Size, Power, and Type II Error

Why Rejection Probabilities Matter

A hypothesis test is a rule for deciding whether observed data are sufficiently inconsistent with a null hypothesis to reject it. That rule can make two kinds of errors: rejecting the null when it is true, or not rejecting it when a particular alternative is true. The probabilities of these outcomes describe how the test behaves under specified models.

The significance level concerns rejection when the null hypothesis holds. Power concerns rejection when an alternative holds. Keeping the two models separate is essential: the same rejection rule can have a small rejection probability under the null and a much larger one under an alternative. These distinctions are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $X$ denote the data and let $R$ be the rejection region: the set of possible data values for which the test rejects $H_0$. For a parameter value $\theta$, write $P_\theta$ for probability calculated under the model with parameter $\theta$. The rejection probability at $\theta$ is $P_\theta(X\in R)$.

Significance Level, Power, and Type II Error

First suppose the hypotheses specify one null parameter value $\theta_0$ and one alternative value $\theta_1$. Under this simple-versus-simple setup, the significance level of the stated test is its probability of rejection when the null model is true. Its power at the specified alternative is the probability of rejection when that alternative model is true.

Definition: For a simple null $H_0:\theta=\theta_0$, the test’s actual significance level is $$ \alpha=P_{\theta_0}(X\in R). $$ For a specified alternative $H_1:\theta=\theta_1$, the power is $$ \pi(\theta_1)=P_{\theta_1}(X\in R), $$ and the Type II error probability at that alternative is $$ \beta(\theta_1)=P_{\theta_1}(X\notin R). $$

For a composite null hypothesis, a test may be described by a maximum rejection probability over all null parameter values. Its size is $\sup_{\theta\in H_0}P_\theta(X\in R)$. A test has level $\alpha$ if this supremum is no greater than $\alpha$. Thus, in a discrete problem, the actual rejection probability under a simple null need not equal a nominal target level; it may be below it. The rejection region and the assumed distribution determine the actual probability.

Power and Type II error at the same alternative are complements. The sample must either fall in $R$, causing rejection, or in its complement $R^c$, causing nonrejection. These events are disjoint and cover all possible outcomes under the alternative model, so their probabilities sum to one.

P_{\theta_1}(X\in R)+P_{\theta_1}(X\notin R)=1 \quad\Longrightarrow\quad \beta(\theta_1)=1-\pi(\theta_1).

This is a direct partition argument, not an approximation. It also gives a practical check on calculations: if the reported power and Type II error probability do not sum to one, at least one is wrong. The complement is taken under the alternative model, not under the null model.

Worked Calculations

For discrete models, calculate the probability of each value in the rejection region and add those probabilities. For continuous models, calculate the probability of the corresponding interval or tail. In every case, use the distribution specified by the parameter value being considered.

Worked Example: A Binomial Inspection Rule

A hypothetical inspection samples five components independently from a large production lot. Let $X$ be the number found defective, with $X\sim\operatorname{Binomial}(5,p)$. The test rejects $H_0:p=0.2$ when at least two components are defective. Find the actual significance level, then find the power and Type II error probability at $p=0.5$.

Method and rejection region. The rule rejects for $R=\{2,3,4,5\}$. Under a binomial model, $P_p(X=k)=\binom{5}{k}p^k(1-p)^{5-k}$. The significance level uses $p=0.2$; power and Type II error use $p=0.5$.

Significance level. It is convenient to use the complement $X\leq1$:

\begin{aligned} \alpha &=P_{0.2}(X\geq2)\\ &=1-P_{0.2}(X=0)-P_{0.2}(X=1)\\ &=1-(0.8)^5-5(0.2)(0.8)^4\\ &=1-0.32768-0.40960\\ &=0.26272. \end{aligned}

Power at $p=0.5$. Under this alternative, each of the $2^5=32$ possible equally likely success-failure sequences has probability $1/32$. There are one sequence with zero defectives and five with exactly one, so

\begin{aligned} \pi(0.5) &=P_{0.5}(X\geq2)\\ &=1-P_{0.5}(X=0)-P_{0.5}(X=1)\\ &=1-\frac{1}{32}-\frac{5}{32}\\ &=\frac{26}{32}=0.8125. \end{aligned}

Therefore $\beta(0.5)=1-0.8125=0.1875$. To four decimal places, the significance level is $0.2627$, the power is $0.8125$, and the Type II error probability is $0.1875$. The two alternative-model probabilities check: $0.8125+0.1875=1$. Common mistake avoided: calculating the power using $p=0.2$. The null parameter determines $\alpha$; the alternative parameter determines power and $\beta$.

Worked Example: A One-Sided Normal Rejection Rule

A hypothetical sensor produces a measurement $X\sim N(\mu,3^2)$, with known standard deviation $3$. A test rejects $H_0:\mu=10$ when $X>12$. Find its significance level and its power and Type II error probability when $\mu=13$.

Method and assumptions. Use the stated normal model and standardize the boundary $12$ separately under each mean. Here $N(\mu,3^2)$ means variance $9$, so the standard deviation used in standardization is $3$.

Significance level under $H_0$. Under $\mu=10$, the standardized boundary is $(12-10)/3=2/3$. Thus

\alpha=P_{10}(X>12) =1-\Phi\left(\frac{2}{3}\right) \approx 1-0.7475 =0.2525,

where $\Phi$ is the standard normal cumulative distribution function.

