Tutorials › VEE Mathematical Statistics › Power for F Tests and One-Way ANOVA

VEE Mathematical Statistics · Power and Sample Size · Tutorial 666 of 880

Power for F Tests and One-Way ANOVA

Learn to calculate noncentral F power from the test’s degrees of freedom, critical value, and effect under a specified model.

Intermediate 13 min read

What You'll Learn

  • Identify the numerator and denominator degrees of freedom in an F test.
  • Calculate an ANOVA noncentrality parameter from group sizes, population means, and common variance.
  • Convert a significance level and central F critical value into a power calculation.
  • Interpret the noncentral F distribution as a ratio involving a noncentral chi-square variable.
  • Distinguish noncentral F tests of model restrictions from tests of equality of two population variances.

Why F-Test Power Depends on the Alternative

An F test compares a measure of variation explained by a model with variation left unexplained. Under a null hypothesis, the ratio is calibrated to follow a central F distribution. Under an alternative, the numerator tends to be larger, and the test statistic follows a noncentral F distribution. The amount of noncentrality records how far the alternative is from the null relative to the error variation.

This distinction matters in power calculations: the null distribution determines the rejection threshold, while the alternative distribution determines the probability of crossing it. The same framework applies to one-way ANOVA and to many tests of restrictions in a normal linear model.

Definition: Let \(U\) have a noncentral chi-square distribution with \(\nu_1\) degrees of freedom and noncentrality parameter \(\lambda\), and let \(V\) be an independent central chi-square variable with \(\nu_2\) degrees of freedom. Then $$ \frac{U/\nu_1}{V/\nu_2}\sim F_{\nu_1,\nu_2}(\lambda). $$ When \(\lambda=0\), this is the ordinary central F distribution. For an upper-tail test with significance level \(\alpha\), let \(c\) satisfy \(P(F_{\nu_1,\nu_2}(0)>c)=\alpha\). Its power at the specified alternative is \(P(F_{\nu_1,\nu_2}(\lambda)>c)\).

One-Way ANOVA: Model, Degrees of Freedom, and Noncentrality

Suppose there are \(k\) independent groups. In group \(i\), observations \(Y_{ij}\), for \(j=1,\ldots,n_i\), are independent normal variables with mean \(\mu_i\) and a common variance \(\sigma^2\). Observations from different groups are independent as well. Write \(N=\sum_{i=1}^k n_i\). The one-way ANOVA null hypothesis is \(H_0:\mu_1=\cdots=\mu_k\); the upper-tail F test rejects when the between-group mean square is sufficiently large relative to the within-group mean square.

Theorem: Under the stated normal, independence, and common-variance assumptions, the ANOVA statistic has numerator and denominator degrees of freedom \(\nu_1=k-1\) and \(\nu_2=N-k\). Under specified group means and common variance, it follows a noncentral F distribution with $$ \lambda=\frac{\displaystyle\sum_{i=1}^k n_i(\mu_i-\bar{\mu}_w)^2}{\sigma^2}, \qquad \bar{\mu}_w=\frac{\displaystyle\sum_{i=1}^k n_i\mu_i}{N}. $$ Under the null, all the means equal \(\bar{\mu}_w\), so \(\lambda=0\).

The weighted mean \(\bar{\mu}_w\) is the mean across all observations implied by the group means; it is not generally the unweighted average of the group means. The noncentrality parameter grows when group means are farther apart, when sample sizes give more information about those differences, or when the common within-group variance is smaller. If the equal-variance or independence assumptions are not reasonable, this standard noncentral F calculation need not describe the test’s power.

Why the Noncentral F Appears

The numerator in one-way ANOVA is a sum of squared differences between group sample means and the overall sample mean. Under normal sampling, its scaled form has a chi-square distribution. When the group means differ, those squared differences are centered around different population means rather than a common mean; the resulting chi-square is noncentral. The within-group sum of squares, after division by \(\sigma^2\), remains central chi-square because it measures deviations around each group’s own mean. These two components are independent under the model.

