would compute it as df2/df1 * (-1 + 1/(1-qchisq(0.95, df1) / df2)). You will obtain $2.2177$, accurate to three significant figures. Note that the $\chi^2$ parameter is a small integer, indicating it will likely be in the table and available without interpolation. – whuber May 8 '17 at 16:57
• I assume I have missed something here -- I've tried several times to work out which advantage you mean in this improvement over what I did in item 3 (which already treats it as simple function of a chi-squared with small integer d.f., as would be suggested by Slutsky's theorem as df2$\to\infty$). In the example at hand, my approximation is both simpler to carry out and more accurate (e.g. has about 57% of the absolute error). Is this suggestion better at other values of the two df's, or better because it's conservative rather than anti-conservative, ... – Glen_b Apr 10 '19 at 1:52
• ... or is the intent that the errors of the two approaches will be opposite in direction (suggesting perhaps to combine the two?). – Glen_b Apr 10 '19 at 2:07
• I recall I was referring to item 4. – whuber Apr 10 '19 at 12:34
• Ah, that might make more sense. Sorry to be dense. I'll try that again. – Glen_b Apr 10 '19 at 13:12