Why the Direction of the Alternative Matters
A comparison of two population variances is not always a question of whether they differ in either direction. An actuary might instead ask whether the variability of one hypothetical loss process exceeds a specified multiple of the variability of another. That directional question determines which tail of an F distribution is relevant. Using the wrong tail can reverse a test’s conclusion even when the sample variances and degrees of freedom are correct.
The procedures here concern independent samples from normal populations. The population means may be unknown and need not be equal. The key parameter is the ratio of population variances, and the alternative hypothesis states whether that ratio is above or below its specified benchmark.
Set the Hypotheses Before Choosing a Tail
For an upper-tailed test, the research question is whether the first population variance is more than \(\rho_0\) times the second. The hypotheses can be written \(H_0:\sigma_1^2/\sigma_2^2\leq\rho_0\) against \(H_1:\sigma_1^2/\sigma_2^2>\rho_0\). A test stated as \(H_0:\sigma_1^2/\sigma_2^2=\rho_0\) against the same one-sided alternative uses the same boundary distribution and rejection rule.
For a lower-tailed test, the hypotheses are \(H_0:\sigma_1^2/\sigma_2^2\geq\rho_0\) against \(H_1:\sigma_1^2/\sigma_2^2<\rho_0\). In both cases, the sample variances remain in the order \(\text{first sample}/\text{second sample}\). The alternative—not an effort to make the observed statistic large—determines the tail.
Write \(q_p(\nu_1,\nu_2)\) for the \(p\)-quantile of an \(F(\nu_1,\nu_2)\) distribution: \(P(F\leq q_p)=p\). At significance level \(\alpha\), reject in the upper tail when \(T>q_{1-\alpha}(\nu_1,\nu_2)\); reject in the lower tail when \(T<q_\alpha(\nu_1,\nu_2)\). Using a strict or non-strict inequality at a continuous-distribution cutoff does not change the test’s probability of rejection.
| Alternative hypothesis | Rejection region | One-sided p-value |
|---|---|---|
| \(\sigma_1^2/\sigma_2^2>\rho_0\) | \(T>q_{1-\alpha}(\nu_1,\nu_2)\) | \(P(F\geq T_{\mathrm{obs}})\) |
| \(\sigma_1^2/\sigma_2^2<\rho_0\) | \(T<q_{\alpha}(\nu_1,\nu_2)\) | \(P(F\leq T_{\mathrm{obs}})\) |
Here \(\nu_1=n_1-1\) and \(\nu_2=n_2-1\), and the p-value is calculated under \(F(\nu_1,\nu_2)\). A p-value is the probability, assuming the null boundary, of observing a statistic at least as far into the alternative’s tail as the one obtained.
Why the Scaled Statistic Has an F Distribution
Under independent normal sampling, each sample variance satisfies \((n_j-1)S_j^2/\sigma_j^2\sim\chi^2_{n_j-1}\), for \(j=1,2\), and these two chi-square variables are independent. This standard normal-sample result explains why the sample-variance ratio has an F distribution after the population-variance ratio is accounted for.
At the null boundary, \(\sigma_1^2/\sigma_2^2=\rho_0\), so the left side is exactly \(T\). This gives the test’s reference distribution. For the composite upper null, the distribution of \(T\) is scaled by \((\sigma_1^2/\sigma_2^2)/\rho_0\), which is at most 1; the probability of exceeding a fixed upper cutoff is therefore no greater than it is at the boundary. For the composite lower null, that scale is at least 1, so the probability of falling below a fixed lower cutoff is no greater than at the boundary. Thus calibrating the rejection region at equality controls the test’s size for these one-sided null hypotheses.
Worked Tests
Worked Example: Upper-Tailed Test at a Five-Percent Level
Method: Scale the observed variance ratio by the hypothesized ratio, then compare it with the upper-tail F critical value. A hypothetical risk analyst independently samples three observations from each of two normal loss populations. The observed unbiased sample variances are \(S_1^2=25\) and \(S_2^2=1\). Test \(H_0:\sigma_1^2/\sigma_2^2=1\) against \(H_1:\sigma_1^2/\sigma_2^2>1\) at \(\alpha=0.05\).
The specified ratio is \(\rho_0=1\), and the degrees of freedom are \(\nu_1=3-1=2\) and \(\nu_2=3-1=2\). The observed statistic is
For \(F(2,2)\), the cumulative distribution function is \(P(F\leq x)=x/(1+x)\). Its 0.95 quantile solves \(x/(1+x)=0.95\), so \(x=0.95/0.05=19\). The upper-tail rejection region is \(T>19\). Since \(25>19\), reject the null boundary in favor of a variance ratio greater than 1.
The upper-tail p-value is \(P(F\geq25)=1/(1+25)=1/26=0.0385\), to four decimal places. Since \(0.0385<0.05\), this confirms the decision. Common mistake avoided: comparing the raw ratio \(S_1^2/S_2^2\) with a cutoff without first dividing by \(\rho_0\). Here the scaling happens to be 1, but that coincidence does not hold in general.
Worked Example: Lower-Tailed Test and the Lower Critical Value
Method: Keep the sample ratio in its stated order, divide by the null ratio, and use the lower-tail quantile. In a hypothetical comparison, independent normal samples have size 3 each and observed sample variances \(S_1^2=0.25\) and \(S_2^2=4\). Test \(H_0:\sigma_1^2/\sigma_2^2=4\) against \(H_1:\sigma_1^2/\sigma_2^2<4\) at \(\alpha=0.05\).
