Hypothesis Tests as Decision Rules

From Sample Data to a Decision

A hypothesis test is not just a calculation of a statistic. It is a procedure that specifies what to do for every possible sample: reject the null hypothesis for some outcomes, and do not reject it for the others. The observed data determine which part of that procedure is used. This decision-rule view makes the roles of the test statistic, rejection region, and significance level explicit, and is aligned with the SOA VEE Mathematical Statistics syllabus.

For example, a quality analyst might count the number of defective components in a sample and reject a specified production-rate hypothesis if the count is unusually high. The rule must be set before seeing the sample results. Otherwise, the decision could depend on an unplanned choice made after looking at the data.

Definition: A nonrandomized hypothesis test is a rule that assigns one of two decisions to every possible sample outcome: reject \(H_0\), or do not reject \(H_0\). The rejection region is the set of sample outcomes for which the rule rejects \(H_0\). The complementary set is the non-rejection region.

Statistics and Rejection Regions

Write a possible complete sample as \(x\), and let \(T(x)\) be a test statistic computed from it. A test often defines a rejection region by specifying a set \(C\) of statistic values. The corresponding set of sample outcomes is every \(x\) for which \(T(x)\) belongs to \(C\). The decision rule is therefore: reject \(H_0\) if \(T(x)\in C\); otherwise, do not reject \(H_0\).

The statistic compresses the data into a quantity chosen to reflect the distinction of interest. The rejection region states which values of that quantity count as evidence against the null hypothesis in the direction specified by the alternative. A rule can also be described directly in terms of sample outcomes without first writing a statistic, but using a statistic often makes the rule easier to state and apply.

Here is why the probability of rejection is found from the statistic’s null distribution. Under a parameter value \(\theta\), rejection occurs exactly when the random sample \(X\) produces a statistic in \(C\). Thus the rejection event in the sample space is the inverse image of \(C\) under \(T\), and its probability is:

P_\theta(\text{reject }H_0) = P_\theta\bigl(T(X)\in C\bigr).

This identity connects a decision rule to the distribution of its statistic. If \(H_0\) is simple, the null distribution is evaluated at its specified parameter value. If \(H_0\) is composite, the rule’s size is the largest rejection probability over parameter values in the null hypothesis. The significance level \(\alpha\) is a chosen upper bound for that size. A test can have size below \(\alpha\), rather than exactly equal to it; this occurs naturally for some discrete distributions.

Key takeaway: A test rule requires all three of the following: a statistic \(T\), a rejection region \(C\), and a decision instruction. The null distribution of \(T\) determines the probability that the rule rejects when the null hypothesis is true.

Worked Examples

Worked Example: A Discrete Upper-Tail Rule

A hypothetical inspection samples \(n=8\) components independently. Let \(X\) be the number found defective. Under the null hypothesis, each component is defective independently with probability \(p=0.25\); the alternative is \(p>0.25\). Consider the rule that rejects \(H_0\) when \(X\geq 5\). Find its significance level and decide what it does when five defects are observed.

Method and assumptions. Under \(H_0:p=0.25\), \(X\sim\operatorname{Binomial}(8,0.25)\). Independence and a common defect probability are the assumptions that give this binomial model. Since larger counts point in the direction of the alternative, the stated rejection region is \(C=\{5,6,7,8\}\).

Calculate the null rejection probability. For \(k=5,6,7,8\), the binomial probability is \(\binom{8}{k}(0.25)^k(0.75)^{8-k}\). Expressing all terms with denominator \(4^8=65{,}536\) gives:

\begin{aligned} P_{0.25}(X=5)&=\frac{\binom{8}{5}3^3}{4^8} =\frac{56\cdot27}{65{,}536} =\frac{1{,}512}{65{,}536},\\ P_{0.25}(X=6)&=\frac{\binom{8}{6}3^2}{4^8} =\frac{28\cdot9}{65{,}536} =\frac{252}{65{,}536},\\ P_{0.25}(X=7)&=\frac{\binom{8}{7}3}{4^8} =\frac{8\cdot3}{65{,}536} =\frac{24}{65{,}536},\\ P_{0.25}(X=8)&=\frac{1}{65{,}536}. \end{aligned}

Adding the four probabilities gives \(1{,}789/65{,}536=0.027298\ldots\), so the rule’s significance level is \(0.0273\) to four decimal places. This is below a target level of \(0.05\).

Check the boundary. If the rule instead rejected at \(X\geq4\), it would also include:

P_{0.25}(X=4) =\frac{\binom{8}{4}3^4}{4^8} =\frac{70\cdot81}{65{,}536} =\frac{5{,}670}{65{,}536}.

The resulting rejection probability would be \((1{,}789+5{,}670)/65{,}536=7{,}459/65{,}536=0.1138\) to four decimal places, which exceeds \(0.05\). Thus \(X\geq5\) is the more conservative of these two rules at the stated target. With five observed defects, the stated rule rejects \(H_0\).

Common mistake avoided: Do not include \(X=4\) in the rejection probability for a rule that rejects only when \(X\geq5\). The boundary is part of the rule, not a detail to adjust after calculating the tail.

Worked Example: A Two-Sided Rule for a Normal Mean

Suppose \(X_1,\ldots,X_{16}\) are independent observations from \(N(\mu,64)\), where the variance \(64\) is known. Test \(H_0:\mu=120\) against \(H_1:\mu\ne120\) using a two-sided rule with significance level \(0.05\). For an observed sample mean of \(124.5\), state the rule and the decision.

