Normal Variance Tests With Known Mean

Why the Known Mean Determines the Test

A normal sample can provide evidence that a population variance is larger or smaller than a specified value. When the population mean is known, the relevant measure of spread is the sum of squared deviations from that known mean. The likelihood-ratio test compares how well two specified variance models explain those deviations, and that comparison determines which tail of the sampling distribution forms the rejection region.

The distinction between a known mean and an estimated mean matters. If the mean is known to be \(\mu_0\), the squared deviations are taken from \(\mu_0\), not from the sample mean. Under the model below, their scaled sum has a chi-square distribution with \(n\) degrees of freedom.

Model and assumptions: The observations \(X_1,\ldots,X_n\) are independent and identically distributed as \(N(\mu_0,\sigma^2)\), where \(\mu_0\) is known and the variance \(\sigma^2\) is the parameter being tested. The two hypotheses specify distinct positive variance values. Define \(Q=\sum_{i=1}^n(X_i-\mu_0)^2\). Under a variance value \(\sigma^2\), \(Q/\sigma^2\sim\chi^2_n\). The result assumes independent observations from the stated normal model; it does not apply unchanged if the mean is unknown or the observations are dependent.

Deriving the Likelihood-Ratio Rule

Write the hypotheses as \(H_0:\sigma^2=\sigma_0^2\) and \(H_1:\sigma^2=\sigma_1^2\), where both variances are specified and positive. For observed values \(x_1,\ldots,x_n\), the likelihood under a variance value \(\sigma^2\) is the product of the normal densities. Since the mean is fixed at \(\mu_0\), the data enter through \(Q=\sum_{i=1}^n(x_i-\mu_0)^2\).

L(\sigma^2) =(2\pi\sigma^2)^{-n/2} \exp\left(-\frac{Q}{2\sigma^2}\right).

For the Neyman–Pearson likelihood ratio, use the alternative likelihood divided by the null likelihood. The lemma says that, for two simple hypotheses, a most powerful test of a given size rejects for sufficiently large values of this ratio. Taking the ratio gives

\Lambda(x) =\frac{L(\sigma_1^2)}{L(\sigma_0^2)} =\left(\frac{\sigma_0^2}{\sigma_1^2}\right)^{n/2} \exp\left\{\frac{Q}{2} \left(\frac{1}{\sigma_0^2}-\frac{1}{\sigma_1^2}\right)\right\}.

The factor before the exponential does not depend on the data. The sign of the coefficient of \(Q\) determines whether \(\Lambda(x)\) increases or decreases as the observed squared deviations increase. If the alternative variance is larger, that coefficient is positive: large \(Q\) favors the alternative. If the alternative variance is smaller, it is negative: small \(Q\) favors the alternative.

Theorem: For \(H_0:\sigma^2=\sigma_0^2\) versus \(H_1:\sigma^2=\sigma_1^2\) in an independent normal sample with known mean \(\mu_0\), the most powerful test rejects for large \(Q\) when \(\sigma_1^2>\sigma_0^2\), and rejects for small \(Q\) when \(\sigma_1^2<\sigma_0^2\). For a test of size \(\alpha\), use the appropriate \(\chi^2_n\) tail under \(H_0\). The statistic has a continuous distribution, so no randomization at the critical boundary is needed.

For precision, let \(q_{n,p}\) denote the \(p\)-quantile of a \(\chi^2_n\) distribution, so \(P(\chi^2_n\le q_{n,p})=p\). If the alternative variance is larger, the size-\(\alpha\) rule is

\text{Reject }H_0 \quad\text{if}\quad Q>\sigma_0^2q_{n,1-\alpha}.

If the alternative variance is smaller, use the lower tail instead:

\text{Reject }H_0 \quad\text{if}\quad Q<\sigma_0^2q_{n,\alpha}.

These cutoffs have size \(\alpha\) because under the null \(Q/\sigma_0^2\sim\chi^2_n\). “Size” here is the probability of rejecting \(H_0\) when \(H_0\) is true. The tail is chosen by the direction of the specified alternative, not by a general preference for upper-tail or two-sided tests.

Worked Tests in Both Directions

Worked Example: Testing for a Larger Variance

A sensor's readings are modeled as independent \(N(5,\sigma^2)\) observations, with the mean known to be 5. Test \(H_0:\sigma^2=4\) against \(H_1:\sigma^2=9\) at size \(0.05\), using the observations \(0,4,6,10\). Identify the most powerful rejection tail and make the decision.

Method. The alternative variance is larger, so the likelihood ratio increases with \(Q\); this is an upper-tail test. The known mean is 5, so calculate squared deviations from 5:

$$ Q=(0-5)^2+(4-5)^2+(6-5)^2+(10-5)^2 =25+1+1+25=52. $$

Here \(n=4\). Under \(H_0\), \(Q/4\sim\chi^2_4\). The 95th percentile is \(q_{4,0.95}\approx9.4877\), so the critical value for \(Q\) is

$$ 4(9.4877)=37.9508. $$

Because \(52>37.9508\), reject \(H_0\) at size \(0.05\). As a likelihood-ratio check, the coefficient of \(Q\) is \(1/2(1/4-1/9)=5/72>0\), confirming that larger squared deviations favor \(H_1\). This avoids the mistake of using a lower-tail rule simply because the null variance is the value being tested.

Worked Example: Testing for a Smaller Variance

A process produces independent \(N(12,\sigma^2)\) measurements, with known mean 12. Test \(H_0:\sigma^2=9\) against \(H_1:\sigma^2=4\) at size \(0.10\), given observations \(10,12,13\).

