Why the Null F Statistic Has an F Distribution

Why the Null Distribution Matters

A one-way ANOVA statistic compares variation among the observed group means with variation among observations inside the groups. Its formula is a ratio of mean squares, but the formula alone does not say what values the ratio would take if the group means were equal. To make that probability statement, we need the sampling model and the distribution of each sum of squares.

Under independent normal errors with one common variance, and when all population group means are equal, the between-group and within-group sums of squares become independent scaled chi-square variables. Dividing each by its degrees of freedom gives two independent estimates on the same variance scale. Their ratio has a central F distribution. This exact result applies to balanced and unbalanced group sizes.

Model and assumptions: There are \(k\) groups, with \(n_i\) observations in group \(i\), and \(N=\sum_{i=1}^k n_i\). The one-way fixed-effects model is \(Y_{ij}=\mu_i+\varepsilon_{ij}\), where the errors are mutually independent and each has distribution \(N(0,\sigma^2)\), with \(\sigma^2>0\). The null hypothesis is \(H_0:\mu_1=\cdots=\mu_k\). Thus, under the null, all observations have the same mean \(\mu\). The exact central F result requires independent observations, normal errors, and a common variance.

The Two Sums of Squares and Their Degrees of Freedom

Write \(\bar{Y}_{i\cdot}\) for the sample mean in group \(i\) and \(\bar{Y}_{\cdot\cdot}\) for the grand mean across all \(N\) observations. The between-group and within-group sums of squares are

SS_{\text{between}}=\sum_{i=1}^k n_i(\bar{Y}_{i\cdot}-\bar{Y}_{\cdot\cdot})^2, \qquad SS_{\text{within}}=\sum_{i=1}^k\sum_{j=1}^{n_i}(Y_{ij}-\bar{Y}_{i\cdot})^2.

The group means have \(k\) values, but their deviations from the grand mean must sum to zero when weighted by group size. This leaves \(k-1\) degrees of freedom for the between-group sum of squares. Estimating a separate mean in each group uses one degree of freedom per group, leaving \(N-k\) degrees of freedom for the within-group sum of squares.

The corresponding mean squares and ANOVA statistic are

MS_{\text{between}}=\frac{SS_{\text{between}}}{k-1}, \qquad MS_{\text{within}}=\frac{SS_{\text{within}}}{N-k}, \qquad F=\frac{MS_{\text{between}}}{MS_{\text{within}}}.

Definition: A chi-square random variable with \(\nu\) degrees of freedom, written \(\chi^2_\nu\), is a sum of squares of \(\nu\) independent standard normal random variables. If \(U\sim\chi^2_{\nu_1}\) and \(V\sim\chi^2_{\nu_2}\) are independent, then \((U/\nu_1)/(V/\nu_2)\) has a central F distribution with \(\nu_1\) numerator and \(\nu_2\) denominator degrees of freedom, written \(F_{\nu_1,\nu_2}\).

Why the Sums of Squares Are Independent Chi-Square Variables

The reason is geometric. Treat the \(N\) observations as one vector. The vectors that are constant within each group form a \(k\)-dimensional subspace. Inside that subspace, the vectors constant across every observation form a one-dimensional subspace. The between-group variation occupies the part of the group-constant subspace perpendicular to the overall-constant direction, so it has dimension \(k-1\). The within-group variation is perpendicular to every group-constant vector, so it has dimension \(N-k\). These two directions of variation are orthogonal.

Under the null, the observation vector has mean \(\mu\) in every coordinate. Subtracting that mean leaves independent normal coordinates with mean zero and variance \(\sigma^2\). An orthogonal change of coordinates preserves this normal structure: the coordinates in orthogonal directions are independent. The between-group sum of squares is the sum of squares of \(k-1\) such coordinates, and the within-group sum of squares is the sum of squares of the other \(N-k\) coordinates.

Identify the dimensions.
The between-group subspace has dimension \(k-1\), and the within-group subspace has dimension \(N-k\).

Use the normal model.
After subtracting the common null mean and dividing by \(\sigma\), the observation vector has independent standard normal coordinates.

Form the sums of squares.
Each sum is a sum of squared standard normal coordinates in its subspace. Their orthogonality makes the two sums independent.

Scale by the variance.
The resulting distributions are \(SS_{\text{between}}/\sigma^2\sim\chi^2_{k-1}\) and \(SS_{\text{within}}/\sigma^2\sim\chi^2_{N-k}\), independently.