Method and assumptions. Under the null hypothesis, the sample mean has a normal distribution with mean \(120\) and standard error \(8/\sqrt{16}=2\). The known variance and independent normal observations justify this reference distribution. The standard-normal cutoffs for a two-sided \(0.05\) rule are approximately \(-1.96\) and \(1.96\).

Define \(Z=(\bar X-120)/2\). The rejection region is \(Z<-1.96\) or \(Z>1.96\), equivalently \(|Z|>1.96\). In terms of the sample mean, this is:

\bar X < 120-1.96(2)=116.08 \quad\text{or}\quad \bar X > 120+1.96(2)=123.92.

For \(\bar X=124.5\), the observed statistic is \(Z=(124.5-120)/2=2.25\). Since \(2.25>1.96\), the sample mean is in the rejection region, so the rule rejects \(H_0\).

Check. The rule includes both tails because the alternative allows the mean to differ in either direction. Its null rejection probability is approximately \(0.025\) in each tail, or \(0.05\) in total. Rounding the critical value to \(1.96\) gives the usual approximate rule.

Common mistake avoided: Using \(1.645\) as the cutoff would give a one-sided \(0.05\) upper-tail rule, not this two-sided rule. Also, this particular \(Z\) rule uses a known variance; if the variance were estimated, its stated reference distribution and cutoff would need to change.

Worked Example: A Rejection Region for a Variance

A hypothetical laboratory measures \(n=10\) independent observations from a normal population with unknown mean \(\mu\) and variance \(\sigma^2\). Test \(H_0:\sigma^2=9\) against \(H_1:\sigma^2\ne9\) using a two-sided \(0.05\) rule. The observed sample variance, calculated with denominator \(n-1\), is \(S^2=24\). State the rule and the decision.

Method and assumptions. For an independent normal sample under \(H_0\), the statistic \(T=(n-1)S^2/9\) has a chi-square distribution with \(n-1=9\) degrees of freedom. The normal-population assumption matters: this exact chi-square reference distribution for the sample variance is not generally valid for a nonnormal population.

For a two-sided rule with \(0.025\) in each tail, the approximate chi-square critical values for 9 degrees of freedom are \(2.700\) and \(19.023\). Therefore, the rule rejects when \(T<2.700\) or \(T>19.023\).

T_{\text{obs}} =\frac{(10-1)(24)}{9} =\frac{216}{9} =24.

Because \(24>19.023\), the observed statistic is in the rejection region, and the rule rejects \(H_0\). If instead the statistic had been between the two critical values, the rule would not reject. The comparison is made using \(T\), not by comparing \(S^2\) directly with the chi-square cutoffs.

Common mistake avoided: The sample-variance statistic uses \(n-1\), and the null variance \(9\) is the denominator. Replacing either with \(n\) or the observed variance changes the statistic and does not give the stated null reference distribution.

What the Significance Level Does—and Does Not—Specify

The significance level constrains the null rejection probability, but it does not by itself determine a unique test. A complete rule also needs an alternative hypothesis, a statistic or direct sample-space condition, and a rejection region. For example, a \(0.05\) upper-tail rule and a \(0.05\) two-sided rule use different regions because they address different alternatives.

In a continuous model, a cutoff can often be chosen so the rejection probability equals the target level, subject to the stated model and any rounding of critical values. In a discrete model, the possible probabilities may jump over the target. The binomial example illustrates this: the rule \(X\geq5\) has level \(0.0273\), while lowering the boundary to \(X\geq4\) raises it to \(0.1138\). A nonrandomized rule cannot attain \(0.05\) with either of those boundaries.

The decision rule is fixed before observing the data. Once the data are collected, calculate the statistic as prescribed and check whether it belongs to the rejection region. Changing the statistic, tail, or boundary after seeing the outcome creates a different rule from the one whose significance level was assessed.

Practical checklist: State \(H_0\) and \(H_1\); identify the statistic and its null distribution; specify the rejection region and target significance level; then calculate the observed statistic and apply the rule without changing its boundary.

Check Your Understanding

For each problem, identify the rule, calculate or apply its null rejection probability as requested, and check the boundary carefully.

A sample of \(5\) independent items has defect probability \(0.20\) under \(H_0\), with \(H_1:p>0.20\). The rule rejects when \(X\geq3\), where \(X\) is the number of defects. Find its significance level and state the decision if \(X=2\). (Answer: \(0.0579\) to four decimal places; do not reject.)
Independent observations are normal with known standard deviation \(6\). For \(n=9\), test \(H_0:\mu=40\) against \(H_1:\mu>40\) using the \(0.05\) upper-tail critical value \(1.645\). Give the rule in terms of \(\bar X\) and decide for \(\bar X=43\). (Answer: reject if \(\bar X>43.29\); for \(\bar X=43\), \(Z=1.5\), so do not reject.)
For a normal sample of size \(8\), test \(H_0:\sigma^2=4\) against \(H_1:\sigma^2<4\). Use \(T=7S^2/4\) and the \(0.05\) lower-tail chi-square critical value \(2.167\) with \(7\) degrees of freedom. Decide if \(S^2=1.5\). (Answer: \(T=2.625>2.167\), so do not reject.)
Two fair six-sided dice are rolled independently. Under \(H_0\), both dice are fair; the alternative concerns unusually large totals. Let \(T\) be the sum, and use the rule that rejects when \(T\geq11\). Find the significance level and decide for the observed roll \((6,5)\). (Answer: \(3/36=0.0833\) to four decimal places; reject.)

What You'll Learn