Power for Chi-Square Tests of Variance

Why Variance-Test Power Requires a Chi-Square Distribution

A variance test asks whether the spread of a population is consistent with a specified value. In a hypothetical actuarial setting, an analyst might test whether the variability of modeled claim-processing costs exceeds a target level. Planning such a test requires more than setting a significance level: it also requires calculating the chance that the test rejects the null hypothesis when a particular alternative variance is true. That chance is the test’s power.

For a normal population, the sampling distribution of the sample variance is known exactly. This makes it possible to calculate power from chi-square probabilities rather than relying on a large-sample approximation. The calculation depends on the sample size, the null variance, the alternative variance, the significance level, and whether the rejection region is in one tail or both.

Definition: Let \(X_1,\ldots,X_n\) be independent observations from a normal population with mean \(\mu\) and variance \(\sigma^2\). Define the sample variance by \(S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2\), where \(\bar{X}\) is the sample mean. With \(\nu=n-1\) degrees of freedom, \(Q=\nu S^2/\sigma^2\sim\chi^2_\nu\). This exact result requires independent observations from a normal population.

To test \(H_0:\sigma^2=\sigma_0^2\), the test statistic is \(T_0=\nu S^2/\sigma_0^2\). Under the null hypothesis, \(T_0\sim\chi^2_\nu\). Under a specified alternative variance \(\sigma_1^2\), write \(r=\sigma_1^2/\sigma_0^2\). Then \(T_0=rQ\), where \(Q\sim\chi^2_\nu\). The test statistic still uses the null variance in its denominator; the factor \(r\) accounts for the alternative variance.

T_0=\frac{\nu S^2}{\sigma_0^2} =\frac{\sigma_1^2}{\sigma_0^2} \left(\frac{\nu S^2}{\sigma_1^2}\right) =rQ,\qquad Q\sim\chi^2_\nu\ \text{under the alternative.}

Rejection Regions and Power Formulas

Let \(c_p\) denote the \(p\)-quantile of a chi-square distribution with \(\nu\) degrees of freedom, so that \(P(\chi^2_\nu\leq c_p)=p\). An upper-tail test rejects for unusually large sample variance; a lower-tail test rejects for unusually small sample variance. A two-sided test allocates half of the significance level to each tail.

Test alternative	Rejection region for \(T_0\)	Power at variance \(\sigma_1^2\)
\(H_1:\sigma^2>\sigma_0^2\), level \(\alpha\)	\(T_0>c_{1-\alpha}\)	\(P(Q>c_{1-\alpha}/r)\)
\(H_1:\sigma^2<\sigma_0^2\), level \(\alpha\)	\(T_0<c_{\alpha}\)	\(P(Q<c_{\alpha}/r)\)
\(H_1:\sigma^2\ne\sigma_0^2\), level \(\alpha\)	\(T_0<c_{\alpha/2}\) or \(T_0>c_{1-\alpha/2}\)	\(P(Q<c_{\alpha/2}/r)+P(Q>c_{1-\alpha/2}/r)\)

The division by \(r\) follows directly from \(T_0=rQ\). For example, the upper-tail rejection condition \(T_0>c\) becomes \(rQ>c\), or \(Q>c/r\), because \(r>0\). The lower-tail condition is transformed in the same way. For a two-sided test, the two rejection events are disjoint, so their probabilities add.

The power is the probability of rejection under the specified alternative. The probability of failing to reject under that alternative is the Type II error probability, often denoted \(\beta\); therefore, \(\beta=1-\text{power}\). These probabilities depend on the particular alternative variance, not just on the test’s significance level.

Worked Power Calculations

Worked Example: Upper-Tail Test for Increased Variability

A hypothetical insurer studies five independent processing-cost measurements. Assume the measurements are normally distributed, with unknown mean. Test \(H_0:\sigma^2=100\) against \(H_1:\sigma^2>100\) at level \(0.05\). Calculate power if the true variance is \(225\), and then find the Type II error probability. There is no interest-rate or mortality assumption in this variance-testing problem.

Method and assumptions. The distribution is normal with alternative variance \(\sigma_1^2=225\); \(n=5\), so \(\nu=4\). Independence and normality make the chi-square sampling distribution exact. For a one-sided upper-tail test with four degrees of freedom, use \(c_{0.95}=9.488\).

The alternative-to-null variance ratio is \(r=225/100=2.25\). Under the alternative, \(T_0=2.25Q\), where \(Q\sim\chi^2_4\). Thus, the rejection condition \(T_0>9.488\) is equivalent to \(Q>9.488/2.25=4.2169\) (to four decimals).

For four degrees of freedom, the chi-square upper-tail probability at \(x\) is \(e^{-x/2}(1+x/2)\). Therefore,

\begin{aligned} \text{Power} &=P(Q>4.2169)\\ &=e^{-4.2169/2}\left(1+\frac{4.2169}{2}\right)\\ &=e^{-2.10845}(3.10845)\\ &\approx 0.3774. \end{aligned}

The power is \(0.3774\), or \(37.74\%\). The Type II error probability is \(1-0.3774=0.6226\), or \(62.26\%\). As a check, the two probabilities sum to \(1.0000\). Common mistake avoided: the displayed exponential expression is an upper-tail probability, so its value is \(0.3774\), not its complement.

