Unknown Variance and Noncentral T Power

Why Unknown Variance Changes the Power Calculation

When a population standard deviation is known, a one-sample test of a mean can be based on a standard normal statistic. In many applications, however, the standard deviation is unknown and must be estimated from the same sample. The resulting test statistic has a random denominator, so its distribution under an alternative is not generally normal. It is a noncentral t distribution.

This distinction matters in power calculations. A normal critical value paired with a normal alternative calculation ignores the extra variation from estimating the standard deviation. For a one-sample t test, power must account for both the shift in the sample mean and the randomness of the sample standard deviation.

Model and assumptions: Let \(X_1,\ldots,X_n\) be an independent random sample from \(N(\mu,\sigma^2)\), where \(\sigma>0\) is unknown. Define the sample mean \(\bar X\) and sample standard deviation \(S\) in the usual way. The one-sample t statistic for testing \(H_0:\mu=\mu_0\) is \(T=(\bar X-\mu_0)/(S/\sqrt n)\). The exact distribution results below require independent observations from a normal population.

The hypothesis test is calibrated at the null value \(\mu_0\), but power is calculated at a specified true mean \(\mu\). Under that alternative, the statistic’s distribution depends on how far \(\mu\) is from \(\mu_0\), measured in units of the true standard error \(\sigma/\sqrt n\).

Deriving the Noncentral t Distribution

Let \(\nu=n-1\), the degrees of freedom associated with the sample variance. Under the stated normal model,

Z=\frac{\bar X-\mu}{\sigma/\sqrt n}\sim N(0,1), \qquad V=\frac{\nu S^2}{\sigma^2}\sim \chi^2_{\nu},

and \(Z\) and \(V\) are independent. These are the standard sampling-distribution results for a normal sample. Write the t statistic as a standardized random numerator divided by a standardized random denominator:

\begin{aligned} T &=\frac{\bar X-\mu_0}{S/\sqrt n}\\ &=\frac{(\bar X-\mu)/(\sigma/\sqrt n)+(\mu-\mu_0)/(\sigma/\sqrt n)} {\sqrt{S^2/\sigma^2}}\\ &=\frac{Z+\lambda}{\sqrt{V/\nu}}, \qquad \lambda=\frac{\sqrt n(\mu-\mu_0)}{\sigma}. \end{aligned}

The final equality follows because \(S^2/\sigma^2=V/\nu\). It shows why the distribution is noncentral t: the numerator is a standard normal variable shifted by \(\lambda\), and the denominator is the square root of an independent chi-square variable divided by its degrees of freedom.

Definition: A noncentral t random variable with \(\nu\) degrees of freedom and noncentrality parameter \(\lambda\) has the representation \(T=(Z+\lambda)/\sqrt{V/\nu}\), where \(Z\sim N(0,1)\), \(V\sim\chi^2_\nu\), and \(Z\) and \(V\) are independent. For the one-sample mean statistic under true mean \(\mu\), \(\nu=n-1\) and \(\lambda=\sqrt n(\mu-\mu_0)/\sigma\).

The sign of \(\lambda\) is the sign of \(\mu-\mu_0\). A positive shift moves probability toward an upper-tail rejection region; a negative shift moves it toward a lower-tail rejection region. The magnitude \(|\lambda|\) measures the standardized shift, but it does not by itself determine power: the test’s rejection region must also match the direction of the shift.

The exact noncentral t result does not apply if the observations are not independent normal observations. In other settings a t procedure may sometimes be used as an approximation, but its power is not given exactly by this noncentral t model. The significance level, the true \(\sigma\), and the alternative mean must also be specified to calculate power.

Critical Values and Power

Let \(F_{\nu,\lambda}(x)\) denote the cumulative distribution function of a noncentral t variable with degrees of freedom \(\nu\) and noncentrality parameter \(\lambda\). The test’s critical values are chosen under the null, where \(\lambda=0\); power is evaluated under the alternative, using the alternative’s \(\lambda\).

Test	Rejection region at significance level \(\alpha\)	Power at true mean \(\mu\)
Upper-tailed, \(H_1:\mu>\mu_0\)	\(T>t_{1-\alpha,\nu}\)	\(1-F_{\nu,\lambda}(t_{1-\alpha,\nu})\)
Lower-tailed, \(H_1:\mu<\mu_0\)	\(T<-t_{1-\alpha,\nu}\)	\(F_{\nu,\lambda}(-t_{1-\alpha,\nu})\)
Two-sided, \(H_1:\mu\ne\mu_0\)	\(\|T\|>t_{1-\alpha/2,\nu}\)	\(F_{\nu,\lambda}(-t_{1-\alpha/2,\nu})+1-F_{\nu,\lambda}(t_{1-\alpha/2,\nu})\)

Here \(t_{q,\nu}\) is the \(q\)-quantile of the central t distribution with \(\nu\) degrees of freedom. The two-sided power is the sum of the probabilities in the two disjoint rejection tails. Noncentral t probabilities are ordinarily evaluated numerically, using a statistical calculator or software that supports the noncentral t distribution.

