# F-test

An F-test is any statistical test in which the test statistic has an F-distribution under the null hypothesis. It is most often used when comparing statistical models that have been fitted to a data set, in order to identify the model that best fits the population from which the data were sampled. Exact F-tests mainly arise when the models have been fitted to the data using least squares. The name was coined by George W. Snedecor, in honour of Sir Ronald A. Fisher. Fisher initially developed the statistic as the variance ratio in the 1920s.[1]

## Common examples of F-tests

Examples of F-tests include:

• The hypothesis that the means of several normally distributed populations, all having the same standard deviation, are equal. This is perhaps the best-known F-test, and plays an important role in the analysis of variance (ANOVA).
• The hypothesis that a proposed regression model fits the data well. See Lack-of-fit sum of squares.
• The hypothesis that a data set in a regression analysis follows the simpler of two proposed linear models that are nested within each other.
• Scheffé's method for multiple comparisons adjustment in linear models.

### F-test of the equality of two variances

Main article: F-test of equality of variances

This F-test is sensitive to non-normality.[2][3] In the analysis of variance (ANOVA), alternative tests include Levene's test, Bartlett's test, and the Brown–Forsythe test. However, when any of these tests are conducted to test the underlying assumption of homoscedasticity (i.e. homogeneity of variance), as a preliminary step to testing for mean effects, there is an increase in the experiment-wise Type I error rate.[4]

## Formula and calculation

Most F-tests arise by considering a decomposition of the variability in a collection of data in terms of sums of squares. The test statistic in an F-test is the ratio of two scaled sums of squares reflecting different sources of variability. These sums of squares are constructed so that the statistic tends to be greater when the null hypothesis is not true. In order for the statistic to follow the F-distribution under the null hypothesis, the sums of squares should be statistically independent, and each should follow a scaled chi-squared distribution. The latter condition is guaranteed if the data values are independent and normally distributed with a common variance.

### Multiple-comparison ANOVA problems

The F-test in one-way analysis of variance is used to assess whether the expected values of a quantitative variable within several pre-defined groups differ from each other. For example, suppose that a medical trial compares four treatments. The ANOVA F-test can be used to assess whether any of the treatments is on average superior, or inferior, to the others versus the null hypothesis that all four treatments yield the same mean response. This is an example of an "omnibus" test, meaning that a single test is performed to detect any of several possible differences. Alternatively, we could carry out pairwise tests among the treatments (for instance, in the medical trial example with four treatments we could carry out six tests among pairs of treatments). The advantage of the ANOVA F-test is that we do not need to pre-specify which treatments are to be compared, and we do not need to adjust for making multiple comparisons. The disadvantage of the ANOVA F-test is that if we reject the null hypothesis, we do not know which treatments can be said to be significantly different from the others — if the F-test is performed at level α we cannot state that the treatment pair with the greatest mean difference is significantly different at level α.

The formula for the one-way ANOVA F-test statistic is

$F = \frac{\text{explained variance}}{\text{unexplained variance}} ,$

or

$F = \frac{\text{between-group variability}}{\text{within-group variability}}.$

The "explained variance", or "between-group variability" is

$\sum_i n_i(\bar{Y}_{i\cdot} - \bar{Y})^2/(K-1)$

where $\bar{Y}_{i\cdot}$ denotes the sample mean in the ith group, ni is the number of observations in the ith group,$\bar{Y}$ denotes the overall mean of the data, and K denotes the number of groups.

The "unexplained variance", or "within-group variability" is

$\sum_{ij} (Y_{ij}-\bar{Y}_{i\cdot})^2/(N-K),$

where Yij is the jth observation in the ith out of K groups and N is the overall sample size. This F-statistic follows the F-distribution with K − 1, N −K degrees of freedom under the null hypothesis. The statistic will be large if the between-group variability is large relative to the within-group variability, which is unlikely to happen if the population means of the groups all have the same value.