Worked Example: Two-Sided Test for a Different Variance

A hypothetical quality team takes \(n=3\) independent, normally distributed measurements of a process output. The null variance is \(\sigma_0^2=4\), and the team tests \(H_0:\sigma^2=4\) against \(H_1:\sigma^2\ne4\) at level \(0.10\). Find power when the true variance is \(8\). No interest-rate or mortality assumption is relevant.

Here \(\nu=n-1=2\), and \(r=8/4=2\). For a two-sided test with \(\alpha=0.10\), the critical values are \(c_{0.05}=0.1026\) and \(c_{0.95}=5.9915\). Under the alternative, \(T_0=2Q\) for \(Q\sim\chi^2_2\), so the rejection region in terms of \(Q\) is \(Q<0.1026/2=0.0513\) or \(Q>5.9915/2=2.99575\).

For two degrees of freedom, \(P(Q<x)=1-e^{-x/2}\) and \(P(Q>x)=e^{-x/2}\). Calculate the lower-tail contribution:

P(Q<0.0513)=1-e^{-0.0513/2} =1-e^{-0.02565} \approx 0.0253.

Calculate the upper-tail contribution and add the disjoint probabilities:

\begin{aligned} P(Q>2.99575)&=e^{-2.99575/2} =e^{-1.497875}\approx0.2236,\\ \text{Power}&=0.0253+0.2236=0.2489. \end{aligned}

The power is \(0.2489\), or \(24.89\%\), to four decimal places. The lower-tail probability is included even though the specified alternative variance is larger than the null variance: a two-sided test rejects in either tail. Common mistake avoided: using only the upper-tail probability would omit a valid part of the two-sided rejection region.

Worked Example: Lower-Tail Test for Reduced Variability

A hypothetical analyst records three independent, normally distributed measurements of a process spread. The null variance is \(\sigma_0^2=9\), and the test is \(H_0:\sigma^2=9\) against \(H_1:\sigma^2<9\), at level \(0.05\). Find power if the true variance is \(3\). No interest-rate or mortality assumption applies.

There are \(\nu=3-1=2\) degrees of freedom, and \(r=3/9=1/3\). The lower-tail critical value is \(c_{0.05}=0.1026\). Under the alternative, the test rejects when \(T_0<0.1026\), equivalently when \(Q<0.1026/(1/3)=0.3078\), with \(Q\sim\chi^2_2\).

Using the two-degree-of-freedom lower-tail formula gives

\begin{aligned} \text{Power} &=P(Q<0.3078)\\ &=1-e^{-0.3078/2}\\ &=1-e^{-0.1539}\\ &\approx 0.1426. \end{aligned}

The power is \(0.1426\), or \(14.26\%\), to four decimal places. The Type II error probability is \(1-0.1426=0.8574\); the check \(0.1426+0.8574=1.0000\) confirms the complement. Common mistake avoided: for a lower-tail test, power is the probability below the transformed critical value, not the upper-tail probability above it.

Interpreting the Calculation and Avoiding Traps

A power calculation must match the test’s direction and the variance assumed under the alternative. The same test can have different power values at different alternative variances. In particular, the ratio \(r=\sigma_1^2/\sigma_0^2\) changes the cutoff on \(Q\), even though the null critical value itself is unchanged.

Do not use the alternative variance in the test statistic’s denominator. The statistic for this test is \(T_0=\nu S^2/\sigma_0^2\). The alternative variance appears through \(r\), when expressing the distribution under the alternative.
Do not confuse an upper-tail probability with a lower-tail probability. If power is \(P(Q>c)\), its complement \(P(Q\leq c)\) is not power for that upper-tail rejection event.
For a two-sided test, include both tails. Add the lower and upper rejection probabilities; do not double one tail unless the two probabilities are known to be equal under the alternative.
Check the assumptions. The exact chi-square distribution for the sample variance requires independent observations from a normal population. Without normality, this exact calculation need not apply.

This distinction matters when choosing a sample size or assessing whether a variance test can detect a practically important change: a test with a small significance level does not automatically have high power at the alternative of interest. Power quantifies detection probability for that specified variance under the stated model. The calculation is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, use the stated normal model and chi-square critical values. Give probabilities to four decimal places.

A normal population is sampled independently with \(n=3\). For an upper-tail test of \(H_0:\sigma^2=5\) at level \(0.05\), use \(c_{0.95}=5.9915\). Find power if \(\sigma_1^2=10\). (Answer: \(0.2236\))
For \(n=3\), a lower-tail test of \(H_0:\sigma^2=12\) has level \(0.05\), with \(c_{0.05}=0.1026\). Assuming independent normal observations, find power if \(\sigma_1^2=6\). (Answer: \(0.0975\))
For \(n=3\), a two-sided test of \(H_0:\sigma^2=7\) has level \(0.10\), with critical values \(0.1026\) and \(5.9915\). Find power if \(\sigma_1^2=14\), assuming independent normal observations. (Answer: \(0.2489\))
For \(n=5\), an upper-tail test of \(H_0:\sigma^2=40\) has level \(0.05\), with \(c_{0.95}=9.488\). Assuming independent normal observations, find power if \(\sigma_1^2=60\). (Answer: \(0.1761\))
For \(n=3\), a two-sided test of \(H_0:\sigma^2=8\) has level \(0.10\), with critical values \(0.1026\) and \(5.9915\). Assuming independent normal observations, find power if \(\sigma_1^2=24\). (Answer: \(0.3854\))

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