fTest Formula
A test statistic that has an Fdistribution under the null hypothesis is called an F test. Here, when we use Ftest we would be referring to the FTest to compare two variances. Ftest formula is used to compare the variances (σ1 and σ2) of two different sets of values by dividing them. Ftest was termed so in the honor of Sir Ronald A. Fisher. Let us learn more about the Ftest formula in detail.
What Is the fTest Formula?
A Statistical F Test uses an F Statistic to compare two variances (σ1 and σ2) by dividing them. Since the variance is positive, the result will always be positive. F value thus obtained is used to support or reject the null hypothesis. FTest formula to compare two variances is given as:
F value = σ1^{2}/σ2^{2}
\(\sigma^{2}=\frac{\sum(x\bar{x})^{2}}{n1}\)
where,
 σ^{2} = Variance
 x = Values given in a set of data
 \(\bar{x}\) = Mean of the data
 n = Total number of values
Let's solve a few solved examples based on the f test formula.
Solved Examples Using f test Formula

Example 1: Run an FTest on the following samples using the ftest formula:
Sample1 with variance = 109, sample size = 41.
Sample2 with Variance = 65, sample size = 21.
Solution:
Step1: Firstly, write the hypothesis statements as:
\(\mathrm{H}_{} 0\) : No difference in variances.
H_a: Difference in variances.
Step2: Calculate the Fcritical value. Here take the highest variance as the numerator and the lowest variance as the denominator:
\(\text { F value} =\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}\)
\(=\frac{109}{65}\)
\(F\) Value \(=1.68\)
Step3: Calculating the degrees of freedom:
The degrees of freedom in the table = sample size 1, so for sample 1 it is 40 and for sample 2 it is 21.Step4: Choose the alpha level. Since no alpha level is not given in the question, we will use the standard level of \(0.05\). This would be halved for the test, so use \(0.025\).
Step5: Finding the critical FValue using the FTable. Use the table with 0.025. At alpha, the criticalF for \((40,20)\) at alpha \((0.025)\) is \(2.287\).Step6: Comparing the calculated value to the standard table value. If the calculated value is higher than the table value, then the null hypothesis is rejected. Here, \(1.68 < 2.287\). So, the null hypothesis is not rejected.
Answer: Null hypothesis is not rejected.