Variance Ratio F Test Calculator
The variance ratio F test asks whether two independent samples come from populations with equal variances. The null hypothesis is σ₁² = σ₂² and the alternative is that the variances differ. The test statistic is the ratio of the two sample variances, and under H₀ it follows an F(n₁−1, n₂−1) distribution; the further the ratio moves away from 1, the stronger the evidence that the variances differ.
Paste two data sets and this calculator returns the F value, the degrees of freedom, the critical F bounds, the p-value, and a confidence interval for the ratio σ₁²/σ₂². It also shows how far the result depends on the normality assumption and runs the Brown–Forsythe (median-centered Levene) test on the same data so that the two can be read side by side. It is used in quality-control work comparing measurement consistency, in laboratory analyses comparing the variability of two methods, and in checking assumptions before a t test.
Variance Ratio F Test Formulas
Hypotheses: H₀: σ₁² = σ₂² H₁: σ₁² ≠ σ₂² Sample variance: s² = Σ(xᵢ − x̄)² / (n − 1) Test statistic: F = s₁² / s₂² Degrees of freedom: df₁ = n₁ − 1 , df₂ = n₂ − 1 Two-tailed p: p = 2 · min[ P(F ≤ F₀) , P(F ≥ F₀) ] Critical region: F < F₍α/₂₎(df₁,df₂) or F > F₍₁₋α/₂₎(df₁,df₂) Confidence interval: F / F₍₁₋α/₂₎ ≤ σ₁²/σ₂² ≤ F / F₍α/₂₎
Because the F distribution is not symmetric, the two-tailed p-value is found by doubling the one-tail probability, and the lower and upper critical values are linked by a reciprocal relation: F₍α/₂₎(df₁,df₂) = 1 / F₍₁₋α/₂₎(df₂,df₁). The test is sensitive to the normality assumption; its robust alternative is the Brown–Forsythe test.
How to Calculate
- Paste the raw data of the two samples into the separate boxes; each needs at least 3 values.
- Choose deliberately which sample goes in the denominator: since F = s₁²/s₂², the order is critical for a one-tailed hypothesis.
- Set the hypothesis direction and the significance level; the confidence interval is always produced two-sided at the 1 − α level.
- Compare the F value with the critical F bounds and check whether the confidence interval for the variance ratio contains 1.
- Read the Shapiro–Wilk summaries in the results section: if normality does not hold, the p-value of the F test is not trustworthy.
- If normality is in doubt, base your decision on the Brown–Forsythe (Levene) row and state clearly in your report which test you used.
Worked Examples
Variability of two measurement methods
The variance of the first method is 4.9000 and that of the second is 26.9444; F = 0.1819 with df = (9, 9). The two-tailed p = 0.018, so the variances differ at α = 0.05: the 95% confidence interval for σ₁²/σ₂², 0.045 – 0.732, does not contain 1. The second method is about 5.50 times more variable.
F statistic (s₁² / s₂²): 0.1819 · Degrees of freedom — numerator (df₁): 9 · Degrees of freedom — denominator (df₂): 9
A case where the variances come out equal
Although the means are very different (50.625 and 60.875), the variances are close: 7.1250 and 8.4107, giving F = 0.8471 with df = (7, 7) and p = 0.832. Homogeneity of variance is not rejected; the 95% confidence interval, 0.170 – 4.231, contains 1 comfortably. This shows that the pooled-variance t test is appropriate for comparing the means.
F statistic (s₁² / s₂²): 0.8471 · Degrees of freedom — numerator (df₁): 7 · Degrees of freedom — denominator (df₂): 7
One-tailed hypothesis — is the new process more consistent?
The variance of the new process is 0.0770 and that of the old one is 0.9579; F = 0.0804 with df = (11, 9). With a left-tail p < 0.001, the variance of the new process is significantly smaller — the old process is about 12.45 times more variable. The critical value is 0.345, and the observed F falls far below it.
F statistic (s₁² / s₂²): 0.0804 · Degrees of freedom — numerator (df₁): 11 · Degrees of freedom — denominator (df₂): 9