Note that when there are only two groups for the one-way ANOVA F-test, F = t2 where t is the Student's t statistic.

### Regression problems

Consider two models, 1 and 2, where model 1 is 'nested' within model 2. Model 1 is the Restricted model, and Model 2 is the Unrestricted one. That is, model 1 has p1 parameters, and model 2 has p2 parameters, where p2 > p1, and for any choice of parameters in model 1, the same regression curve can be achieved by some choice of the parameters of model 2. (We use the convention that any constant parameter in a model is included when counting the parameters. For instance, the simple linear model y = mx + b has p = 2 under this convention.) The model with more parameters will always be able to fit the data at least as well as the model with fewer parameters. Thus typically model 2 will give a better (i.e. lower error) fit to the data than model 1. But one often wants to determine whether model 2 gives a significantly better fit to the data. One approach to this problem is to use an F test.

If there are n data points to estimate parameters of both models from, then one can calculate the F statistic, given by

$F=\frac{\left(\frac{\text{RSS}_1 - \text{RSS}_2 }{p_2 - p_1}\right)}{\left(\frac{\text{RSS}_2}{n - p_2}\right)} ,$

where RSSi is the residual sum of squares of model i. If your regression model has been calculated with weights, then replace RSSi with χ2, the weighted sum of squared residuals. Under the null hypothesis that model 2 does not provide a significantly better fit than model 1, F will have an F distribution, with (p2 − p1, n − p2) degrees of freedom. The null hypothesis is rejected if the F calculated from the data is greater than the critical value of the F-distribution for some desired false-rejection probability (e.g. 0.05). The F-test is a Wald test.

## One-way ANOVA example

Consider an experiment to study the effect of three different levels of a factor on a response (e.g. three levels of a fertilizer on plant growth). If we had 6 observations for each level, we could write the outcome of the experiment in a table like this, where a1, a2, and a3 are the three levels of the factor being studied.

a1 a2 a3
6 8 13
8 12 9
4 9 11
5 11 8
3 6 7
4 8 12

The null hypothesis, denoted H0, for the overall F-test for this experiment would be that all three levels of the factor produce the same response, on average. To calculate the F-ratio:

Step 1: Calculate the mean within each group:

$\begin{align} \overline{Y}_1 & = \frac{1}{6}\sum Y_{1i} = \frac{6 + 8 + 4 + 5 + 3 + 4}{6} = 5 \\ \overline{Y}_2 & = \frac{1}{6}\sum Y_{2i} = \frac{8 + 12 + 9 + 11 + 6 + 8}{6} = 9 \\ \overline{Y}_3 & = \frac{1}{6}\sum Y_{3i} = \frac{13 + 9 + 11 + 8 + 7 + 12}{6} = 10 \end{align}$

Step 2: Calculate the overall mean:

$\overline{Y} = \frac{\sum_i \overline{Y}_i}{a} = \frac{\overline{Y}_1 + \overline{Y}_2 + \overline{Y}_3}{a} = \frac{5 + 9 + 10}{3} = 8$
where a is the number of groups.

Step 3: Calculate the "between-group" sum of squares:

$\begin{align} S_B & = n(\overline{Y}_1-\overline{Y})^2 + n(\overline{Y}_2-\overline{Y})^2 + n(\overline{Y}_3-\overline{Y})^2 \\[8pt] & = 6(5-8)^2 + 6(9-8)^2 + 6(10-8)^2 = 84 \end{align}$

where n is the number of data values per group.

