The F test is a statistical test used most commonly in the ANOVA (Analysis of Variance) and ANCOVA (Analysis of Covariance) procedures to determine whether there is any significant difference between sample means.

Let us understand this by means of an example. Suppose we are interested in studying the effect of four different diets on the weight loss of individuals. Then we shall divide our sample into four groups and assign the four different diets to each of the groups.

As our null hypothesis we shall assume that there is no significant difference between sample means, that is, all four diets have the same effect. We then run the one-way ANOVA procedure and obtain the value of the F statistic.

**Calculation of the F-statistic**:

The formula for the F statistic is,

F=MSST/MSSE where,

MSST= Mean sum of squares which can be calculated by dividing the sum of squares due to treatment by k-1. (Note that k is the number of treatments)

MSSE= Mean sum of squares due to error which can be calculated by dividing the sum of squares due to error by N-k (where N is the total number of observations)

**Decision criteria**:

Then we compare the value of the F statistic with the critical F value. If the F statistic exceeds the critical value then the null hypothesis is rejected and we conclude that the diets have different effects on weight loss. If the F statistic does not exceed the critical value then we conclude that the null hypothesis is true, that is, there is no significant difference between any of the four diets.

**In short the procedure to carry F test in one way ANOVA is as follows:**

- Formulate the null hypothesis that there is no significant difference between sample means.
- Calculate the value of the F statistic using above formula and compare the result with the critical F value.
- If the F statistic exceeds the critical value then null hypothesis is rejected.
- If the F statistic does not exceed the critical value then we conclude that null hypothesis is true.

**Example**: Consider the given data about the annual yield of crops for three different fertilizers. Run the F statistic value test to check whether there is any statistical difference in yield due to the fertilizers.

Fertilizer 1: 12, 2, 3, 4

Fertilizer 2: 5, 7, 8, 6

Fertilizer 3: 12, 14, 5, 19

**Solution**: We calculate the within-group and between-group sum of squares. We then calculate the mean sum of squares and the value of the F statistic.

The critical F value at the 5% level of significance is 4.256. Since the F statistic is less than the critical value, we accept the null hypothesis. This means that there is no significant difference between the fertilizers.

**Assumptions behind F test:**

- The observations are independent.
- The errors are distributed normally.
- The variances of the several normal populations (corresponding to different treatments) are equal. This property is called homoscedasticity.