The observed sample-variance ratio is \(0.25/4=0.0625\). With \(\rho_0=4\), the scaled statistic is
The degrees of freedom are \((2,2)\). For this distribution, the lower 0.05 quantile is found from \(x/(1+x)=0.05\), giving \(x=0.05/0.95=1/19\approx0.05263\). Because \(0.015625<0.05263\), reject the null boundary in favor of a ratio below 4.
The lower-tail p-value is \(0.015625/(1+0.015625)=1/65=0.0154\), to four decimal places. Since this is less than 0.05, the p-value gives the same decision. Common mistake avoided: treating the upper 0.95 critical value 19 as the lower 0.05 critical value. The lower cutoff is \(1/19\), approximately 0.05263—not 19.
Worked Example: A Non-Unit Null Ratio Changes the Statistic
Method: Apply the null ratio as a scale factor before using the F cutoff. For two independent normal samples of size 3, suppose the sample variances are \(S_1^2=15\) and \(S_2^2=1\). Test \(H_0:\sigma_1^2/\sigma_2^2=3\) against \(H_1:\sigma_1^2/\sigma_2^2>3\) at \(\alpha=0.20\).
Here \(\rho_0=3\), the observed raw variance ratio is \(15/1=15\), and both degrees of freedom equal 2. Therefore,
For \(F(2,2)\), the 0.80 quantile solves \(x/(1+x)=0.80\), giving \(x=0.80/0.20=4\). The upper-tail rejection region is \(T>4\); because \(5>4\), reject the null boundary. The upper-tail p-value is \(1/(1+5)=1/6=0.1667\), to four decimal places, which is below 0.20.
Check: The p-value and critical-value methods agree. Common mistake avoided: comparing the raw ratio 15 with the cutoff 4. The test statistic is 5, not 15, because the hypothesized ratio is 3.
Practical Checks and Exam Traps
Before making a decision, verify the sample order, null ratio, degrees of freedom, alternative direction, and tail probability. The first degrees of freedom belong to the variance in the numerator of \(T\); the second belong to the denominator. If the sample order is reversed, both the statistic and degrees of freedom must be adjusted accordingly. Do not reverse the ratio merely to make a statistic greater than 1.
Record which population variance is in the numerator and whether the alternative is greater than or less than \(\rho_0\).
The exact classical test requires independent samples from normal populations. It is not an exact F test for paired or dependent samples, or generally for non-normal populations.
Find \(S_1^2/S_2^2\), then divide by \(\rho_0\). Use \((n_1-1,n_2-1)\) degrees of freedom.
For a greater-than alternative, use \(q_{1-\alpha}\); for a less-than alternative, use \(q_\alpha\). Decide by the rejection region or the matching one-sided p-value.
A useful check for \(F(2,2)\) is its simple quantile formula \(q_p=p/(1-p)\). At \(\alpha=0.05\), the upper critical value is \(q_{0.95}=0.95/0.05=19\), while the lower critical value is \(q_{0.05}=0.05/0.95\approx0.05263\). These values are not interchangeable. More generally, F distributions are not symmetric around 1, so do not obtain a lower cutoff by subtracting an upper cutoff from 1.
These tests are aligned with the SOA VEE Mathematical Statistics syllabus. The assumptions are part of the method, not an optional qualification: normality and independence support the exact reference distribution, and the direction of the alternative determines which tail answers the question.
Check Your Understanding
For each problem, assume independent samples from normal populations unless stated otherwise. Determine the requested test result using the given sample sizes, significance level, and null ratio.
- Samples of size 3 have \(S_1^2=18\) and \(S_2^2=3\). Test \(H_0:\sigma_1^2/\sigma_2^2=2\) against \(H_1:\sigma_1^2/\sigma_2^2>2\) at \(\alpha=0.05\). Give the statistic and decision. (Answer: \(T=3\), \(F(2,2)\); do not reject, since \(3<19\).)
- Samples of size 3 have \(S_1^2=0.5\) and \(S_2^2=2.5\). Test \(H_0:\sigma_1^2/\sigma_2^2=5\) against a lower-tailed alternative at \(\alpha=0.05\). Give the statistic and decision. (Answer: \(T=0.04\); reject, since \(0.04<1/19\approx0.05263\).)
- Samples of size 3 have \(S_1^2=10\) and \(S_2^2=1\). Test a null ratio of 2 against the upper-tailed alternative at \(\alpha=0.20\). Find the statistic, critical value, and decision. (Answer: \(T=5\), upper critical value \(4\); reject.)
- Independent samples have sizes \(n_1=4\) and \(n_2=3\). The alternative is \(\sigma_1^2/\sigma_2^2<1/3\), and the significance level is \(\alpha\). State the statistic and degrees of freedom, and express the rejection rule using a quantile. (Answer: \(T=(S_1^2/S_2^2)/(1/3)\sim F(3,2)\) at the null boundary; reject if \(T<q_\alpha(3,2)\).)
- Samples of size 3 have \(S_1^2=1/16\) and \(S_2^2=1\). Test a null ratio of 1 against the lower alternative at \(\alpha=0.05\). Find the statistic and decision. (Answer: \(T=0.0625\); do not reject, since \(0.0625>0.05263\).)