Problem (f test with two samples)
An f test with two samples compares two population variances by using the ratio of sample variances. Consider two independent samples summarized by:
| Sample | Size \(n\) | Sample variance \(s^2\) |
|---|---|---|
| 1 | 12 | 25 |
| 2 | 10 | 9 |
Goal: Test whether the population variances are equal at \(\alpha=0.05\).
Assumptions for the classical two-sample F test: the two samples are independent, and each population is approximately normal.
\(H_0:\sigma_1^2=\sigma_2^2\)
\(H_1:\sigma_1^2\ne\sigma_2^2\)
Step 1: Form the F statistic
For a two-sided test, the ratio is commonly built with the larger sample variance in the numerator so that \(F \ge 1\).
\[ F=\frac{s_{\max}^2}{s_{\min}^2}=\frac{25}{9}=2.777\ldots \approx 2.78. \]
Step 2: Degrees of freedom
The F distribution degrees of freedom follow the variances used in the ratio: \[ df_1=n_{\text{(numerator)}}-1=12-1=11,\qquad df_2=n_{\text{(denominator)}}-1=10-1=9. \]
Under \(H_0\), the statistic has distribution \(F \sim F(11,9)\).
Step 3: p-value for a two-sided variance test
With \(F \ge 1\) constructed using the larger variance on top, the two-sided p-value is computed by doubling the right-tail probability: \[ p = 2\cdot P\!\left(F(11,9)\ge 2.78\right). \]
Numerically, \[ P\!\left(F(11,9)\ge 2.78\right)\approx 0.06799 \quad\Rightarrow\quad p \approx 2\cdot 0.06799 = 0.13598 \approx 0.136. \]
Step 4: Decision at \(\alpha=0.05\)
Since \(p \approx 0.136 > 0.05\), the null hypothesis is not rejected.
Conclusion: The data do not provide sufficient evidence (at the 5% level) that the population variances differ.
Visualization: F distribution and the observed variance ratio
The curve is the \(F(11,9)\) density. The solid vertical line marks the observed \(F\). The dashed line marks the upper critical value for a two-sided \(\alpha=0.05\) test (\(\alpha/2\) in the right tail).
Reference: critical-value form (optional check)
For a two-sided test with \(\alpha=0.05\) and the larger variance placed in the numerator, rejection occurs when \(F\) exceeds the upper critical value \(F_{1-\alpha/2}(df_1,df_2)\).
\[ F_{0.975}(11,9)\approx 3.912,\quad \text{and } 2.78 < 3.912 \Rightarrow \text{do not reject } H_0. \]