Explanation
Snedecor's F distribution arises as the ratio \( F = (\chi^2_{d_1}/d_1) / (\chi^2_{d_2}/d_2) \) between two independent chi-square variables normalized by their respective degrees of freedom \( d_1 \) and \( d_2 \). It only takes positive values and is right-skewed; with large degrees of freedom it tends to approach a normal distribution. Its mean (when \( d_2 > 2 \)) is \( d_2/(d_2-2) \), which approaches 1 as \( d_2 \) grows.
It is used mainly in three contexts: (1) ANOVA, to test whether the means of several groups are equal by comparing between-group variance with within-group variance; (2) equality-of-variances tests (Levene's test or the classic F test), where \( F = S_1^2/S_2^2 \) follows an F distribution under the null hypothesis of equal variances; and (3) linear regression, to assess whether the overall model is significantly better than the null model. Use this calculator to get right-tail p-values or the critical values needed to make decisions in these tests.
Formula
$$ f(x)= \frac{1}{B\left(\frac{d_1}{2},\frac{d_2}{2}\right)} \left(\frac{d_1}{d_2}\right)^{d_1/2} x^{d_1/2-1} \left(1+\frac{d_1}{d_2}x\right)^{-(d_1+d_2)/2} $$
Parameters
- d₁: numerator degrees of freedom.
- d₂: denominator degrees of freedom.
Worked example
Situation: An agronomic trial compares crop yield under 4 different treatments (\(\nu_1 = 4 - 1 = 3\) numerator degrees of freedom). Each treatment is applied to replicated plots, with a total of 24 observations (\(\nu_2 = 24 - 4 = 20\) denominator degrees of freedom). The ANOVA F statistic measures whether the variation between treatments exceeds the variation within groups.
Question 1: What is the cumulative probability \(P(F \leq 3.10)\) with \(\nu_1 = 3\) and \(\nu_2 = 20\)?
Solution: Looking up the CDF of \(F_{3,20}\) at \(x = 3.10\): \[ P(F \leq 3.10) \approx 0.95 \] The right tail is \(P(F > 3.10) = 0.05\). If the observed F statistic were exactly 3.10, the p-value of the test would be exactly 5%.
Question 2: What is the critical value \(F^*\) for the ANOVA test at level \(\alpha = 0.05\)?
Solution: We look for the 0.95 quantile of \(F_{3,20}\): \[ F^*_{0.05;\,3,\,20} = 3.10 \] If the statistic \(F_{\text{obs}} > 3.10\), we reject the null hypothesis of equal means at the 5% significance level.
Interpretation: An \(F_{\text{obs}} = 3.10\) statistic means that the variability between treatments is 3.10 times greater than the variability within groups. With \(\nu_1 = 3\) and \(\nu_2 = 20\), this value falls right at the 5% threshold, signaling statistically significant differences between treatments.