Type II Error at a Specified Alternative

Why the Alternative Distribution Matters

A hypothesis test uses observed data to choose between a rejection decision and a non-rejection decision. The probability of a wrong decision depends on which hypothesis is true. If the null hypothesis is true, rejecting it is a Type I error. If a specified alternative is true, failing to reject the null is a Type II error.

To calculate a Type II error probability, first identify which outcomes lead to non-rejection. Then calculate the probability of those outcomes under the specified alternative. This last condition is essential: the null distribution describes behavior when the null is true, not the probability of a Type II error at an alternative.

Definition: Let $C$ be the rejection region of a test, so that the test rejects $H_0$ when the observed outcome lies in $C$. At a specified parameter value $\theta_1$ in the alternative hypothesis, the Type II error probability is $$ \beta(\theta_1)=P_{\theta_1}(\text{do not reject }H_0) =P_{\theta_1}(T\notin C), $$ when the decision rule is based on test statistic $T$. If the rule is based on the full data, use the probability of the non-rejection outcomes under that data distribution instead.

The value $\theta_1$ must be specified because the probability of non-rejection can vary across alternatives. A statement such as “the Type II error probability is 0.20” is incomplete unless the alternative, sampling model, and decision rule are understood. In a test whose rejection probability at $\theta_1$ is called its power, the Type II error probability is one minus that rejection probability.

Calculate Probability on the Non-Rejection Region

The calculation follows directly from the definition. For a discrete test statistic, add the probabilities of every non-rejection value under the alternative. For a continuous statistic with density $f_T(t;\theta_1)$, integrate that density over the non-rejection region. If the non-rejection region has multiple pieces, include all of them.

\beta(\theta_1) =\sum_{t\notin C}P_{\theta_1}(T=t) \quad\text{(discrete)},\qquad \beta(\theta_1) =\int_{t\notin C}f_T(t;\theta_1)\,dt \quad\text{(continuous)}.

These formulas hold because the event “do not reject” is exactly the event that the statistic is outside $C$. The probabilities in the sum or integral must therefore come from the distribution of $T$ when $\theta=\theta_1$. If a decision rule rejects for $T\ge c$, for example, its non-rejection event is $T<c$; for a discrete statistic, the boundary matters and should be written explicitly.

Write the decision rule.
State the rejection region and, just as carefully, its complement.

Specify the alternative.
Identify the parameter value at which the Type II error probability is being requested.

Use the alternative distribution.
Find the distribution of the test statistic under that parameter value.

Sum or integrate over non-rejection.
Include every outcome that leads to not rejecting, taking account of strict and non-strict boundaries.

Worked Examples

Worked Example: A One-Sided Test for a Normal Mean

Suppose independent observations $X_1,\ldots,X_{25}$ come from a normal distribution with unknown mean $\mu$ and known standard deviation $10$. Test $H_0:\mu=50$ against $H_1:\mu>50$ using the rule “reject $H_0$ if $\bar X>53$.” Find the Type II error probability at $\mu=57$. Assume the stated normal sampling model and independence; no interest-rate or mortality assumptions apply to this statistical model.

Method. Under the specified alternative, $\bar X$ has a normal distribution with mean $57$ and standard deviation $10/\sqrt{25}=2$. Non-rejection occurs when $\bar X\le 53$, including the boundary because the rule rejects only when $\bar X>53$.

\begin{aligned} \beta(57) &=P_{\mu=57}(\bar X\le 53)\\ &=P\left(\frac{\bar X-57}{2}\le\frac{53-57}{2}\right)\\ &=\Phi(-2)\\ &\approx 0.0228. \end{aligned}

Thus, when the true mean is $57$, this test fails to reject about 2.28% of the time. The calculation is checked by noting that the cutoff $53$ is two standard errors below the alternative mean, so a small lower-tail probability is expected. The common mistake to avoid is using the null mean $50$ to standardize the cutoff: that would calculate a null probability, not the Type II error probability at $\mu=57$.

Worked Example: A Binomial Test With a Discrete Boundary

A laboratory records the number $X$ of positive results among 12 independent trials. Suppose $X\sim\operatorname{Binomial}(12,p)$, with $H_0:p=0.20$ and $H_1:p>0.20$. The rule rejects $H_0$ if $X\ge 5$. Calculate the Type II error probability at the specified alternative $p=0.50$. The trials are assumed to have the same success probability and to be independent.