Method. Since \(4<9\), small deviations from the known mean favor the alternative, so use the lower tail. First calculate

$$ Q=(10-12)^2+(12-12)^2+(13-12)^2 =4+0+1=5. $$

There are \(n=3\) observations. Under \(H_0\), \(Q/9\sim\chi^2_3\). The 10th percentile is \(q_{3,0.10}\approx0.5844\), giving the critical value

$$ 9(0.5844)=5.2596. $$

Since \(5<5.2596\), reject \(H_0\) at size \(0.10\). The result is close to the cutoff, but the comparison is strict and unambiguous. The likelihood-ratio coefficient of \(Q\) is \(1/2(1/9-1/4)=-5/72<0\), which independently confirms the lower-tail direction. A common error is to see “variance test” and automatically use an upper tail; the alternative's variance determines the direction.

Power and the Role of Sample Size

Power is the probability that the test rejects \(H_0\) when the alternative is true. The same chi-square scaling used to set the null cutoff gives power under the specified alternative. For an upper-tail test with critical value \(c\), the power at variance \(\sigma_1^2\) is \(P(Q>c\mid\sigma^2=\sigma_1^2)=P(\chi^2_n>c/\sigma_1^2)\). For a lower-tail test, replace the upper-tail probability with \(P(\chi^2_n<c/\sigma_1^2)\). These expressions rely on the alternative being a specified normal variance and on the mean remaining known and fixed.

Worked Example: Computing Power for a Larger-Variance Test

Independent observations follow \(N(20,\sigma^2)\), with known mean 20. Test \(H_0:\sigma^2=1\) against \(H_1:\sigma^2=4\), using \(n=2\) and size \(0.10\). Find the rejection cutoff, decide using observations \(18,22\), and calculate the power.

Method. The alternative variance is larger, so this is an upper-tail test. For two degrees of freedom, \(q_{2,0.90}\approx4.6052\). Since the null variance is 1, the cutoff for \(Q\) is \(1(4.6052)=4.6052\).

The observed squared deviations are

$$ Q=(18-20)^2+(22-20)^2=4+4=8. $$

Because \(8>4.6052\), reject \(H_0\). To calculate power, scale the same cutoff by the alternative variance: under \(H_1\), \(Q/4\sim\chi^2_2\). Thus

\text{Power} =P\left(\chi^2_2>\frac{4.6052}{4}\right) =P(\chi^2_2>1.1513) =\exp(-1.1513/2) \approx0.5623.

The last equality uses the upper-tail probability \(P(\chi^2_2>x)=\exp(-x/2)\). The power is \(0.5623\) to four decimal places. This is greater than the size \(0.10\), as expected for an alternative that differs from the null. The calculation also avoids confusing the variance ratio with a standard-deviation ratio: the scaling is by the variances 1 and 4.

Common Mistakes and Interpretation

The test is most useful when the direction of the alternative is stated carefully. A claim that the variance is larger and a claim that it is smaller lead to different rejection regions. The test is not automatically two-sided merely because variance can differ in either direction; the hypotheses here specify one simple alternative. A test against a composite alternative, such as all variances unequal to the null value, is a different problem.

Centering at the wrong value. Use \(Q=\sum(X_i-\mu_0)^2\), with the known mean. Replacing \(\mu_0\) with the sample mean changes the statistic and its distribution.
Using \(n-1\) degrees of freedom. Here \(Q/\sigma^2\sim\chi^2_n\), because the mean is known. The \(n-1\) degrees of freedom arise in the usual sample-variance result when the mean is estimated.
Choosing a tail from the null alone. The direction follows from whether \(\sigma_1^2\) is greater or less than \(\sigma_0^2\).
Substituting a standard deviation for a variance. If the hypotheses specify standard deviations, square them before using the variance formulas.
Reversing the ratio without changing the decision rule. The rule here uses \(L_1/L_0\) and rejects for large values. If the ratio is inverted, the inequality must also be reversed.

A useful check is to ask whether the data look more plausible under the alternative: unusually large squared deviations support a larger variance; unusually small squared deviations support a smaller one. The chi-square cutoff then calibrates that intuitive comparison to the required size. This likelihood-ratio construction is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, use the known-mean model and state the relevant rejection direction or decision.

A sample has known mean 8 and tests \(H_0:\sigma^2=4\) against \(H_1:\sigma^2=9\). Which tail of \(Q\) gives the most powerful test? (Answer: Upper tail.)
For \(n=4\), known mean 0, test \(H_0:\sigma^2=16\) against \(H_1:\sigma^2=9\) at size \(0.05\). Using \(q_{4,0.05}\approx0.7107\), give the cutoff for \(Q\) and the rejection direction. (Answer: Reject for \(Q<11.3712\), a lower-tail test.)
With known mean 20, test \(H_0:\sigma^2=1\) against \(H_1:\sigma^2=4\) for \(n=2\) at size \(0.05\). The observations are 23 and 17, and \(q_{2,0.95}\approx5.9915\). Decide. (Answer: \(Q=18\), the cutoff is \(5.9915\), so reject \(H_0\).)
For \(n=3\), known mean 0, test \(H_0:\sigma^2=4\) against \(H_1:\sigma^2=1\) at size \(0.10\). Given \(q_{3,0.10}\approx0.5844\), use observations \(1,0,1\) to decide. (Answer: The lower-tail cutoff is \(4(0.5844)=2.3376\); \(Q=2\), so reject \(H_0\).)
For \(n=2\), observed \(Q=10\), and hypotheses \(H_0:\sigma^2=4\) versus \(H_1:\sigma^2=1\), calculate \(L_1/L_0\). Can this ratio alone determine a test decision without a specified size or critical ratio? (Answer: \(L_1/L_0=4e^{-3.75}\approx0.0941\); no, a decision requires a critical ratio or an equivalent size-based rule.)

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