Why Estimate a Variance Ratio?
A confidence interval for a ratio of variances describes the range of relative variability compatible with two samples. For example, a hypothetical laboratory may compare the variability of measurement errors from two procedures. An interval around one would indicate that equal population variances are compatible with the data; an interval well above one would indicate that the first population variance may be larger. The interval quantifies uncertainty rather than proving that the variances are equal or different.
The exact interval developed here applies to two independent random samples from normal populations. Its construction uses the F distribution and the sample variances. The order of the two populations matters: an interval for the first variance divided by the second is not the same as an interval for the reciprocal ratio.
For a probability \(p\), write \(F_{p;\nu_1,\nu_2}\) for the lower-tail \(p\)-quantile of an F distribution with numerator degrees of freedom \(\nu_1\) and denominator degrees of freedom \(\nu_2\). Thus, if \(W\sim F(\nu_1,\nu_2)\), then \(P(W\leq F_{p;\nu_1,\nu_2})=p\). This is a lower-tail convention: \(F_{0.025;\nu_1,\nu_2}\) is the small quantile, and \(F_{0.975;\nu_1,\nu_2}\) is the large one.
The Confidence-Interval Formula
Under the stated model, the scaled sample-variance ratio has an F distribution:
For a two-sided confidence level \(1-\alpha\), the central probability statement places probability \(\alpha/2\) in each tail. Rearranging that statement gives the interval for \(\theta\).
Why the Endpoints Have This Order
The F quantile statement is initially about a random variable, not directly about the unknown variance ratio. To obtain an interval for the ratio, solve the probability inequality while remembering that every quantity is positive. This positivity is why dividing reverses which quantile appears at each endpoint.
Set \(Q=(S_1^2/S_2^2)/\theta\). Then \(P(F_{\alpha/2;\nu_1,\nu_2}<Q<F_{1-\alpha/2;\nu_1,\nu_2})=1-\alpha\).
For the confidence procedure, the observed value \(R=S_1^2/S_2^2\) is fixed, so the inequality is \(F_{\alpha/2;\nu_1,\nu_2}<R/\theta<F_{1-\alpha/2;\nu_1,\nu_2}\).
The right inequality gives \(\theta<R/F_{\alpha/2;\nu_1,\nu_2}\), while the left gives \(\theta>R/F_{1-\alpha/2;\nu_1,\nu_2}\). Together these are the stated interval.
The lower endpoint divides by the larger F quantile, and the upper endpoint divides by the smaller F quantile. Because the lower quantile is smaller, the resulting lower endpoint cannot exceed the upper endpoint. This is a useful arithmetic check.
In the examples below, each sample has size three, so both degrees of freedom equal two. For these particular degrees of freedom, \(F(2,2)\) has lower-tail quantile \(F_{p;2,2}=p/(1-p)\). For instance, the 0.025 quantile is \(0.025/0.975=1/39\), and the 0.975 quantile is \(0.975/0.025=39\). This convenient identity is specific to \(F(2,2)\); in other cases, use the quantiles for the correct degrees of freedom.
Worked Example: A 95% Interval for Procedure A Versus Procedure B
Model and data. In a hypothetical study, three independent measurements are taken with each of two procedures. Assume each procedure’s measurement errors are normally distributed, and assume the two samples are independent. The sample variance is \(S_A^2=18\) for Procedure A and \(S_B^2=8\) for Procedure B. Construct a 95% confidence interval for \(\sigma_A^2/\sigma_B^2\).
Method. Use the two-sample variance-ratio interval. The target ratio is A divided by B, so the sample ratio must be formed in that same order:
For a 95% interval, \(\alpha=0.05\), so the required lower-tail quantiles are \(F_{0.025;2,2}=1/39\) and \(F_{0.975;2,2}=39\). Substitute them in the interval formula:
Rounding both endpoints to four decimal places gives \((0.0577,\ 87.7500)\). Under the model, this procedure has 95% coverage for the population variance ratio. The interval is very wide, which is unsurprising with only three observations per procedure. Check: the estimate \(2.25\) lies between the endpoints, and the lower endpoint is below the upper endpoint. The common mistake avoided here is dividing the lower endpoint by the smaller F quantile.
Worked Example: Reversing Which Population Is in the Numerator
Model and data. A hypothetical reliability team compares the variation in cycle times from two independent processes. Each sample contains three observations. Assume normal populations and independent samples. Process A has sample variance \(S_A^2=3\), and Process B has sample variance \(S_B^2=12\). Find a 90% interval for \(\sigma_B^2/\sigma_A^2\), not for the reverse ratio.