More explicitly, if \(SS_{\text{between}}\) and \(SS_{\text{within}}\) are the between-group and within-group sums of squares, then under the specified alternative, $$ \frac{SS_{\text{between}}}{\sigma^2}\sim\chi^2_{k-1}(\lambda), \qquad \frac{SS_{\text{within}}}{\sigma^2}\sim\chi^2_{N-k}. $$ The two mean squares divide these sums by \(k-1\) and \(N-k\), respectively. Their ratio therefore has the noncentral F distribution defined above. This is why power uses the noncentral F tail, but the cutoff still comes from the central F distribution under \(H_0\).

Calculating ANOVA Power

For a planned upper-tail ANOVA test, specify the significance level, group sizes, common variance, and the group means under the alternative of interest. Then calculate the weighted grand mean, \(\lambda\), and both degrees of freedom. Find the central F critical value for the chosen significance level; finally, evaluate the upper tail of the corresponding noncentral F distribution at that cutoff.

1
Set up the model and alternative.
State the independent normal groups, common variance, sample sizes, and the means at which power is being evaluated.
2
Find degrees of freedom and noncentrality.
Use \(\nu_1=k-1\), \(\nu_2=N-k\), and the weighted sum of squared mean deviations divided by \(\sigma^2\).
3
Find the null critical value.
Choose \(c\) so that the central F upper-tail probability is \(\alpha\).
4
Evaluate power under the alternative.
Calculate \(P(F_{\nu_1,\nu_2}(\lambda)>c)\), using a noncentral F table or a statistical calculator.

Worked Example: Power for Three ANOVA Groups

A hypothetical materials laboratory plans a one-way ANOVA comparing three formulations. Assume independent normal measurements with a common variance of \(\sigma^2=1\). There will be five observations per formulation. For planning, suppose the true means are \(-1,0,1\). At significance level \(\alpha=0.05\), calculate power. Use the central critical value \(c=3.885\) for degrees of freedom \((2,12)\).

Method and assumptions. This is an upper-tail one-way ANOVA power calculation under the common-variance normal model. There are \(k=3\) groups, each of size 5, so \(N=15\).

Degrees of freedom. The numerator degrees of freedom are \(k-1=3-1=2\). The denominator degrees of freedom are \(N-k=15-3=12\).

Weighted mean and noncentrality. Equal group sizes give \(\bar{\mu}_w=(5(-1)+5(0)+5(1))/15=0/15=0\). Thus \(\lambda=[5(-1-0)^2+5(0-0)^2+5(1-0)^2]/1=(5+0+5)/1=10\).

Power. The cutoff is the central F critical value given in the question. Under the specified alternative, the statistic follows \(F_{2,12}(10)\). Hence power is \(P(F_{2,12}(10)>3.885)\), approximately \(0.703\), or about \(70.3\%\). The numerical tail is evaluated from the noncentral F distribution, not the central F distribution.

Check and mistake avoided. The noncentrality is positive because the specified means are not all equal, and both degrees of freedom match the group count and total sample size. Using the central F upper tail at 3.885 would return the null rejection probability, approximately \(0.05\), not power under these means.

Tests of Restrictions as Variance-Ratio Tests

One-way ANOVA is a special case of a broader class of F tests for restrictions in a normal linear model. For example, a model may compare a restricted fit with a full fit. Under the assumptions of independent normal errors with common variance, the statistic is the reduction in residual sum of squares per restriction, divided by the full model’s residual mean square. Under the null restrictions it has a central F distribution; under a specified alternative it has a noncentral F distribution.