This argument is the key justification for the null F distribution: it gives both the chi-square degrees of freedom and the independence needed to form the F ratio. In particular, the result is

\frac{SS_{\text{between}}}{\sigma^2}\sim\chi^2_{k-1}, \qquad \frac{SS_{\text{within}}}{\sigma^2}\sim\chi^2_{N-k}, \qquad \frac{SS_{\text{between}}}{\sigma^2} \ \text{and}\ \frac{SS_{\text{within}}}{\sigma^2} \ \text{are independent}.

Dividing each scaled sum by its degrees of freedom produces two independent quantities with the common factor \(\sigma^2\) when expressed as mean squares. That factor cancels in their ratio:

F = \frac{SS_{\text{between}}/(k-1)}{SS_{\text{within}}/(N-k)} = \frac{\left(SS_{\text{between}}/\sigma^2\right)/(k-1)} {\left(SS_{\text{within}}/\sigma^2\right)/(N-k)} \sim F_{k-1,N-k} \quad\text{under }H_0.

Theorem: In the one-way fixed-effects model with independent \(N(\mu_i,\sigma^2)\) observations and a common variance, the ANOVA statistic has a central \(F_{k-1,N-k}\) distribution under \(H_0:\mu_1=\cdots=\mu_k\). “Central” indicates that the null model has no group-mean differences.

Worked Examples

Worked Example: Interpreting a Small Balanced ANOVA

A hypothetical laboratory compares a measurement from three independent treatment groups, with three observations per group. The observations are group 1: 2, 4, 6; group 2: 3, 5, 7; and group 3: 4, 6, 8. Assume independent normal observations with a common variance. Calculate the observed F statistic and state its exact null distribution.

Method. Calculate the group means, sums of squares, and degrees of freedom, then apply the null-distribution result. Here \(k=3\), \(n_i=3\), and \(N=9\).

The group means are \(\bar{Y}_{1\cdot}=4\), \(\bar{Y}_{2\cdot}=5\), and \(\bar{Y}_{3\cdot}=6\). The grand mean is \(\bar{Y}_{\cdot\cdot}=(12+15+18)/9=5\). Therefore,

SS_{\text{between}}=3(4-5)^2+3(5-5)^2+3(6-5)^2 =3+0+3=6.

Within group 1, the squared deviations from 4 sum to \((2-4)^2+(4-4)^2+(6-4)^2=4+0+4=8\). The corresponding sum for group 2 is 8, and for group 3 it is also 8. Thus \(SS_{\text{within}}=8+8+8=24\). The degrees of freedom are \(k-1=2\) and \(N-k=6\), so \(MS_{\text{between}}=6/2=3\), \(MS_{\text{within}}=24/6=4\), and \(F=3/4=0.75\).

Under the equal-means null and the stated model, the statistic has distribution \(F_{2,6}\). The observed value \(0.75\) is a calculation from these data; the distributional statement describes the statistic’s sampling distribution if the null and assumptions hold.

Check. The total sum of squares is \(6+24=30\), which also equals the sum of squared deviations of all observations from the grand mean 5. A common mistake is to report \(SS_{\text{between}}/SS_{\text{within}}=6/24\) as the ANOVA statistic. The correction is to divide both sums by their respective degrees of freedom first; the statistic is \(0.75\), not \(0.25\).

Worked Example: Unequal Group Sizes Still Give the Same Form

A hypothetical materials study records a continuous response in three independent groups: group 1 has observations 2 and 4, group 2 has 3, 5, and 7, and group 3 has 4, 6, 8, and 10. Assume independent normal observations with a common variance. Find the observed F statistic and its null degrees of freedom.

Method. Use the group-size-weighted grand mean and the general degrees of freedom \(k-1\) and \(N-k\). The group means are 3, 5, and 7, and the total sample size is \(N=2+3+4=9\). The grand mean is \((6+15+28)/9=49/9\).

The between-group sum of squares is

SS_{\text{between}} =2(3-49/9)^2+3(5-49/9)^2+4(7-49/9)^2 =\frac{968+48+784}{81} =\frac{200}{9}.

The within-group sums of squares are \( (2-3)^2+(4-3)^2=2\), \((3-5)^2+(5-5)^2+(7-5)^2=8\), and \((4-7)^2+(6-7)^2+(8-7)^2+(10-7)^2=20\). Hence \(SS_{\text{within}}=2+8+20=30\).

