Deriving Power for a One-Sample Z Test

Why Derive the Power Function?

A one-sided Z test is designed to detect a change in a specified direction. Its significance level determines how often it rejects at the null boundary, but that alone does not tell us how likely it is to reject when the true mean differs from the boundary. Power answers that second question: it is the probability of rejection under a particular true parameter value.

For a one-sample Z test with known variance, the power calculation follows directly from the test's rejection rule. The key is to express that rule as an event involving the sample mean, then calculate the probability of that event using the sampling distribution under the true mean.

Model, Hypotheses, and Rejection Rule

Suppose $X_1,\ldots,X_n$ are independent observations from a normal distribution with mean $\mu$ and known variance $\sigma^2$, where $\sigma>0$. The sample mean then has distribution $\bar{X}\sim N(\mu,\sigma^2/n)$. Independence and the normal model make this distribution exact for any positive integer $n$; the variance must be known for the Z statistic below to have the stated null distribution.

Definition: For an upper-tailed test of $H_0:\mu\leq\mu_0$ against $H_1:\mu>\mu_0$, the power function at true mean $\mu$ is $\pi(\mu)=P_\mu(\text{reject }H_0)$. It gives the rejection probability for each possible true value of the mean.

Write $\Phi(z)=P(Z\leq z)$ for the standard normal cumulative distribution function when $Z\sim N(0,1)$. Let $z_{1-\alpha}$ be its $1-\alpha$ quantile, so $\Phi(z_{1-\alpha})=1-\alpha$. For a test with significance level $\alpha$, the upper-tailed rejection rule is

\text{Reject }H_0 \quad\text{if}\quad \frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}>z_{1-\alpha}.

Under the boundary value $\mu=\mu_0$, the statistic has a standard normal distribution, so the probability of rejection is $\alpha$. For every $\mu<\mu_0$, the rejection probability is no larger than $\alpha$. Thus the test has size $\alpha$ for the composite null $H_0:\mu\leq\mu_0$.

Deriving the Upper-Tailed Power Function

To calculate power, keep the rejection rule fixed but evaluate its probability when the true mean is $\mu$, not necessarily $\mu_0$. Since the standard error $\sigma/\sqrt{n}$ is known, the event can be rewritten as a threshold event for $\bar{X}$.

Rewrite the rejection event.
Multiplying by the positive quantity $\sigma/\sqrt{n}$ and adding $\mu_0$ gives $\bar{X}>\mu_0+z_{1-\alpha}\sigma/\sqrt{n}$.

Use the distribution under the true mean.
When the true mean is $\mu$, $\bar{X}\sim N(\mu,\sigma^2/n)$, so standardizing the threshold gives a standard normal probability.

Express the upper-tail probability.
Subtract the probability of being at or below the threshold from one.

In detail, the rejection probability at true mean $\mu$ is

\begin{aligned} \pi(\mu) &=P_\mu\left(\bar{X}>\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}}\right)\\ &=1-\Phi\left( \frac{\mu_0+z_{1-\alpha}\sigma/\sqrt{n}-\mu}{\sigma/\sqrt{n}} \right)\\ &=1-\Phi\left(z_{1-\alpha}-\frac{\sqrt{n}(\mu-\mu_0)}{\sigma}\right). \end{aligned}

The quantity $\sqrt{n}(\mu-\mu_0)/\sigma$ is the distance between the true mean and the null boundary, measured in standard errors of $\bar{X}$. This shows why power depends on the alternative mean, the sample size, the known standard deviation, and the chosen significance level.

Theorem: For the upper-tailed one-sample Z test under the stated normal model, the power function is $$ \pi(\mu)=1-\Phi\left(z_{1-\alpha}-\frac{\sqrt{n}(\mu-\mu_0)}{\sigma}\right). $$ At $\mu=\mu_0$, this equals $\alpha$. For $\mu>\mu_0$, power increases as $\mu$ increases, as $n$ increases, or as $\sigma$ decreases, with the other quantities fixed.

The monotonicity follows because increasing $\mu$, increasing $n$, or decreasing $\sigma$ increases $\sqrt{n}(\mu-\mu_0)/\sigma$. The argument of $\Phi$ therefore decreases, making $1-\Phi(\cdot)$ increase. At means far below $\mu_0$, the rejection probability is small; at means far above it, the probability approaches one.

Worked Example: Power at a Specified Increase in Mean

A hypothetical laboratory measures a chemical concentration in independent samples from a normal population. Assume the population variance is known to be $\sigma^2=100$, so $\sigma=10$. For $n=25$ measurements, test $H_0:\mu\leq100$ against $H_1:\mu>100$ at $\alpha=0.05$. Find the power when the true mean is $\mu=106$.

Method. Use the upper-tailed power function. The standard error is $\sigma/\sqrt{n}=10/5=2$, and the upper-tail critical value is $z_{0.95}=1.6449$.

The rejection threshold for the sample mean is

$$ 100+1.6449(2)=103.2898. $$

Under the specified alternative, $\bar{X}\sim N(106,2^2)$. Standardizing the threshold under this distribution gives

\frac{103.2898-106}{2}=-1.3551.

Therefore, the power is

P_{106}(\bar{X}>103.2898) =1-\Phi(-1.3551) =\Phi(1.3551) \approx 0.9123.

The test has about $91.23\%$ power at a true mean of 106. As a check, the threshold is below the alternative mean, and the probability of exceeding it is greater than one half. Common mistake avoided: using the null distribution to calculate this rejection probability would give $0.05$, the boundary significance level, rather than the power under $\mu=106$.

Worked Example: A Different Significance Level and Standard Error

A hypothetical process has independent normal measurements with known variance $\sigma^2=64$. A quality analyst tests $H_0:\mu\leq50$ against $H_1:\mu>50$ using $n=16$ observations and significance level $\alpha=0.10$. Calculate the power if the true mean is 54.

