Sample Size And Precision For Variance Intervals

Why Sample Size Matters for Variance Intervals

A confidence interval for a population variance is built from chi-square quantiles. Its endpoints depend both on the observed sample variance and on the degrees of freedom, which are determined by the sample size. Planning the sample size therefore means asking how much precision the interval is expected to provide, not simply choosing a large number of observations without reference to a target.

There are two useful ways to express precision. Absolute width is the upper endpoint minus the lower endpoint, in squared measurement units. Relative width compares the endpoints, often through their ratio. Absolute width depends on the scale of the variance; the endpoint ratio does not depend on the observed sample variance. A planning calculation must say which meaning of precision is intended.

Model and notation: Let \(X_1,\ldots,X_n\) be independent observations from a normal population with mean \(\mu\) and variance \(\sigma^2\), where both parameters are unknown. Let \(S^2\) be the usual unbiased sample variance and let \(\nu=n-1\). Write \(q_{p,\nu}\) for the lower-tail \(p\) quantile of a chi-square distribution with \(\nu\) degrees of freedom, so \(P(\chi^2_\nu\le q_{p,\nu})=p\). For confidence level \(1-\alpha\), the exact two-sided interval for \(\sigma^2\) is \(\left(\nu S^2/q_{1-\alpha/2,\nu},\ \nu S^2/q_{\alpha/2,\nu}\right)\).

The exact interval requires independent observations from a normal population. Its precision statements are conditional on that model. If the observations are not normally distributed, the chi-square pivot used to obtain this interval is generally not exact; a larger sample alone does not make the stated exact coverage automatically valid.

How to Measure the Interval's Precision

For observed \(S^2=s^2\), the interval width is

W_\nu = \nu s^2\left(\frac{1}{q_{\alpha/2,\nu}}-\frac{1}{q_{1-\alpha/2,\nu}}\right).

The lower endpoint uses the larger chi-square quantile, while the upper endpoint uses the smaller one. As the degrees of freedom increase, the chi-square distribution becomes relatively more concentrated around its mean \(\nu\). For a fixed confidence level, this usually makes the interval narrower relative to the variance being estimated. Increasing the confidence level has the opposite effect: more probability must be covered, so the interval generally widens.

The ratio of the upper endpoint to the lower endpoint is

R_\nu = \frac{\nu s^2/q_{\alpha/2,\nu}}{\nu s^2/q_{1-\alpha/2,\nu}} = \frac{q_{1-\alpha/2,\nu}}{q_{\alpha/2,\nu}}.

This ratio cancels both \(\nu s^2\) and the units of measurement. A requirement such as \(R_\nu\le 2.50\) is therefore a relative-precision requirement. It does not say that the interval has a particular width in the original units.

Worked Example: Comparing 90% and 95% Precision

A hypothetical laboratory measures a continuous quality characteristic on \(n=10\) independent items. Assume the measurements are normally distributed, and suppose the observed sample variance is \(s^2=16\) squared units. Compare the widths of the exact 90% and 95% intervals for the population variance. Use \(q_{0.05,9}=3.3251\), \(q_{0.95,9}=16.9190\), \(q_{0.025,9}=2.7004\), and \(q_{0.975,9}=19.0228\).

Method. Use \(\nu=n-1=9\), multiply each reciprocal-quantile difference by \(\nu s^2=9(16)=144\), and compare the resulting widths. This calculation avoids comparing intervals from different sample sizes or confusing a higher confidence level with a narrower interval.

For the 90% interval, \(\alpha=0.10\), so the lower endpoint is \(144/16.9190\) and the upper endpoint is \(144/3.3251\). Thus the endpoints are approximately \(8.51\) and \(43.31\), and the width calculated from the unrounded endpoints is

144\left(\frac{1}{3.3251}-\frac{1}{16.9190}\right) \approx 34.80.

For the 95% interval, \(\alpha=0.05\). The endpoints are \(144/19.0228\approx 7.57\) and \(144/2.7004\approx 53.33\). Its width, again calculated before rounding endpoints, is

144\left(\frac{1}{2.7004}-\frac{1}{19.0228}\right) \approx 45.76.

