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189 LearnersLast updated on November 24, 2025

In statistics, the F-test is used to compare the variances of two samples when evaluating hypotheses. This test checks whether the data come from populations that have equal variability. The F-test is used in quality checks, education, finance, and many other sectors to compare variations. In this article, we will explore the F-test and its applications in detail.
In mathematics, the F-test is a statistical test used to compare two variances. We use this test to determine whether the differences between the variances are statistically significant. The criteria for performing an F-test are that the data should come from a population that follows an F-distribution, and the samples compared should be independent of each other. When conducting the hypothesis test, if the findings of the F-test are statistically significant, the null hypothesis can be rejected; otherwise, it cannot.
The F-tests possess certain properties that make them unique:
The F-test is a hypothesis testing tool used to verify if variances are equal. The F-test formula used in various hypothesis tests
Left-Tailed Test:
Null Hypothesis:\(
H_0: \sigma_1^2 = \sigma_2^2
\)
Alternate Hypothesis:\(
H_1: \sigma_1^2 < \sigma_2^2
\)
Decision Criteria: If the F-statistic is less than the critical value, then reject the null hypothesis
Right-Tailed test:
Null Hypothesis: \(
H_0: \sigma_1^2 = \sigma_2^2
\)
Alternate Hypothesis: \(
H_1: \sigma_1^2 < \sigma_2^2
\)
Decision Criteria: Reject H0 if the calculated F-value is greater than the right critical value.
Two-Tailed test:
Null Hypothesis: \(H0: σ21 = σ22\)
Alternate Hypothesis: \(
H_1: \sigma_1^2 \ne \sigma_2^2
\)
Decision Criteria: If the F-test statistic > f test critical value, then the null hypothesis is rejected.
F-test statistic
The F-test statistic for larger samples is mathematically represented as: \(
F = \frac{\sigma_1^2}{\sigma_2^2}
\)
Where:
\(
\sigma_1^2
\) = variance of the first population
\(
\sigma_2^2
\) = variance of the second population
For small samples:
\(
F = \frac{S_1^2}{S_2^2}
\)
Where:
\(
S_1^2
\) = variance of the first sample
\(
S_2^2
\)= variance of the second sample
The F-test is important for comparing the variability of two or more groups and checking if their spreads are similar. It ensures that further statistical tests, like t-tests or ANOVA, are reliable and that conclusions from data are valid.
The F-test helps to compare the variances of two or more groups.
It shows whether the spread or variability of data across groups is similar or different.
Before performing other tests such as the t-test or ANOVA, it is essential to check the assumption of equal variances.
It helps decision-making in experiments and research by indicating whether differences in the data are statistically significant.
Overall, it ensures that statistical test results are reliable and valid.


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The F-test statistic is calculated by dividing one sample variance by the other, and it is always positive since variances cannot be negative. A larger F-value indicates a greater difference between the two variances. This statistic follows an F-distribution determined by the degrees of freedom of each sample and helps assess whether the difference in variances is statistically significant. The F-value is then compared with a critical value or used to calculate a p-value. If the F-value exceeds the critical value, or if the p-value is very small, it suggests that the variances are likely not equal.
When doing an F-test, you use an F-distribution table to find the critical value, which depends on the degrees of freedom and the chosen significance level. For example, at a 5% significance level with certain degrees of freedom, the table tells you the value to compare against your calculated F-value. If your F-value is bigger than the critical value, the result is considered statistically significant; if not, it isn’t. This comparison helps you decide whether the difference between the variances is meaningful.
The F test is an important hypothesis test in statistics. However, students often face some difficulties in understanding the concept. We will now look at a few common mistakes and their solutions:
F-tests are extensively used in various fields to compare variances, such as in finance, research, and more. Here are a few real-life applications of the F-test:
A researcher wants to compare the variability in test scores of two different classes. Class A: Variance = 20, Sample Size = 15 Class B: Variance = 10, Sample Size = 12 Test at the 5% significance level whether the variances are significantly different.
There is no significant difference between the variances.
Hypotheses:
F-ratio (larger variance / smaller variance):
\(
F = \frac{S_1^2}{S_2^2} = \frac{20}{10} = 2
\)
Degrees of Freedom:
Critical Value:
Using F-table at \(xα = 0.05, df1 = 14, df2 = 11\)
Critical value \(
F_{0.025, 14, 11} \approx 3.29
\)
Decision:
Since \(2 < 3.29\), we fail to reject the null hypothesis.
The variances of pizza delivery times in two different cities are being compared. City 1: Sample size = 28, Variance = 38 City 2: Sample size = 25, Variance = 83 At a significance level of 0.05, test whether the variance in delivery times for City 1 is significantly less than that of City 2.
There is sufficient evidence at the 5% significance level to conclude that the delivery times in City 1 have significantly lower variance compared to City 2.
So we reject the null hypothesis.
Set the Hypotheses
This is a left-tailed F-test for comparing variances.
Identify Given Values
Degrees of freedom:
Compute the F-statistic
\( F = \frac{S_1^2}{S_2^2} = \frac{38}{83} \approx 0.4578 \)
Determine the Critical Value
Since this is a left-tailed test at \(α=0.05\), we compute:
\(
F_{0.95, 27, 24} = \frac{1}{F_{0.05, 24, 27}} = \frac{1}{1.93} \approx 0.5181
\)
For a left-tailed test, the critical value\(
F_{0.025, 23, 35} = \frac{1}{F_{0.05, 24, 27}} \approx 0.5181
\)
Step 5: Decision
Compare the calculated F-value with the critical value:
\(0.4578< 0.5181\)
Since the F-statistic falls in the rejection region, we reject the null hypothesis.
A tech company is testing the battery life consistency of two different brands of wireless earbuds. From Brand X, 36 samples were taken, and the variance in battery life was found to be 95 hours². From Brand Y, 24 samples were taken, and the variance was 60 hours². At a 0.05 significance level, test whether there is a significant difference in the variances of battery life between the two brands.
Fail to reject the null hypothesis.
Set the Hypotheses
Identify Given Values
Compute the F-statistic
\(
F = \frac{S_1^2}{S_2^2} = \frac{95}{60} \approx 1.5833
\)
Determine the Critical Value
This is a two-tailed test, so divide\( α=0.05α=0.05\) into two tails:
\(α/2 = 0.025\)
From the F-table:
\(F0.025,35,23 ≈ 2.24(approx)\)
Also check the lower bound:
\(F0.975,35,23 = 1/F0.025,23,35 ≈ 1/2.30 ≈ 0.4348\)
Lower critical value = \(1 / F0.025,23,35 ≈ 0.4348\)
Decision
\( 0.4348< F =1.5833< 2.24\)
Since the F-value falls within the acceptance region, we fail to reject H0.
There is not enough evidence at the 5% level to conclude that the variances of battery life between Brand X and Brand Y are significantly different.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






