From a Rejection Rule to Power
A one-sided z test has a clear rejection rule: reject when the test statistic falls far enough into the tail specified by the alternative hypothesis. Power asks a different question from the test itself: if the population mean actually has a particular alternative value, how often would this rule reject the null hypothesis? The answer follows by applying the normal sampling distribution under that alternative to the test’s rejection boundary.
The calculation depends on the test’s direction, significance level, sample size, known population standard deviation, and the alternative mean being considered. Power is therefore not a single property of a test independent of the actual effect. It is evaluated at a specified alternative.
Model and Critical Values
Consider a one-sample test about a population mean \(\mu\), with null boundary \(\mu_0\). Let \(X_1,\ldots,X_n\) be an independent random sample, and suppose the population standard deviation \(\sigma\) is known. The population is normally distributed, so the sample mean has an exact normal sampling distribution. If the population is not normal, a normal approximation for the sample mean may be appropriate for a sufficiently large sample, but that is an approximation rather than the exact model.
Under a population mean \(\mu\), the sample mean has standard error \(\sigma/\sqrt{n}\) and sampling distribution \(N(\mu,\sigma^2/n)\). The test statistic is \(Z=(\bar{x}-\mu_0)/(\sigma/\sqrt{n})\). Write \(z_{1-\alpha}\) for the standard normal value with probability \(1-\alpha\) to its left; equivalently, if \(Z_0\sim N(0,1)\), then \(P(Z_0\leq z_{1-\alpha})=1-\alpha\).
The hypotheses determine which tail defines rejection. A power calculation must use that same tail; changing the sign or using the other tail would calculate the probability of a different decision rule.
Worked Example: Finding an Upper-Tailed Rejection Boundary
Suppose a hypothetical quality-control test assesses whether the mean fill volume exceeds \(100\) mL. Assume a random independent sample of \(n=36\) containers, a normally distributed fill volume, and a known population standard deviation of \(\sigma=15\) mL. The test is \(H_0:\mu=100\) mL versus \(H_a:\mu>100\) mL, at \(\alpha=0.05\). Find the boundary for \(\bar{x}\) above which the test rejects.
The standard error under the test model is \(\sigma/\sqrt{n}=15/\sqrt{36}=15/6=2.5\) mL. For an upper-tailed test with \(\alpha=0.05\), \(z_{1-\alpha}=z_{0.95}\approx1.6449\). The rejection rule \(Z>1.6449\) is equivalent to:
Thus the test rejects when the sample mean is greater than about \(104.1\) mL. The assumptions used here are part of the hypothetical setup: the observations are independent, the sampling distribution is normal, and \(\sigma\) is known.
Deriving Power for an Upper-Tailed Test
For an upper-tailed test, keep the rejection boundary fixed and evaluate the chance of crossing it when the true mean is \(\mu_1>\mu_0\). Under that alternative, the sample mean is centered at \(\mu_1\), not at \(\mu_0\). Standardizing the boundary using the alternative sampling distribution gives the power formula.
The test rejects when \(\bar{x}>\mu_0+z_{1-\alpha}\sigma/\sqrt{n}\).
At \(\mu=\mu_1\), power is the probability that \(\bar{x}\) exceeds that same boundary.
Subtract \(\mu_1\) and divide by \(\sigma/\sqrt{n}\), then express the upper-tail probability using the standard normal distribution.
For a complete derivation, let \(SE=\sigma/\sqrt{n}\). Under the alternative \(\mu=\mu_1\), the variable \((\bar{x}-\mu_1)/SE\) has the standard normal distribution. Therefore:
Here \(Z_0\sim N(0,1)\), and \(\Phi(a)=P(Z_0\leq a)\) is the standard normal cumulative distribution function. The last equality follows from \(P(Z_0>a)=1-\Phi(a)=\Phi(-a)\). This derivation shows why power increases when the specified alternative mean moves farther above the null boundary: the rejection boundary becomes less extreme relative to the alternative sampling distribution.
Worked Example: Power at a Specified Larger Mean
Use the fill-volume test above, and suppose the true mean is \(108\) mL. Calculate the test’s power at this alternative.
The test rejects when \(\bar{x}>104.11225\) mL. Under the specified alternative, the sample mean has distribution \(N(108,15^2/36)\), with standard error \(2.5\) mL. The power is the chance, under this distribution, that the sample mean exceeds the rejection boundary:
At a true mean fill volume of \(108\) mL, this test has about \(94.0\%\) probability of rejecting \(H_0\). Its Type II error probability at this same alternative is \(1-0.9400=0.0600\), or \(6.00\%\). This is a probability about the test’s repeated-sampling behavior when \(\mu=108\) mL; it is not the probability that the alternative hypothesis is true.