Specify the test and alternative.
Identify whether the rejection region is upper-tailed, lower-tailed, or two-sided, and record the significance level \(\alpha\).

Find the degrees of freedom and noncentrality.
Use \(\nu=n-1\) and \(\lambda=\sqrt n(\mu-\mu_0)/\sigma\), evaluated at the true alternative values.

Find the null critical value.
Use a central t quantile with \(\nu\) degrees of freedom, not a standard normal quantile.

Calculate the probability of rejection.
Use the appropriate noncentral t tail probability or sum of tail probabilities, then check that the direction and degrees of freedom match the test.

Worked Power Calculations

Worked Example: Two-Sided Power With an Unknown Standard Deviation

A hypothetical laboratory measures a normally distributed response from \(n=16\) independent units. The null mean is \(\mu_0=10\), and the test is two-sided at \(\alpha=0.05\). For planning, suppose the true mean is \(\mu=11\) and the true standard deviation is \(\sigma=2\). Calculate the power. No interest-rate or mortality assumption applies to this statistical measurement example.

Method and model. The population model is \(N(11,2^2)\) under the specified alternative. Because the variance is unknown and estimated by \(S^2\), use the one-sample t statistic and its noncentral t distribution.

Degrees of freedom. With \(n=16\), \(\nu=n-1=15\).

Noncentrality parameter. The mean shift is \(11-10=1\), so

\lambda=\frac{\sqrt{16}(11-10)}{2} =\frac{4(1)}{2}=2.

Critical value. For a two-sided test at level \(0.05\), each tail has probability \(0.025\) under the null. The central t critical value is \(t_{0.975,15}\approx2.1314\), so reject when \(T<-2.1314\) or \(T>2.1314\).

Noncentral tail probabilities. Numerical evaluation of the noncentral t CDF gives \(F_{15,2}(-2.1314)\approx0.00006\) and \(F_{15,2}(2.1314)\approx0.5350\). The lower-tail probability is about \(0.00006\), or \(0.0001\) when rounded to four decimal places.

Power. Add the lower-tail probability to the upper-tail probability:

\begin{aligned} \text{Power} &=F_{15,2}(-2.1314)+1-F_{15,2}(2.1314)\\ &\approx0.00006+(1-0.5350)\\ &=0.46506\approx0.4651. \end{aligned}

The estimated power is \(0.4651\), or \(46.51\%\) (about \(46.5\%\)). The two terms are the probabilities of entering the lower and upper rejection tails, respectively. The result is checked by confirming that both tails, rather than just the tail in the direction of the mean shift, were included.

Common mistake avoided: Rounding the lower-tail probability to \(0.000\) before adding it is unnecessary, and treating it as \(0.001\) overstates it. Retaining the numerical value \(0.00006\) gives power about \(0.4651\), not \(0.466\).

Worked Example: A One-Sided Test Against a Shift in the Wrong Direction

A hypothetical device’s response is normally distributed. A one-sample upper-tailed test uses \(n=16\), \(H_0:\mu=10\), and \(\alpha=0.025\). Suppose the true alternative parameters are \(\mu=9\) and \(\sigma=2\). Calculate the power and interpret its size. No financial or mortality assumptions apply.

Method and distribution. Use the one-sample t statistic with \(\nu=n-1=15\). The specified alternative has population distribution \(N(9,2^2)\).

Noncentrality parameter. The shift is negative:

\lambda=\frac{\sqrt{16}(9-10)}{2} =\frac{4(-1)}{2}=-2.

Critical value. An upper-tailed test rejects for \(T>t_{0.975,15}\). The critical value is approximately \(2.1314\), so the power is \(1-F_{15,-2}(2.1314)\).

Evaluate the tail. The noncentral t distribution has the symmetry relation \(F_{\nu,-\lambda}(-c)=1-F_{\nu,\lambda}(c)\). Therefore,

\begin{aligned} 1-F_{15,-2}(2.1314) &=F_{15,2}(-2.1314)\\ &\approx0.00006. \end{aligned}

Thus the power is about \(0.0001\) to four decimal places, or about \(0.006\%\) using the unrounded value. This is a small probability because the true mean is below the null mean, while the test rejects only for a sufficiently large positive t statistic. The calculation is checked by confirming that the negative noncentrality parameter and upper-tail rejection region point in opposite directions.

Common mistake avoided: Replacing \(\lambda=-2\) by \(|\lambda|=2\) and then calculating an upper-tail probability would describe a shift in the opposite direction. The sign is part of the noncentral t specification and must be retained.

Worked Example: Checking Power at the Null Mean

A hypothetical normal population has unknown standard deviation, and a two-sided one-sample t test uses \(n=11\), \(H_0:\mu=20\), and \(\alpha=0.10\). Evaluate the test’s rejection probability when the true parameters are \(\mu=20\) and \(\sigma=4\). No interest or mortality model is involved.