Power under $\mu=13$. Under the alternative, standardizing the same boundary gives $(12-13)/3=-1/3$. Therefore

\pi(13)=P_{13}(X>12) =1-\Phi\left(-\frac{1}{3}\right) =\Phi\left(\frac{1}{3}\right) \approx 0.6306.

The equality uses the symmetry of the standard normal distribution. The Type II error probability is

\beta(13)=P_{13}(X\leq12) =\Phi\left(-\frac{1}{3}\right) \approx 0.3694.

Rounded to four decimal places, $\alpha=0.2525$, power is $0.6306$, and $\beta(13)=0.3694$; the last two sum to $1.0000$. Common mistake avoided: standardizing the cutoff with the null mean when finding power. For power, the distribution has mean $13$, not $10$.

Worked Example: A Poisson Count Threshold

A hypothetical monitoring system counts unusual events during one fixed shift. Assume $X\sim\operatorname{Poisson}(\lambda)$, where $\lambda$ is the mean count per shift. The test rejects $H_0:\lambda=1$ when $X\geq4$. Find the significance level and the power and Type II error probability at $\lambda=3$.

Method. For a Poisson variable, $P_\lambda(X=k)=e^{-\lambda}\lambda^k/k!$. Since the rejection region is $X\geq4$, its probability is one minus the sum of the probabilities at $0,1,2,3$.

Significance level. Under the null $\lambda=1$:

\begin{aligned} \alpha &=P_1(X\geq4)\\ &=1-e^{-1}\left(1+1+\frac{1}{2}+\frac{1}{6}\right)\\ &=1-\frac{8}{3e}\\ &\approx 0.0190. \end{aligned}

Power at $\lambda=3$. Under the alternative, the probabilities of counts zero through three sum to

\begin{aligned} P_3(X\leq3) &=e^{-3}\left(1+3+\frac{3^2}{2!}+\frac{3^3}{3!}\right)\\ &=e^{-3}(1+3+4.5+4.5)\\ &=13e^{-3}. \end{aligned}

Consequently,

\pi(3)=1-13e^{-3}\approx0.3528, \qquad \beta(3)=13e^{-3}\approx0.6472.

To four decimal places, the significance level is $0.0190$, power is $0.3528$, and Type II error probability is $0.6472$. The alternative probabilities sum to one. Common mistake avoided: treating the null rejection probability as power. Here $0.0190$ is calculated at $\lambda=1$; the power $0.3528$ is calculated at $\lambda=3$.

What Power Does—and Does Not—Mean

In the usual frequentist interpretation, the parameter is fixed, while the observed data vary across repeated samples from the model. Thus, power at $\theta_1$ means: if the specified alternative model with parameter $\theta_1$ were true and the same sampling and decision procedure were repeated, the long-run fraction of samples falling in the rejection region would be $P_{\theta_1}(X\in R)$.

Power is not the probability that the alternative hypothesis is true, nor is $\beta$ the probability that the null hypothesis is true after observing the data. Those statements concern uncertainty about hypotheses or parameters and do not follow from the rejection probability alone. Similarly, failure to reject $H_0$ does not prove that $H_0$ is true; it records that the observed data did not meet the test’s rejection rule.

Key takeaway: To calculate these quantities, identify the rejection region first. Evaluate its probability under the null model to obtain the actual significance level, then evaluate it under the specified alternative to obtain power. The probability of nonrejection under that same alternative is Type II error, equal to one minus power.

Check Your Understanding

For each problem, use the stated rejection rule and model. Give probabilities to four decimal places where needed.

A hypothetical audit samples four records independently, and $X$, the number with a specified error, has distribution $\operatorname{Binomial}(4,p)$. The test rejects $H_0:p=0.25$ when $X\geq2$. Find the actual significance level, power at $p=0.6$, and Type II error probability at $p=0.6$. (Answer: $\alpha=0.2617$, power $=0.8208$, $\beta=0.1792$.)
A hypothetical count per shift follows $\operatorname{Poisson}(\lambda)$. A test rejects $H_0:\lambda=1.5$ when $X\geq3$. Find the significance level, power at $\lambda=3$, and Type II error probability at $\lambda=3$. (Answer: $\alpha=0.1912$, power $=0.5768$, $\beta=0.4232$.)
A measurement has distribution $X\sim N(\mu,5^2)$. A test rejects $H_0:\mu=100$ when $X>108$. Find the significance level and the power and Type II error probability at $\mu=112$. (Answer: $\alpha=0.0548$, power $=0.7881$, $\beta=0.2119$.)
Let $X\sim\operatorname{Binomial}(2,p)$, and suppose the rejection region is $R=\{0\}$. Under $H_0:p=0.5$, find the actual significance level. Then find power and Type II error probability at $p=0.2$. (Answer: $\alpha=0.2500$, power $=0.6400$, $\beta=0.3600$.)
A test has power $0.72$ at a particular alternative parameter value. State its Type II error probability at that alternative and explain what the power does not represent. (Answer: $\beta=0.28$; power is the probability of rejection over repeated samples if that alternative model holds, not the probability that the alternative is true.)

What You'll Learn