Method. Under $p=0.50$, non-rejection means $X=0,1,2,3,$ or $4$. Since $X$ is discrete, sum the five binomial probabilities. Here $\binom{12}{k}$ is the number of ways to obtain $k$ positive results.

\begin{aligned} \beta(0.50) &=P_{p=0.50}(X\le 4)\\ &=\sum_{k=0}^{4}\binom{12}{k}(0.50)^k(0.50)^{12-k}\\ &=\frac{\binom{12}{0}+\binom{12}{1}+\binom{12}{2}+\binom{12}{3}+\binom{12}{4}}{2^{12}}\\ &=\frac{1+12+66+220+495}{4096}\\ &=\frac{794}{4096}\\ &=0.19384765625\\ &\approx 0.1938. \end{aligned}

The result is 0.1938 to four decimal places. As a check, the numerator includes every non-rejection count, including 4, while the rejection region begins at 5. A common error is to calculate $P_{p=0.20}(X\le4)$, which is about $0.9274$. That is the probability of non-rejection under the null, not the Type II error probability at $p=0.50$. The alternative probability must be used in every term of the sum.

Worked Example: A Two-Sided Test and a Corrected Normal-Tail Calculation

Let $X_1,\ldots,X_{16}$ be independent observations from a normal distribution with mean $\mu$ and known standard deviation $8$. Test $H_0:\mu=0$ against $H_1:\mu\ne0$ by rejecting when $|\bar X|>4$. Find the Type II error probability at $\mu=3$. Assume the normal model and independent observations.

Method. The non-rejection region is $-4\le\bar X\le4$. Under $\mu=3$, the sample mean has standard deviation $8/\sqrt{16}=2$. Standardizing each endpoint gives a lower standardized value of $(-4-3)/2=-3.5$ and an upper standardized value of $(4-3)/2=0.5$.

\begin{aligned} \beta(3) &=P_{\mu=3}(-4\le\bar X\le4)\\ &=\Phi(0.5)-\Phi(-3.5)\\ &\approx 0.691462461274-0.000232629079\\ &\approx 0.691229832195\\ &\approx 0.691230\quad\text{to six decimal places}\\ &\approx 0.6912\quad\text{to four decimal places.} \end{aligned}

The alternative mean is inside the non-rejection interval and not far from its upper endpoint, so a fairly large probability of non-rejection is plausible. The subtraction is $0.691229832195$, which rounds to $0.691230$ to six decimal places. Rounding the two cumulative probabilities too early can change the last reported digit. The common mistake is to calculate only the upper-tail probability above $4$; a two-sided test also has a lower rejection tail, but the Type II error here is the probability between both rejection boundaries.

Common Interpretations and Checks

A Type II error probability is tied to a particular alternative and a particular test rule. It is not the probability that the null hypothesis is true after seeing the data, and it does not describe what happened in one observed sample. It describes the long-run chance of non-rejection over repeated samples when the specified alternative model is true.

Before accepting a numerical answer, check the following:

Decision region: Did you identify all outcomes that lead to non-rejection, including the correct boundary?
Alternative: Did you substitute the specified alternative parameter into the distribution?
Direction: For a one-sided test, did you use the correct tail? For a two-sided test, did you include the full interval between rejection tails?
Scale: For a sample mean, did you use the standard error, such as $\sigma/\sqrt n$, rather than the standard deviation of one observation?
Probability range: Is the result between 0 and 1, and is its size consistent with the alternative distribution’s location relative to the rejection boundary?

If the statistic’s distribution under the alternative is not the same as under the null, that is expected: the two distributions describe different parameter values. The null distribution is used to assess the test’s behavior when the null is true. The alternative distribution is used here to find the chance that the observed statistic nevertheless falls in the non-rejection region. Keeping those roles separate is central to calculations aligned with the SOA VEE Mathematical Statistics syllabus.

Key takeaway: At a specified alternative, a Type II error is non-rejection when that alternative is true. Find the non-rejection region, use the alternative distribution, and sum or integrate over that region.

Check Your Understanding

For each problem, identify the non-rejection event and use the distribution at the stated alternative.

Let $X\sim\operatorname{Binomial}(8,p)$, and reject $H_0$ when $X\ge5$. Find the Type II error probability at $p=0.60$. (Answer: $0.4059$)
Independent observations have a normal distribution with mean $\mu$ and known standard deviation $15$. For $n=25$, reject $H_0$ when $\bar X>86$. Find the Type II error probability at $\mu=89$. (Answer: $0.1587$)
A count $X$ has a Poisson distribution with mean $\lambda$, and the test rejects when $X\ge4$. Find the Type II error probability at $\lambda=2$. (Answer: $e^{-2}(1+2+2+8/6)\approx0.8571$)
Suppose $\bar X$ has a normal distribution with mean $\mu$ and standard error $2$. A test rejects $H_0:\mu=0$ when $|\bar X|>4$. Find the Type II error probability at $\mu=2$. (Answer: $\Phi(1)-\Phi(-3)\approx0.8400$)
A test rejects when a statistic $T$ is at least 10. At a specified alternative, $T$ has a continuous distribution with cumulative distribution function $F_{\theta_1}$. Express the Type II error probability using that distribution. (Answer: $P_{\theta_1}(T<10)=F_{\theta_1}(10)$)

What You'll Learn