Method and model. The observations are independent and distributed as \(N(20,4^2)\). The test uses the sample standard deviation, so the statistic has \(\nu=11-1=10\) degrees of freedom.

Noncentrality parameter. Since the true mean equals the null mean,

\lambda=\frac{\sqrt{11}(20-20)}{4}=0.

Critical values. A two-sided test at level \(0.10\) places \(0.05\) in each tail. The critical value is \(t_{0.95,10}\approx1.8125\), giving rejection region \(T<-1.8125\) or \(T>1.8125\).

Power at the null. When \(\lambda=0\), the noncentral t distribution is the central t distribution. The two rejection tails therefore have total probability \(0.10\) by the definition of the critical values:

F_{10,0}(-1.8125)+1-F_{10,0}(1.8125)=0.05+0.05=0.10.

The rejection probability is \(0.1000\). This check confirms that the test has size \(\alpha\) at the null mean. It does not imply that power at a non-null alternative is also \(0.10\).

Common Traps and Interpretation

The noncentral t calculation combines three choices that are easy to mix up: the null critical value, the alternative’s noncentrality parameter, and the test’s rejection direction. A sound calculation keeps those choices separate until the final tail probability is formed.

Using a normal critical value. When \(\sigma\) is estimated, use a central t critical value with \(n-1\) degrees of freedom to define the test’s rejection region.
Using the wrong degrees of freedom. For a one-sample t statistic, the degrees of freedom are \(n-1\), not \(n\).
Dropping the sign of \(\lambda\). A lower mean gives \(\lambda<0\); a higher mean gives \(\lambda>0\). Taking an absolute value can reverse the interpretation of a one-sided test.
Using the null distribution to calculate alternative power. The critical value comes from the central t distribution, but the rejection probability under a specified alternative uses the noncentral t distribution.
Forgetting one tail in a two-sided test. Add the probabilities below the negative critical value and above the positive critical value.

Power is the probability of rejection at a particular true parameter value. For this model, that value affects power through \(\lambda=\sqrt n(\mu-\mu_0)/\sigma\), while the critical value reflects the chosen significance level and degrees of freedom. The calculation is aligned with the SOA VEE Mathematical Statistics syllabus.

Key takeaway: Under an independent normal sample with unknown variance, the one-sample t statistic has a noncentral t distribution under an alternative, with \(\nu=n-1\) and \(\lambda=\sqrt n(\mu-\mu_0)/\sigma\). Obtain critical values from the central t distribution, then use the noncentral t tails matching the test’s rejection region to calculate power.

Check Your Understanding

For each problem, identify the degrees of freedom and noncentrality parameter before writing the relevant rejection probability.

A hypothetical normal measurement has \(n=9\), null mean \(\mu_0=5\), true mean \(\mu=7\), and true standard deviation \(\sigma=3\). For an upper-tailed test at \(\alpha=0.05\), state the degrees of freedom, noncentrality parameter, critical value, and power expression. (Answer: \(\nu=8,\ \lambda=2,\ t_{0.95,8}\approx1.8595,\ \text{power}=1-F_{8,2}(1.8595)\).)
A hypothetical normal response has \(n=25\), \(\mu_0=12\), true \(\mu=11\), and \(\sigma=2\). A lower-tailed test uses \(\alpha=0.05\). State the degrees of freedom, noncentrality parameter, critical value, and power expression. (Answer: \(\nu=24,\ \lambda=-2.5,\ \text{critical value}=-t_{0.95,24}\approx-1.7109,\ \text{power}=F_{24,-2.5}(-1.7109)\).)
For a two-sided test with \(n=16\), \(\mu_0=30\), true \(\mu=30.5\), and \(\sigma=1\), use \(\alpha=0.05\). Given \(F_{15,2}(-2.1314)\approx0.00006\) and \(F_{15,2}(2.1314)\approx0.5350\), calculate the power. (Answer: \(\nu=15,\ \lambda=2,\ \text{power}\approx0.00006+1-0.5350=0.46506\approx0.4651\).)
A hypothetical normal population is tested with a lower-tailed one-sample t test, \(n=12\), \(\mu_0=8\), true \(\mu=9\), and \(\sigma=3\). For \(\alpha=0.05\), state the degrees of freedom, noncentrality parameter, critical value, and power expression. Is the shift toward or away from the rejection region? (Answer: \(\nu=11,\ \lambda=\sqrt{12}/3\approx1.155,\ \text{critical value}=-t_{0.95,11}\approx-1.7959,\ \text{power}=F_{11,1.155}(-1.7959)\); the positive shift is away from the lower-tail rejection region.)
A two-sided one-sample t test has significance level \(0.01\). At the null mean, the population is normal with unknown standard deviation and the sample size is \(n=20\). Give the degrees of freedom, noncentrality parameter, and rejection probability at the null. (Answer: \(\nu=19,\ \lambda=0,\ \text{rejection probability}=0.0100\).)

What You'll Learn