For \(q\) linear restrictions \(R\beta=r\), with design matrix \(X\), coefficient vector \(\beta\), and error variance \(\sigma^2\), the noncentrality parameter is $$ \lambda=\frac{(R\beta-r)^\mathsf{T} \left[R(X^\mathsf{T}X)^{-1}R^\mathsf{T}\right]^{-1} (R\beta-r)}{\sigma^2}. $$ The numerator degrees of freedom are \(q\), and the denominator degrees of freedom are \(n-p\), where \(n\) is the number of observations and \(p\) is the number of fitted coefficients in the full model. This form makes the connection to variance ratios explicit: the F statistic compares variation attributable to the tested restrictions with residual variation. The assumptions include a full-rank model and an invertible restriction covariance matrix.

Worked Example: Power for a One-Parameter Regression Test

A hypothetical process study fits a simple linear regression with an intercept to \(n=16\) independent observations. Assume the errors are normal with common variance \(\sigma^2=1\). The eight predictor values equal \(-1\), and the other eight equal \(1\). The planned test is \(H_0:\beta_1=0\) against a nonzero slope; evaluate power when the true slope is \(\beta_1=0.5\). Use the upper-tail critical value \(c=4.600\) at \(\alpha=0.05\), with degrees of freedom \((1,14)\).

Method and assumptions. The usual regression F test for one slope restriction is equivalent to squaring the corresponding t statistic. Under the independent normal, common-variance model, its alternative distribution is noncentral F.

Degrees of freedom. There is one tested restriction, so \(\nu_1=1\). The full model estimates an intercept and a slope, so \(p=2\) and \(\nu_2=n-p=16-2=14\).

Noncentrality. The predictor mean is \(\bar{x}=((-1)(8)+(1)(8))/16=0\). Therefore \(\sum(x_j-\bar{x})^2=8(1)^2+8(1)^2=16\). For the slope test, \(\lambda=\beta_1^2\sum(x_j-\bar{x})^2/\sigma^2=(0.5)^2(16)/1=4\).

Power. Under the specified slope, the statistic follows \(F_{1,14}(4)\). Thus power is \(P(F_{1,14}(4)>4.600)\), approximately \(0.461\), or about \(46.1\%\).

Check and mistake avoided. The alternative slope is nonzero, so \(\lambda=4\), not zero. The value \(4.600\) is the central-F null cutoff; it is not itself a probability or a noncentral critical value.

A Necessary Distinction: Comparing Two Population Variances

Not every statistic called an F test has a noncentral F distribution under its alternative. For two independent normal populations with sample variances \(S_1^2\) and \(S_2^2\), the ratio \(S_1^2/S_2^2\), when scaled by the population variance ratio \(\sigma_1^2/\sigma_2^2\), follows a central F distribution with the two sample-variance degrees of freedom. Under a specified variance ratio, the unscaled statistic is a scaled central F variable. It is not generally noncentral F.

The noncentral F calculation in this tutorial applies when the tested numerator is noncentral chi-square under the alternative, as in ANOVA and normal linear-model tests of restrictions. Applying a noncentrality parameter to the ordinary two-sample test of equality of population variances is a common modeling error. Identify the sampling distribution under the stated alternative before calculating power.

Worked Example: Checking a Proposed Noncentrality for Unequal Groups

A hypothetical study compares two treatment protocols. The group sizes are 5 and 10, the specified means are 8 and 10, and the common variance is 4. Calculate the weighted mean and the ANOVA noncentrality parameter. This calculation concerns the two-group ANOVA test of equal means, not a test of equality of population variances.

Method and assumptions. Assume independent normal observations with the stated common variance. Use the one-way ANOVA formula, with \(k=2\) groups and total sample size \(N=15\).

Weighted mean. The overall mean implied by the alternative is \(\bar{\mu}_w=(5(8)+10(10))/15=(40+100)/15=140/15=28/3\), approximately \(9.3333\).

Noncentrality. The deviations are \(8-28/3=-4/3\) and \(10-28/3=2/3\). The weighted squared-deviation sum is \(5(-4/3)^2+10(2/3)^2=5(16/9)+10(4/9)=80/9+40/9=120/9=40/3\). Dividing by the common variance gives \(\lambda=(40/3)/4=10/3\), approximately \(3.3333\).