Method. The standard deviation is $\sigma=8$, and the standard error is $8/\sqrt{16}=2$. The critical value for an upper-tail probability of $0.10$ is $z_{0.90}=1.2816$. Thus the sample-mean rejection threshold is

$$ 50+1.2816(2)=52.5632. $$

At the specified true mean, $\bar{X}\sim N(54,2^2)$. The standardized threshold under this distribution is

\frac{52.5632-54}{2}=-0.7184.

The rejection probability is therefore

P_{54}(\bar{X}>52.5632) =1-\Phi(-0.7184) =\Phi(0.7184) \approx 0.7637.

The power is $0.7637$, or $76.37\%$. The threshold is below 54, so a power greater than $0.5$ is reasonable. Common mistake avoided: the upper-tail critical value is $z_{0.90}$, not $z_{0.10}$; the latter is a negative quantile and would place the rejection boundary in the wrong direction.

Lower-Tailed Tests and Sign Checks

For a lower-tailed test $H_0:\mu\geq\mu_0$ against $H_1:\mu<\mu_0$, reject when the standardized statistic is less than $-z_{1-\alpha}$. The same standardization method gives the lower-tail power function. Writing $z_\alpha$ for the $\alpha$ quantile, where $z_\alpha=-z_{1-\alpha}$, the result is

\pi_{\mathrm{lower}}(\mu) =\Phi\left(z_\alpha-\frac{\sqrt{n}(\mu-\mu_0)}{\sigma}\right).

For a true mean below $\mu_0$, the standardized difference $\sqrt{n}(\mu-\mu_0)/\sigma$ is negative, so lower-tail power increases as the true mean moves farther below the null boundary. The direction of the inequality matters: upper-tail tests use a right-tail probability, while lower-tail tests use a left-tail probability.

Worked Example: Power for a Lower-Tailed Test

A hypothetical sensor produces independent normal readings with known variance $\sigma^2=144$. With $n=36$ readings, a technician tests $H_0:\mu\geq100$ against $H_1:\mu<100$ at $\alpha=0.025$. Find the power if the true mean is 96.

Method. Here $\sigma=12$, so the standard error is $12/\sqrt{36}=2$. The lower-tail critical value is $z_{0.025}=-1.96$. The rejection rule for the sample mean is

\bar{X}<100+(-1.96)(2)=96.08.

When $\mu=96$, the sample mean has distribution $N(96,2^2)$. Standardizing the rejection threshold under that distribution gives

\frac{96.08-96}{2}=0.04.

Thus the power is

P_{96}(\bar{X}<96.08)=\Phi(0.04)\approx0.5160.

The power is $51.60\%$. Because the true mean is only slightly below the boundary, a probability just above one half is plausible. Common mistake avoided: using the upper-tail formula unchanged would reverse the relevant event and give an incorrect probability.

Interpreting Power and Avoiding Exam Traps

Power is a function, not a single property of a test independent of the alternative. A statement such as “the test has power 0.8” is incomplete unless it specifies the true parameter value (or a particular alternative). At the null boundary, the upper-tailed test's power is $\alpha$, not zero: a test at level $0.05$ rejects with probability $0.05$ when $\mu=\mu_0$.

A useful check is to derive the cutoff in the original units before standardizing under the alternative. For an upper-tailed test, that cutoff must be above $\mu_0$, and a true mean above the cutoff should generally yield power above one half. For a lower-tailed test, the cutoff must be below $\mu_0$, and a true mean below it should generally yield power above one half. These checks often expose a sign error before it affects the final answer.

Wrong distribution: using the null distribution for an alternative's rejection probability returns the null rejection probability, not the desired power.
Wrong sign: for an upper-tail test, the power expression is $1-\Phi(z_{1-\alpha}-\sqrt{n}(\mu-\mu_0)/\sigma)$. Reversing the difference can make power decrease as the alternative mean moves in the tested direction.
Wrong standard error: the sampling standard deviation is $\sigma/\sqrt{n}$, not $\sigma/n$. The square root is essential.
Wrong tail quantile: an upper-tail test of size $\alpha$ uses $z_{1-\alpha}$, while a lower-tail test uses $z_\alpha$.

Key takeaway: To derive power, keep the test's critical boundary fixed, substitute the true mean into the sampling distribution, and calculate the probability of falling in the rejection region.

Check Your Understanding

For each problem, use the stated normal model and known variance. Report probabilities to four decimal places.

A hypothetical fill process has independent normal measurements with known $\sigma=6$. For $n=9$, test $H_0:\mu\leq30$ against $H_1:\mu>30$ at $\alpha=0.05$. Find power at $\mu=34$. (Answer: 0.6386)
A hypothetical material-strength measurement is normal with known $\sigma=5$. For $n=25$, test $H_0:\mu\leq20$ against $H_1:\mu>20$ at $\alpha=0.01$. Find power at $\mu=23$. (Answer: 0.7500)
A hypothetical device's independent normal errors have known $\sigma=14$. For $n=49$, test $H_0:\mu\leq80$ against $H_1:\mu>80$ at $\alpha=0.05$. Find power at $\mu=82$. (Answer: 0.2595)
For a hypothetical normal population with known $\sigma=4$, a test uses $n=16$ to test $H_0:\mu\leq10$ against $H_1:\mu>10$ at $\alpha=0.05$. Find the true mean that gives power $0.8000$. (Answer: approximately 12.4865)
A hypothetical process has independent normal observations with known $\sigma=6$. For $n=36$, test $H_0:\mu\geq40$ against $H_1:\mu<40$ at $\alpha=0.05$. Find power at $\mu=38$. (Answer: 0.6386)

What You'll Learn