The 95% interval is wider: \(45.76>34.80\). This is expected because the higher confidence level requires wider coverage. As a check, the 90% interval's upper endpoint exceeds its lower endpoint, and both intervals are centered in neither the ordinary arithmetic nor the reciprocal-quantile sense; they need not be symmetric about \(s^2=16\).

Why the Chi-Square Formula Gives These Endpoints

Under the stated independent normal model, the standardized sample variance has a chi-square distribution: \(\nu S^2/\sigma^2\sim\chi^2_\nu\). The central probability statement is

P\left(q_{\alpha/2,\nu}\le \frac{\nu S^2}{\sigma^2}\le q_{1-\alpha/2,\nu}\right)=1-\alpha.

Every quantity in the event is positive. Dividing the first inequality by its positive quantile and rearranging gives \(\sigma^2\le \nu S^2/q_{\alpha/2,\nu}\). Rearranging the second gives \(\sigma^2\ge \nu S^2/q_{1-\alpha/2,\nu}\). Combining them yields the two-sided interval in the definition. The reversal in the endpoint order occurs because \(\sigma^2\) is in the denominator of the pivot: the smaller chi-square quantile produces the larger variance endpoint.

This derivation also shows why the sample size enters through \(\nu=n-1\), not \(n\). One degree of freedom is used in estimating the unknown mean when calculating the sample variance. The formula is exact under normality; it is not a distribution-free interval for any population variance.

Choosing a Sample Size for Relative Precision

Suppose the design requirement is that the interval's upper endpoint be no more than \(c\) times its lower endpoint, where \(c>1\). Because the endpoint ratio is independent of \(s^2\), a candidate sample size can be checked directly from the chi-square quantiles:

\frac{q_{1-\alpha/2,n-1}}{q_{\alpha/2,n-1}}\le c.

For each candidate integer \(n\ge2\), use \(\nu=n-1\). The suitable design is the smallest \(n\) that satisfies the inequality. In practice, one obtains quantiles from a chi-square table or software and checks integer degrees of freedom. The criterion must be checked at the intended confidence level: a higher confidence level typically requires a larger sample size to attain the same endpoint ratio.

Worked Example: Finding the Minimum Sample Size for an Endpoint Ratio

A hypothetical study plans a central 90% interval for a normal population variance. The design requirement is that the upper endpoint be at most 2.50 times the lower endpoint. Quantile calculations give \(q_{0.05,26}=15.379\), \(q_{0.95,26}=38.885\), \(q_{0.05,27}=16.151\), and \(q_{0.95,27}=40.113\). Find the smallest sample size meeting the requirement.

Method. For a central 90% interval, the endpoint ratio is \(q_{0.95,\nu}/q_{0.05,\nu}\). Check candidate degrees of freedom in increasing order, remembering that \(n=\nu+1\).

At \(\nu=26\), the ratio is

\frac{38.885}{15.379}\approx 2.529.

This exceeds 2.50, so \(n=27\) does not meet the target. At \(\nu=27\), the ratio is

\frac{40.113}{16.151}\approx 2.484.

This is below 2.50, so \(n=28\) meets the target. Checking the preceding integer degrees of freedom shows that 27 is the first qualifying degree of freedom; therefore the smallest suitable sample size is \(n=28\). The ratio is dimensionless and does not require a guess for the population variance. The common error of using \(n=\nu\) would report a sample size one too small.

Choosing a Sample Size for Absolute Width

An absolute-width requirement, such as \(W\le W_0\), is different. The width formula contains \(s^2\), which is not known when the study is being planned. A practical planning calculation therefore specifies a plausible planning variance \(s_p^2\) and substitutes it for the future sample variance. This gives a planning estimate of width, not a guarantee: the realized sample variance is random and may be larger or smaller than the planning value.

Worked Example: Meeting a Planned Absolute-Width Target

A hypothetical instrument study will estimate the variance of a normally distributed measurement. The planning variance is \(s_p^2=25\) squared units, the desired central confidence level is 90%, and the target interval width is at most 50 squared units. Compare \(n=10\) and \(n=11\). Use \(q_{0.05,9}=3.3251\), \(q_{0.95,9}=16.9190\), \(q_{0.05,10}=3.9403\), and \(q_{0.95,10}=18.3070\).

Method. Substitute \(s_p^2=25\) into the interval-width formula for each candidate sample size. The 90% width uses the difference \(1/q_{0.05,\nu}-1/q_{0.95,\nu}\).