The degrees of freedom are \(k-1=2\) and \(N-k=6\). Therefore \(MS_{\text{between}}=(200/9)/2=100/9\), \(MS_{\text{within}}=30/6=5\), and \(F=(100/9)/5=20/9\), approximately 2.2222. Under the null, its distribution is \(F_{2,6}\).

Check. The weighted deviations from the grand mean sum to zero: \(2(3-49/9)+3(5-49/9)+4(7-49/9)=0\). A common mistake is to average the three group means equally to obtain the grand mean. That would ignore the unequal group sizes; the correct grand mean is \(49/9\), not 5.

Worked Example: Scaling Chi-Square Realizations

In a hypothetical four-group study, there are 2, 3, 4, and 5 independent observations, for \(N=14\). Assume the one-way normal model with common variance and equal means. For an illustrative realization, suppose \(SS_{\text{between}}/\sigma^2=6\) and \(SS_{\text{within}}/\sigma^2=5\). Find the statistic and identify the null distribution.

Method. Under the null, the first scaled sum has \(k-1=3\) degrees of freedom and the second has \(N-k=10\). They are independent chi-square variables. Divide each by its degrees of freedom:

F=\frac{6/3}{5/10}=\frac{2}{0.5}=4.

Thus the corresponding ratio has a central \(F_{3,10}\) distribution under the null. The unequal group sizes do not change this conclusion: the dimensions of the between-group and within-group subspaces depend on \(k\) and \(N\), not on whether each group has the same size.

Check. The denominator degrees of freedom are \(14-4=10\), not \(14-1=13\), because one group mean is estimated for each of the four groups. The supplied scaled sums are illustrative chi-square realizations, not actual study data; their ratio demonstrates the degrees-of-freedom scaling.

When the Central F Result Does Not Apply

The exact central result is conditional on the model and the null hypothesis. If group means differ, the between-group sum of squares includes a systematic mean-difference component; the ratio is generally noncentral F rather than central F. If the errors are dependent or group variances differ, the independent, common-scale chi-square argument no longer applies as stated. If the errors are not normal, the exact finite-sample chi-square result is also not guaranteed.

This does not mean that every departure makes ANOVA unusable. It means only that the exact central F distribution derived here cannot be asserted from the stated theorem without its assumptions. The sampling design and model assumptions must be checked before attaching that distribution to a computed ratio.

Do not infer the distribution from the ratio’s appearance. A quotient of two mean squares is not automatically F-distributed.
Do not omit independence. Orthogonality of the sums of squares gives independent chi-square quantities under the normal model; dependence in the observations can invalidate that reasoning.
Do not confuse the null with the model. Normality, independence, and common variance describe the error model; equal means is the additional null condition that makes the F distribution central.

Key takeaway: Under independent normal errors with common variance and equal group means, the two ANOVA sums of squares divided by \(\sigma^2\) are independent chi-square variables with \(k-1\) and \(N-k\) degrees of freedom. Their standardized ratio is therefore \(F_{k-1,N-k}\).

Check Your Understanding

For each problem, state the reasoning needed to identify the null distribution or the condition that prevents using it.

A one-way study has 3 groups with 6 observations in each group. Under independent normal errors with common variance and equal means, what is the exact null distribution of the ANOVA statistic? (Answer: \(F_{2,15}\).)
There are 4 groups and 20 observations in total. Under the model assumptions and equal-means null, identify the degrees of freedom for the between-group and within-group chi-square sums. (Answer: 3 and 16 degrees of freedom, respectively.)
Independent scaled sums have hypothetical values \(U=8\) and \(V=12\), with \(U\sim\chi^2_4\) and \(V\sim\chi^2_{18}\) independently. Calculate \((U/4)/(V/18)\) and state its distribution. (Answer: 3; the ratio has distribution \(F_{4,18}\).)
A study has group sizes 3, 5, and 7, with independent normal observations and a common variance. Under equal group means, state the degrees of freedom of the central F distribution. (Answer: \(F_{2,12}\).)
A repeated-measures study treats several measurements from each participant as if they were independent observations in separate groups. Can the exact central F result be justified by the independent-error derivation above? (Answer: No; measurements from the same participant need not be independent, so the stated derivation does not establish a central F distribution.)

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