The between-group degrees of freedom is one less than the number of groups

$f_b = 3-1 = 2$

so the between-group mean square value is

$MS_B = 84/2 = 42$

Step 4: Calculate the "within-group" sum of squares. Begin by centering the data in each group

a1 a2 a3
6 − 5 = 1 8 − 9 = -1 13 − 10 = 3
8 − 5 = 3 12 − 9 = 3 9 − 10 = -1
4 − 5 = -1 9 − 9 = 0 11 − 10 = 1
5 − 5 = 0 11 − 9 = 2 8 − 10 = -2
3 − 5 = -2 6 − 9 = -3 7 − 10 = -3
4 − 5 = -1 8 − 9 = -1 12 − 10 = 2

The within-group sum of squares is the sum of squares of all 18 values in this table

$S_W = 1 + 9 + 1 + 0 + 4 + 1 + 1 + 9 + 0 + 4 + 9 + 1 + 9 + 1 + 1 + 4 + 9 + 4 = 68$

The within-group degrees of freedom is

$f_W = a(n-1) = 3(6-1) = 15$

Thus the within-group mean square value is

$MS_W = S_W/f_W = 68/15 \approx 4.5$

Step 5: The F-ratio is

$F = \frac{MS_B}{MS_W} \approx 42/4.5 \approx 9.3$

The critical value is the number that the test statistic must exceed to reject the test. In this case, Fcrit(2,15) = 3.68 at α = 0.05. Since F = 9.3 > 3.68, the results are significant at the 5% significance level. One would reject the null hypothesis, concluding that there is strong evidence that the expected values in the three groups differ. The p-value for this test is 0.002.

After performing the F-test, it is common to carry out some "post-hoc" analysis of the group means. In this case, the first two group means differ by 4 units, the first and third group means differ by 5 units, and the second and third group means differ by only 1 unit. The standard error of each of these differences is $\sqrt{4.5/6 + 4.5/6} = 1.2$. Thus the first group is strongly different from the other groups, as the mean difference is more times the standard error, so we can be highly confident that the population mean of the first group differs from the population means of the other groups. However there is no evidence that the second and third groups have different population means from each other, as their mean difference of one unit is comparable to the standard error.

Note F(x, y) denotes an F-distribution with x degrees of freedom in the numerator and y degrees of freedom in the denominator.

## ANOVA's robustness with respect to Type I errors for departures from population normality

The oneway ANOVA can be generalized to the factorial and multivariate layouts, as well as to the analysis of covariance. None of these F-tests, however, are robust when there are severe violations of the assumption that each population follows the normal distribution, particularly for small alpha levels and unbalanced layouts.[5] Furthermore, if the underlying assumption of homoscedasticity is violated, the Type I error properties degenerate much more severely.[6] For nonparametric alternatives in the factorial layout, see Sawilowsky.[7] For more discussion see ANOVA on ranks.

## References

1. Lomax, Richard G. (2007) Statistical Concepts: A Second Course, p. 10, ISBN 0-8058-5850-4
2. Box, G.E.P. (1953). "Non-Normality and Tests on Variances". Biometrika 40 (3/4): 318–335. JSTOR 2333350.  More than one of `|number=` and `|issue=` specified (help)
3. Markowski, Carol A; Markowski, Edward P. (1990). "Conditions for the Effectiveness of a Preliminary Test of Variance". 44 (4): 322–326. doi:10.2307/2684360. JSTOR 2684360.  More than one of `|number=` and `|issue=` specified (help)
4. Sawilowsky, S. (2002). "Fermat, Schubert, Einstein, and Behrens-Fisher:The Probable Difference Between Two Means When σ12 ≠ σ22". Journal of Modern Applied Statistical Methods, 1(2), 461–472.
5. Blair, R. C. (1981). "A reaction to 'Consequences of failure to meet assumptions underlying the fixed effects analysis of variance and covariance.'" Review of Educational Research, 51, 499-507.
6. Randolf, E. A., & Barcikowski, R. S. (1989, November). "Type I error rate when real study values are used as population parameters in a Monte Carlo study". Paper presented at the 11th annual meeting of the Mid-Western Educational Research Association, Chicago.
7. Sawilowsky, S. (1990). Nonparametric tests of interaction in experimental design. Review of Educational Research, 25(20-59).
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