For \(n=10\), \(\nu=9\), and \(\nu s_p^2=9(25)=225\). The planned width is

225\left(\frac{1}{3.3251}-\frac{1}{16.9190}\right) \approx 54.37.

Since \(54.37>50\), this candidate does not meet the target. For \(n=11\), \(\nu=10\), and \(\nu s_p^2=10(25)=250\). The planned width is

250\left(\frac{1}{3.9403}-\frac{1}{18.3070}\right) \approx 49.79.

Since \(49.79\le50\), \(n=11\) meets the planned target. Among these two adjacent candidates it is the smaller qualifying sample size. This conclusion is conditional on the planning variance and normal model; it does not promise that the realized interval from every sample of size 11 will have width at most 50.

Common Mistakes and Practical Interpretation

Precision calculations are especially vulnerable to errors in degrees of freedom, tail labels, and rounding. Keep the quantile definition explicit, because some tables use upper-tail probabilities rather than the lower-tail convention used here. If a table reports the upper-tail area, translate it before substituting a value.

Using \(n\) degrees of freedom. The sample variance interval uses \(\nu=n-1\). After finding the required degrees of freedom, add one to obtain the sample size.
Reversing the endpoints. The lower variance endpoint divides by \(q_{1-\alpha/2,\nu}\), while the upper endpoint divides by \(q_{\alpha/2,\nu}\).
Confusing confidence level and width. At a fixed sample size, increasing the confidence level generally widens the interval. It does not make the estimate more precise in the sense of a shorter interval.
Treating a planning width as guaranteed. An absolute-width calculation based on \(s_p^2\) is a design estimate. Actual width varies with the observed \(S^2\).
Rounding too early. Calculate endpoint differences or widths from unrounded quantiles and endpoints, then round the final reported result. Subtracting already rounded endpoints can change the reported width.

For exam-style calculations, state the normality and independence assumptions, identify the confidence level, calculate \(\nu=n-1\), and write the precision criterion before looking up quantiles. This makes the distinction between an absolute-width target and an endpoint-ratio target clear. These procedures are aligned with the SOA VEE Mathematical Statistics syllabus.

Key takeaway: For an exact normal-variance interval, sample size acts through the chi-square degrees of freedom. Use the quantile ratio for relative precision, and use a clearly stated planning variance for an estimated absolute width.

Check Your Understanding

For each problem, use the stated normal-model assumptions and lower-tail chi-square quantiles. Round final endpoints and widths to two decimals and ratios to three decimals.

A hypothetical calibration study has \(n=12\), \(s^2=9\), and seeks a central 90% interval. Use \(q_{0.05,11}=4.575\) and \(q_{0.95,11}=19.675\). Find the endpoints and width. (Answer: endpoints approximately \(5.03\) and \(21.64\); width approximately \(16.61\).)
A hypothetical production sample has \(n=10\) and \(s^2=4\). For a central 95% interval, use \(q_{0.025,9}=2.7004\) and \(q_{0.975,9}=19.0228\). Calculate the variance endpoints and explain which quantile gives the upper endpoint. (Answer: approximately \(1.89\) and \(13.33\); the smaller quantile, \(2.7004\), gives the upper endpoint.)
A central 90% variance interval must have endpoint ratio no greater than 2.90. For \(\nu=19\), use \(q_{0.05,19}=10.117\) and \(q_{0.95,19}=30.144\); for \(\nu=20\), use \(q_{0.05,20}=10.851\) and \(q_{0.95,20}=31.410\). Which of these two consecutive sample sizes first meets the target? (Answer: \(n=21\), since the ratios are approximately \(2.979\) at \(\nu=19\) and \(2.895\) at \(\nu=20\).)
For a central 90% interval with \(n=10\), suppose the planning variance is 10 squared units. Use \(q_{0.05,9}=3.3251\) and \(q_{0.95,9}=16.9190\). Find the planned width. (Answer: approximately \(21.75\) squared units.)
Two studies use the same normal model and confidence level, but one has a larger sample size. State how the endpoint ratio is expected to change, and whether the realized absolute width must decrease. (Answer: the endpoint ratio generally decreases as degrees of freedom increase; realized absolute width is not guaranteed to decrease because the observed sample variance also changes.)

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