Why a Test and an Interval Give the Same Decision
A two-sided test asks whether a specified population variance ratio is consistent with the observed sample variances. A confidence interval asks which population variance ratios remain consistent with those same observations. When the test and interval use matching tail probabilities and the same model, these are two forms of the same calculation: the test rejects a proposed ratio exactly when that ratio falls outside the interval.
The connection is useful in both directions. After constructing an interval, it lets you check a test decision without repeating the tail comparison. Conversely, a test’s critical values can be inverted to give an interval. The equivalence depends on using the same equal-tail convention, degrees of freedom, and assumptions in both procedures.
For these results, assume that each sample is a random sample from its stated normal population and that the two samples are independent. The exact F distribution result is
Here \(F_{\nu_1,\nu_2}\) denotes an F random variable with numerator degrees of freedom \(\nu_1\) and denominator degrees of freedom \(\nu_2\). This exact result is what allows us to compare the observed ratio with critical values. If the samples are not independent or the population distributions are not normal, the stated exact F result need not apply.
Test and Interval Notation
Write \(q_p(\nu_1,\nu_2)\) for the \(p\)-quantile of \(F_{\nu_1,\nu_2}\), defined by \(P(F_{\nu_1,\nu_2}\leq q_p(\nu_1,\nu_2))=p\). In a two-sided equal-tail test at significance level \(\alpha\), each tail has probability \(\alpha/2\). For a proposed value \(\theta_0>0\), the test statistic is \(R/\theta_0\), not \(R\) by itself.
The quantile in the lower endpoint is the upper-tail quantile \(q_{1-\alpha/2}\); the quantile in the upper endpoint is the lower-tail quantile \(q_{\alpha/2}\). This order is a consequence of dividing by a positive random ratio, and is the main detail to keep straight.
Why the Duality Holds
The derivation begins with the test statistic under the null hypothesis. If \(H_0:\theta=\theta_0\) is true, then \(R/\theta_0\sim F_{\nu_1,\nu_2}\). The equal-tail test does not reject when that statistic lies between the two critical values:
To find every value of \(\theta_0\) that would not be rejected, solve this inequality for \(\theta_0\). All quantities are positive, so taking reciprocals reverses the inequality signs. Equivalently, divide through carefully, first using the upper critical value for the lower bound:
The values not rejected are therefore exactly the values in the stated confidence interval. The same result also verifies the interval’s coverage: when the true value is \(\theta\), the event that \(\theta\) lies in the interval is equivalent to the event that \(R/\theta\) lies between the two F quantiles. That event has probability \(1-\alpha\).
This is a duality for the equal-tail test and its matching equal-tail interval. The confidence level must be \(1-\alpha\), and the test must use significance level \(\alpha\). Changing the tail convention or the level changes the corresponding interval.
Worked Examples
For transparent arithmetic, the following examples use samples of size 3 from each population, so \(\nu_1=\nu_2=2\). For these degrees of freedom, the F cumulative distribution function is \(F(x)=x/(1+x)\) for \(x>0\). Solving \(p=x/(1+x)\) gives \(q_p=p/(1-p)\). Thus, at \(\alpha=0.10\), \(q_{0.05}=1/19\) and \(q_{0.95}=19\). Each example uses this same matching 90% interval and 10% test.
Worked Example: A Proposed Ratio Inside the Interval
In a hypothetical instrument study, independent samples of three readings each come from normal populations with variances \(\sigma_1^2\) and \(\sigma_2^2\). The observed sample variances are \(S_1^2=6\) and \(S_2^2=4\). At the 10% significance level, test \(H_0:\theta=2\), where \(\theta=\sigma_1^2/\sigma_2^2\), and construct the matching 90% confidence interval.
Method and assumptions. Use the equal-tail F test and its inverted interval. The independent normal-sample model is stipulated; the degrees of freedom are \(\nu_1=\nu_2=3-1=2\).
Calculate the observed ratio. \(R=S_1^2/S_2^2=6/4=1.5\). Under the null, the test statistic is \(R/\theta_0=1.5/2=0.75\).
Make the test decision. The critical values are \(1/19\approx0.0526\) and \(19\). Since \(0.0526\leq0.75\leq19\), do not reject \(H_0\) at the 10% level.
Construct the interval. The matching interval is \([R/19,\;R/(1/19)]=[1.5/19,\;1.5(19)]\). The lower endpoint is \(3/38\approx0.0789\), and the upper endpoint is \(28.5\). The interval is approximately \([0.0789,28.5000]\).
Check the duality. The proposed value \(2\) lies in this interval, agreeing with the test’s decision not to reject. The common mistake avoided here is comparing \(R=1.5\) directly with F critical values; the test statistic under the proposed ratio is \(R/\theta_0\).
Worked Example: A Proposed Ratio Outside the Interval
A hypothetical quality-control study compares the variances of two independent normal measurement processes. Each process has three observations, and the sample variances are \(S_1^2=80\) and \(S_2^2=2\). Test \(H_0:\theta=2\) against a two-sided alternative at level \(0.10\), then use the matching interval to check the result.
Method and assumptions. Use the equal-tail F test with \(\nu_1=\nu_2=2\), since each sample has size 3. Independence and normality are part of the hypothetical model.
Calculate and test. The observed ratio is \(R=80/2=40\). Under \(H_0\), \(R/\theta_0=40/2=20\). The critical values are \(1/19\approx0.0526\) and \(19\); because \(20>19\), reject \(H_0\) at the 10% level.