Power for a Lower-Tailed Test
The reasoning is parallel for a lower-tailed test, but its rejection region is on the low side of the null boundary. At a specified \(\mu_1<\mu_0\), the rejection event is \(\bar{x}<\mu_0-z_{1-\alpha}SE\). Standardizing that boundary under the distribution centered at \(\mu_1\) gives the following result.
To see the standardization directly, the power is \(P_{\mu_1}(\bar{x}<\mu_0-z_{1-\alpha}SE)\). Subtracting \(\mu_1\) and dividing by \(SE\) makes the cutoff \((\mu_0-\mu_1)/SE-z_{1-\alpha}\), so the probability is \(\Phi\) of that cutoff.
Worked Example: Power for a Lower-Tailed Test
Suppose a hypothetical instrument is being checked for a mean reading below \(50\) units. A random independent sample of \(n=25\) readings is taken. Assume normally distributed readings and a known population standard deviation \(\sigma=10\) units. Test \(H_0:\mu=50\) units versus \(H_a:\mu<50\) units at \(\alpha=0.10\). Find the power if the true mean is \(46\) units.
The standard error is \(10/\sqrt{25}=2\) units. Since \(z_{0.90}\approx1.2816\), the lower rejection boundary for the sample mean is:
Under the specified alternative \(\mu_1=46\), the sample mean is centered at \(46\) units, with standard error \(2\) units. Thus:
If the true mean reading is \(46\) units, the probability that this test rejects is about \(76.4\%\). The Type II error probability at that alternative is about \(1-0.7637=0.2363\), or \(23.6\%\). Independence, normality, and known \(\sigma\) were stipulated in the example, so the normal sampling calculation applies.
How the Distance to the Boundary Matters
Power is governed by how far the specified alternative lies from the null boundary in standard-error units, relative to the critical value. For an upper-tailed test, the relevant distance is \((\mu_1-\mu_0)/SE\); for a lower-tailed test, it is \((\mu_0-\mu_1)/SE\). If that distance is close to \(z_{1-\alpha}\), the alternative distribution is centered near the rejection boundary and power can be near one half. If it is much larger, most of the alternative sampling distribution falls in the rejection region.
Worked Example: A Small Shift Above the Null Boundary
A hypothetical upper-tailed test examines whether a process mean exceeds \(200\) units. Assume a random independent sample of \(n=64\), a normally distributed population, and known \(\sigma=12\) units. The hypotheses are \(H_0:\mu=200\) units versus \(H_a:\mu>200\) units, with \(\alpha=0.025\). Find power at \(\mu_1=203\) units.
The standard error is \(12/\sqrt{64}=1.5\) units, and \(z_{0.975}\approx1.960\). The upper rejection boundary is:
Under the alternative, standardize this cutoff using mean \(203\) and standard error \(1.5\):
At a true process mean of \(203\) units, the probability of rejection is about \(51.6\%\). Although the test has a relatively large sample, the alternative mean is only \(3/1.5=2\) standard errors above the null boundary, just slightly beyond the critical value \(1.960\). The resulting power is therefore only a little above one half.
Common Mistake and Interpretation
A frequent mistake is to calculate power by using the null distribution for both the rejection boundary and the probability. The null distribution determines the critical value, because that is how the significance level is set. But power is evaluated under the specified alternative, so the sampling distribution must be centered at \(\mu_1\). For example, in the first calculation, \(104.11225\) mL is the same fixed rejection boundary under either mean, but the chance of exceeding it is calculated from the distribution centered at \(108\) mL.
Another error is to report power without naming the alternative. A statement such as “the test has 94% power” is incomplete unless the conditions and the value of \(\mu_1\) are clear. A suitable interpretation says that, under the stated model and at the specified alternative mean, the test would reject \(H_0\) in about the stated proportion of repeated samples of the same size.
Power is useful when assessing how likely a test is to detect an effect of a particular size, but it does not establish that the effect exists. A computed power value is conditional on the model, design quantities, and specified alternative. In particular, it is not a probability that \(H_a\) is true after observing data.
Check Your Understanding
Use the rejection-boundary method and identify the distribution used for each probability.
- For an upper-tailed test, what is the rejection rule in terms of \(Z\), and what is the corresponding boundary in terms of \(\bar{x}\)?
- Why is the critical value chosen using the null distribution, while power is calculated using the alternative distribution?
- A lower-tailed test has null boundary \(\mu_0\), standard error \(SE\), and significance level \(\alpha\). Write its rejection boundary for \(\bar{x}\).
- If power at a specified alternative is \(0.82\), what is the Type II error probability at that alternative, and what does it mean in context?
- For an upper-tailed test, explain what happens to the power formula when the specified alternative mean moves farther above \(\mu_0\), with the other test settings fixed.