Check and mistake avoided. The weighted deviations sum to \(5(-4/3)+10(2/3)=0\), as they must. The noncentrality is \(10/3\), not 5; using an unweighted grand mean or an incorrect weighted sum of squared deviations produces the wrong power.

Numerical Evaluation and Exam Traps

Noncentral F probabilities are usually evaluated with a noncentral F table or statistical software. One useful representation expresses the noncentral F upper tail as a Poisson-weighted sum of central F upper tails: $$ P(F_{\nu_1,\nu_2}(\lambda)>c) = \sum_{j=0}^{\infty} e^{-\lambda/2}\frac{(\lambda/2)^j}{j!} P(F_{\nu_1+2j,\nu_2}(0)>c). $$ The weights are the probabilities for a Poisson variable with mean \(\lambda/2\). This identity explains how the alternative distribution differs from the central one and provides a computational route; in practice, software evaluates the sum or an equivalent numerical method.

Several checks prevent common errors:

  • Use the correct distribution for each stage. Obtain the rejection cutoff from the central F null distribution, then evaluate power with the noncentral F alternative distribution.
  • Keep the ANOVA denominator degrees of freedom straight. They are \(N-k\), not \(N-1\), because each group mean is estimated.
  • Use the weighted mean for unequal group sizes. The grand mean implied by the alternative weights each group mean by its sample size.
  • Divide the weighted squared deviations by the common variance. A larger variance reduces \(\lambda\), and hence generally reduces power for a fixed design and effect.
  • Do not assume every variance-ratio statistic is noncentral F under the alternative. The exact distribution depends on the model and the tested hypothesis.

Power is the probability that a specified design rejects the null under a particular alternative. It is not a property of the test alone: changing the group sizes, mean pattern, error variance, significance level, or model assumptions can change it. The noncentrality parameter gives a compact summary of the alternative’s signal relative to error variation, while the degrees of freedom describe the test and its residual information.

Check Your Understanding

For each problem, identify the model and use the stated critical value where one is supplied. Answers give the key calculations and, when appropriate, the exact noncentral F tail for checking.

  1. A one-way ANOVA has three groups of size 4, common variance 2, and specified means \(1,1,3\). Find the weighted mean, degrees of freedom, and noncentrality parameter. (Answer: \(\bar{\mu}_w=5/3\), \((\nu_1,\nu_2)=(2,9)\), and \(\lambda=16/3\).)
  2. For a two-group ANOVA, each group has size 6, the common variance is 9, and the specified means are 10 and 13. Find \(\lambda\) and the degrees of freedom. (Answer: \(\lambda=3\), \((\nu_1,\nu_2)=(1,10)\).)
  3. A four-group ANOVA has group sizes \(4,4,8,8\), means \(2,2,5,5\), and common variance 9. Find the weighted mean, \(\lambda\), and degrees of freedom. (Answer: \(\bar{\mu}_w=4\), \(\lambda=8/3\), \((\nu_1,\nu_2)=(3,20)\).)
  4. A normal linear-model test has \(q=2\) restrictions, \(n=18\) observations, and \(p=5\) full-model coefficients. Under the null, the test has significance level \(0.05\); under a specified alternative its noncentrality is \(3\). Express its power using the central null cutoff \(c=F^{-1}_{2,13}(0.95)\). (Answer: \((\nu_1,\nu_2)=(2,13)\), and power is \(P(F_{2,13}(3)>c)\).)
  5. Two independent normal samples are used to test equality of population variances. A candidate proposes to calculate power by assigning a noncentrality parameter to \(S_1^2/S_2^2\). Is that the appropriate standard model? (Answer: No. Under a specified variance ratio, the sample-variance ratio has a scaled central F distribution; the ordinary two-sample variance-ratio test is not generally modeled by a noncentral F distribution.)
Clinical Biostats