Invert the test. The matching interval is \([40/19,\;40(19)]\). Its lower endpoint is approximately \(2.1053\), and its upper endpoint is \(760\), giving \([2.1053,760.0000]\).
Check the duality. The proposed value \(2\) is below the lower endpoint, so the interval excludes it, as the test did. The common mistake avoided is putting the smaller critical value in the lower endpoint’s denominator: doing so would incorrectly produce a lower endpoint of \(40/(1/19)=760\).
Worked Example: One Interval Checks Two Proposals
A hypothetical laboratory compares repeatability in two independent normal populations. Each sample has size 3, with sample variances \(S_1^2=12\) and \(S_2^2=8\). Construct the 90% interval for \(\theta=\sigma_1^2/\sigma_2^2\), then determine the 10% test decision for each proposal: \(\theta_0=0.05\) and \(\theta_0=3\).
Method and assumptions. First form the equal-tail confidence interval, then check each proposed value against it. The independent normal-sample model gives \(\nu_1=\nu_2=2\) and the 10% critical values \(1/19\) and \(19\).
Construct the interval. The observed ratio is \(R=12/8=1.5\). Therefore the interval is \([1.5/19,\;1.5(19)]=[3/38,\;28.5]\), or approximately \([0.0789,28.5000]\).
Check \(\theta_0=0.05\). This value is below \(0.0789\), so it is outside the interval. Directly, \(R/\theta_0=1.5/0.05=30\), which exceeds \(19\); reject this proposal.
Check \(\theta_0=3\). This value lies between \(0.0789\) and \(28.5\). Directly, \(R/\theta_0=1.5/3=0.5\), and \(1/19\leq0.5\leq19\); do not reject this proposal.
Check the duality. Both interval-based conclusions agree with the direct test calculations. This example also shows that a single interval summarizes the decisions for all positive proposed ratios, rather than only one null value.
Common Tail-Reversal Trap
Because F distributions are generally right-skewed, the lower and upper quantiles are not interchangeable. More importantly, the endpoints of the interval use the opposite quantile order from what a reader might guess by looking at the quantile subscripts alone. The lower endpoint divides by the larger quantile; the upper endpoint divides by the smaller one.
A reliable check is that the lower endpoint must be less than or equal to the upper endpoint. Another is to substitute a proposed \(\theta_0\) into \(R/\theta_0\): it should be rejected precisely when this statistic falls below \(q_{\alpha/2}\) or above \(q_{1-\alpha/2}\).
Using the Connection Effectively
For a test-and-interval question, keep the parameter orientation fixed throughout: if \(\theta=\sigma_1^2/\sigma_2^2\), then \(R=S_1^2/S_2^2\) and the numerator degrees of freedom belong to sample 1. Reversing the variance ratio changes both the parameter and the degrees-of-freedom order, so it is not a harmless rearrangement.
Write \(\theta=\sigma_1^2/\sigma_2^2\), \(R=S_1^2/S_2^2\), and confirm the independent normal-sample assumptions.
Use \(\nu_1=n_1-1\), \(\nu_2=n_2-1\), and the same \(\alpha\) for the test and the \(100(1-\alpha)\%\) interval.
Test with \(R/\theta_0\), or build the interval with \(R/q_{1-\alpha/2}\) as the lower endpoint and \(R/q_{\alpha/2}\) as the upper endpoint.
A proposed ratio inside the interval must not be rejected; a proposed ratio outside it must be rejected by the matching equal-tail test.
The equality at an interval endpoint corresponds to equality at a critical value. Since the F distribution is continuous, such a boundary has probability zero under the model; standard test conventions treat rejection as occurring beyond either critical value. The key practical point is not to mix a test at one level with an interval at another.
Check Your Understanding
For each problem, use the stated independent normal-sample model and verify that the test and interval give the same conclusion.
- Two samples each have size 3, and their sample variances are 9 and 3. Find the 90% confidence interval for \(\theta=\sigma_1^2/\sigma_2^2\). (Answer: \([3/19,57]\), approximately \([0.1579,57.0000]\).)
- For samples of size 3 with sample variances 24 and 2, test \(\theta=1\) at level \(0.10\), and state whether the matching 90% interval contains 1. (Answer: \(R=12\), \(R/\theta_0=12>19\); reject, and the interval \([12/19,228]\) excludes 1.)
- For samples of size 3 with sample variances 6 and 3, test \(\theta=0.2\) at level \(0.20\). The corresponding critical values for \(F_{2,2}\) are \(q_{0.10}=1/9\) and \(q_{0.90}=9\). Find the matching interval and decision. (Answer: \(R=2\), interval \([2/9,18]\); \(R/0.2=10>9\), so reject and the interval excludes \(0.2\).)
- For samples of size 3 with sample variances 4 and 4, construct a 95% interval and test \(\theta=0.04\) at level \(0.05\). Use \(q_{0.025}=1/39\) and \(q_{0.975}=39\) for \(F_{2,2}\). (Answer: \(R=1\), interval \([1/39,39]\), approximately \([0.0256,39.0000]\); \(1/0.04=25\) is between the critical values, so do not reject, and the interval contains \(0.04\).)
- For samples of size 3 with sample variances 10 and 2, construct the 90% interval and test both \(\theta=0.25\) and \(\theta=0.5\) at level \(0.10\). (Answer: \(R=5\), interval \([5/19,95]\); \(5/0.25=20>19\), so reject \(0.25\), while \(5/0.5=10\) is between \(1/19\) and \(19\), so do not reject \(0.5\).)