Method. Write the requested order before calculating. The numerator sample variance is therefore the variance from B:
A 90% interval has \(\alpha=0.10\). The required quantiles are \(F_{0.05;2,2}=1/19\) and \(F_{0.95;2,2}=19\). Thus,
Rounded to four decimal places, the interval is \((0.2105,\ 76.0000)\). It estimates the variance of B relative to the variance of A. For example, the sample ratio of 4 is in the interval, as required by this calculation. Check: if the question had instead requested \(\sigma_A^2/\sigma_B^2\), the corresponding interval would be the reciprocal endpoints in reverse order, \((1/76,\ 1/(4/19))=(0.0132,\ 4.7500)\) after rounding to four decimal places. The common mistake avoided is keeping the sample variances in A/B order when the requested parameter is B/A.
Worked Example: A 99% Interval and the Effect of Tail Choice
Model and data. A hypothetical engineering study records three independent observations from each of two normal-population settings. Assume the samples are independent. The first setting has sample variance \(S_1^2=5\), and the second has sample variance \(S_2^2=20\). Construct a 99% interval for \(\sigma_1^2/\sigma_2^2\).
Method. The observed ratio is \(R=5/20=0.25\), and the degrees of freedom are \(\nu_1=\nu_2=2\). For 99% confidence, \(\alpha=0.01\), so the two tail probabilities are 0.005. The quantiles are \(F_{0.005;2,2}=1/199\) and \(F_{0.995;2,2}=199\). Consequently,
Rounded to six decimal places, the interval is \((0.001256,\ 49.750000)\). A 99% interval is wider than a 95% interval from the same data because it must provide greater coverage. The sample ratio \(0.25\) falls within the interval. Check: the upper endpoint is the sample ratio multiplied by 199, while the lower endpoint is the sample ratio divided by 199. The common mistake avoided is treating 0.005 as the upper-tail quantile probability; the formula uses lower-tail quantiles, and the large quantile has probability 0.995.
Interpreting the Interval and Avoiding Traps
A 95% confidence procedure means that, over repeated sampling under the model, 95% of intervals constructed by this method would contain the fixed population variance ratio. Once an interval has been calculated, its endpoints are fixed. The confidence statement does not mean that there is a 95% probability that this particular fixed parameter lies between these particular endpoints.
A ratio interval is also scale-free: both variances have squared measurement units, and those units cancel. Its interpretation is relative. If an interval is \((L,U)\) for \(\sigma_1^2/\sigma_2^2\), it compares the variability of population 1 to that of population 2. An interval containing 1 includes equal variances among the values compatible with the interval; an interval excluding 1 does not. That observation is an interpretation of the interval, not a substitute for checking that the model assumptions apply.
- Do not swap the population order. Write the requested ratio first, then form the sample ratio in the same order.
- Do not swap the degrees of freedom casually. The first sample variance supplies the numerator degrees of freedom; the second supplies the denominator degrees of freedom.
- Do not use the same tail quantile at both endpoints. For a central interval, use \(1-\alpha/2\) in the lower endpoint and \(\alpha/2\) in the upper endpoint.
- Do not describe the confidence level as a probability about a fixed parameter. It describes the long-run coverage of the interval procedure under the model.
- Check positivity and ordering. Both endpoints must be positive, and the lower endpoint must be less than the upper endpoint.
The exactness of this method depends on the model: the samples must be independent random samples from normal populations, with the stated sample sizes and sample variances. If those conditions do not hold, this exact F-based interval is not justified by the result used here. Selecting a method appropriate to a different model is a separate question.
Check Your Understanding
For each problem, assume independent samples from normal populations and use the stated sample sizes and variances. The F quantiles below follow \(F_{p;2,2}=p/(1-p)\).
- Two samples each have size 3 and sample variances \(S_1^2=6\) and \(S_2^2=2\). Find an 80% interval for \(\sigma_1^2/\sigma_2^2\). (Answer: \((0.3333,\ 27.0000)\))
- Two samples each have size 3 and sample variances \(S_1^2=8\) and \(S_2^2=4\). Find a 95% interval for \(\sigma_1^2/\sigma_2^2\). (Answer: \((0.0513,\ 78.0000)\))
- Two samples each have size 3. The sample variance for population A is 2.5 and for population B is 10. Find a 90% interval for \(\sigma_A^2/\sigma_B^2\). (Answer: \((0.0132,\ 4.7500)\))
- Two samples each have size 3 and sample variances \(S_1^2=15\) and \(S_2^2=5\). Find a 99% interval for \(\sigma_1^2/\sigma_2^2\), rounding endpoints to four decimals. (Answer: \((0.0151,\ 597.0000)\))
- Two samples each have size 3. Population A has sample variance 4 and population B has sample variance 16. Find a 95% interval for \(\sigma_B^2/\sigma_A^2\), rounding endpoints to four decimals. (Answer: \((0.1026,\ 156.0000)\))