I am resolving an anova one-way exercise. But I am doubtful because when using the F-table to get the p-value, I am getting a probability around 2.2 (an obvious error), but if I use a calculator the probability is 0.042.

I have an independent variable with 4 levels. The mean and standard deviation are this:

N         Mean         Standard Deviation
10         4.3          2.2
10         3.8          1.6
10         4.9          1.9
10         6.3          2.1

I did these calculations:

Grand mean = 4.8.

Sum of squares between groups = 35.07.

Sum of squares within groups = 138.7.

The degrees of freedom between groups is 3, the degree of freedom within groups is 36.

So with this I get a value of 3.

With this f value I check the f table for a 5% of significance level, using a degree of freedom of 3 in the numerator and 36 in the denominator and the value in the table is around 2.2. So I'm not understanding why I get 2.2 and the calculator gives 0.042. Do you have any idea?

  • $\begingroup$ Using a calculator, I get moreless the same values for everything only the p is different, is 0.042. For example in the calculator the f is also 3.033. $\endgroup$
    – AOn
    Jan 8, 2017 at 0:33

2 Answers 2


The value you're looking up in the table is a critical value, not a p-value. There's no contradiction when they differ since they're not at all the same thing.

(However, at the 5% level you mention, I don't get a critical value of 2.2 when I look up an F with 3 and 36 df. Instead I get 2.866. Did you mistakenly look up the 10% critical value?)

Note that when you looked up the table your data values didn't come into what you looked up at all. Only the degrees of freedom and the significance level. What you looked up is the smallest F value that would be significant at the 5% level, which you compare your F with; if it's at least that large, you have a significant F. (That F value is 2.87. Your F value was 3.03 which was larger than it, so you would reject the null if you did it that way)

This is completely consistent with getting a p-value below 0.05:

Density for F(3,36) showing 5% critical value and calculated F, as well as the shaded area to the right of the calculated F which is the p-value of 0.042

We can see that there are two ways you can get the outcome of the hypothesis test:

(i) see if the F value is greater than the critical F from the table

(ii) see if the p-value is less than the significance level.

You can do whichever one of the two is convenient (done correctly, they give identical reject/don't reject decisions).

In general, if you have a p-value (say from computer output), you will compare that with the significance level, but if you are working "by hand" and have only tables, compare the F with the F critical value. Note that if you're using tables you often need to get values not in the table.

  • $\begingroup$ Thanks. Yes I was checking at 10%, but the table just shows values for df of 30 and then df of 40, I need for a df of 36, but yes itt should be arround 2.8 because 40 it is 2.83. $\endgroup$
    – AOn
    Jan 8, 2017 at 17:59
  • $\begingroup$ But so for example in anova we compare always the f statistics with the critical value? We dont compare the probability with the level of significance, for example compere 0.05 with 0.042. Or we can do both ways? $\endgroup$
    – AOn
    Jan 8, 2017 at 18:04
  • $\begingroup$ see update at end of answer $\endgroup$
    – Glen_b
    Jan 9, 2017 at 1:27

Your table value is 2.2. Compare the computed value of F= 3 with 2.2. This suggests signficant effect at alpha = 5%..You need to check that .042 is some value that you have generated. May be it is p-value. You should know that F-table does not show p-values. It incorporates only F values.

  • $\begingroup$ Thanks for your answer. So we need to compare to f and not with the p value? For example in a two sample t test, when we calculate the test statistic we get a t and we go to the table to check the p that corresponds to the obtained t and then we compare that probability with the significance level. In anova its not like that? In anova we compare that 2.2 or 0.052(in calculator) with f and not with the significance level of 5%? $\endgroup$
    – AOn
    Jan 8, 2017 at 0:50
  • $\begingroup$ If computed F is greater than table value at a specified significance level, it would imply that there is significant effect. May be you have another question in your mind . let it be in question form rather than putting it as a comment. $\endgroup$
    – user10619
    Jan 8, 2017 at 3:17

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