Fundamentals of Statistics contains material of various lectures and courses of H. Lohninger on statistics, data analysis and chemometrics......click here for more.

## Two-Sample F-Test

In order to compare two methods, it is often important to know whether the variabilities for both methods are the same. In order to compare two variances v1, and v2, one has to calculate the ratio of the two variances. This ratio is called the F-statistic (in honor of R.A. Fisher) and follows an F distribution:

F = v1/v2
The null hypothesis H0 assumes that the variances are equal and the ratio F is therefore one. The alternative hypothesis H1 assumes that v1 and v2 are different, and that the ratio deviates from unity. The F-test is based on two assumptions: (1) the samples are normally distributed, and (2) the samples are independent of each other. If these assumptions are fulfilled and H0 is true, the statistic F follows an F-distribution. The following is a decision table for the application of an F-test. In order to calculate the F-quantile, or an associated probability, refer to an F table, or to the distribution calculator.

Remarks:

• When the normality assumption is not fulfilled, one should use a non-parametric method. In general the F-test is more sensitive to deviations from normality than the t-test.
• The F-test can be used to check the equal variance assumption needed for the two sample t-test, but the non-rejection of H0 does not imply that the assumption (of equal variance) is valid, since the probability of the type 2 error is unknown.

 Example: Suppose you have two series of measurements, one with 10 observations, and one with 13 observations. The variance of the first series is 0.88, and the variance of the second series is 1.79. Is the variance of the second series significantly larger than the variance of the first series (at a significance level of 0.05)? In order to check this, we assume the null hypothesis that the variance of the second series is not larger than the variance of the first series. The alternative hypothesis would be that the second variance is indeed larger than the first one. Next we have to calculate the F statistic: F = 1.79/0.88 = 2.034. Now we can compare the F statistic with the critical value at a 5 percent level of significance. By using the distribution calculator we find a critical value of 3.073. Since F is only 2.034 we cannot reject our null hypothesis (the second variance is not significantly larger than the first one).