How to write a hypothesis test conclusion

Q: What if the p-value is exactly equal to alpha?

Reject H₀. The AP rule is written with an inequality that includes equality: if the p-value is less than or equal to α, reject; if it is greater than α, fail to reject. A p-value of exactly 0.05 against α = 0.05 falls on the reject side.

By Jude Wallis · Updated July 31, 2026

A conclusion has two parts. Part 1 compares: because the p-value is less than (or greater than) $\alpha$, we reject (or fail to reject) $H_0$. Part 2 answers in context: there is (or is not) convincing statistical evidence that the alternative is true. Never accept $H_0$ and never say proof.

AP Statistics: Unit 3 (topics 3.7 Carrying Out a Test for a Population Proportion, 4.5 Carrying Out a Test for a Population Mean or Population Mean Difference). Writing the conclusion is the justification step in Unit 3 topic 3.7 and Unit 4 topic 4.5 of the Fall 2026 AP Statistics course, where the formal decision must explicitly compare the p-value to the significance level and the conclusion must be stated in context, in terms of the alternative hypothesis, using non-definitive language.

The two-part conclusion

A complete conclusion is two sentences, and they do different jobs.

Part 1, the formal decision. Compare the p-value to the significance level $\alpha$ (say "alpha") with both numbers on the page, then state the decision about $H_0$ (say "H-naught"). The AP course is explicit here: a formal decision explicitly compares the p-value to $\alpha$ . If the p-value $\leq \alpha$ , reject $H_0$ ; if the p-value $> \alpha$ , fail to reject $H_0$ .

Part 2, the answer in context. State what that decision means about the real question, in terms of the alternative hypothesis $H_a$ , naming the parameter and the population, and using non-definitive language.

A response that gives only Part 1 has done the statistics and skipped the question. A response that gives only Part 2 has answered the question without justification. Both halves are scored, so write both. If you are still fixing which hypothesis is which, see null vs alternative hypothesis.

Two templates, word for word

**When the p-value is at or below $\alpha$ :**

"Because the p-value of [p-value] is less than or equal to the significance level of [alpha], we reject $H_0$ . There is convincing statistical evidence that [the alternative, in words, naming the parameter and the population]."

**When the p-value is above $\alpha$ :**

"Because the p-value of [p-value] is greater than the significance level of [alpha], we fail to reject $H_0$ . There is not convincing statistical evidence that [the alternative, in words, naming the parameter and the population]."

Notice how little changes: the comparison word, the decision phrase, and the single word "not". Everything after "evidence that" stays identical, because both conclusions are stated in terms of $H_a$ . Students who write the failing-to-reject case in terms of $H_0$ almost always end up claiming the null is true.

The boundary case follows the AP rule as written. A p-value exactly equal to $\alpha$ falls in the reject region, since the rule is p-value $\leq \alpha$ . For how the two quantities differ in the first place, see p-value vs alpha.

Accept, fail to reject, prove: the words that decide the score

**Never "accept $H_0$ ".** The AP course flags this directly: students may incorrectly claim that we can accept or have proven the null, but statistical tests do not provide evidence for what can be accepted or proven. They provide evidence only for rejecting or failing to reject.

"Fail to reject" is the phrase. It is deliberately awkward, and the awkwardness is the point. It says the data did not supply enough evidence against $H_0$ , which is not the same as evidence for $H_0$ . Lack of statistical evidence for the alternative is not evidence for the null.

"Evidence that", never "proof". Rejecting $H_0$ means there is convincing statistical evidence to support the alternative hypothesis. It does not prove the alternative, because a Type I error is always possible. Write "convincing statistical evidence that", not "proves that" or "shows conclusively that".

Never conclude the null is true. A large p-value never licenses "there is no difference" or "the mean is 340 grams". Use non-definitive language instead: "the data do not provide convincing evidence that the mean differs from 340 grams". A large p-value only means your data are consistent with $H_0$ ; see what a p-value means.

What has to be in the context half

The AP course asks for a conclusion stated in context, consistent with and in terms of the alternative hypothesis, using non-definitive language, and containing a reference to the parameter and the population. That is four requirements in one sentence, so check them off.

Parameter. Say which quantity you tested: the proportion of tickets, the mean weight, the difference in mean commute times.
Population. Say whose. "All tickets filed last year", not "the tickets".
**In terms of $H_a$ .** If $H_a: p < 0.80$ , the context half says "less than 0.80", not "different from 0.80".
Non-definitive language. "There is convincing statistical evidence that" and "there is not convincing statistical evidence that" both qualify. Bare declaratives do not.

A quick test: could a reader who has not seen the problem tell what was tested and about whom, from your last sentence alone? If not, the context half is incomplete.

Common wrong conclusions, corrected

Wrong	Why it costs the point	Corrected
"We accept $H_0$ ."	A test never provides evidence for the null.	"We fail to reject $H_0$ ."
"We reject $H_0$ ." and nothing else	No numerical comparison, no context, no parameter or population.	"Because the p-value of 0.0170 is less than 0.05, we reject $H_0$ . There is convincing statistical evidence that ..."
"There is no difference in the true mean weight."	Concludes the null is true from a large p-value.	"There is not convincing statistical evidence that the true mean weight differs from 340 grams."
"The p-value proves the claim is false."	Proof is not on offer; a Type I error is always possible.	"There is convincing statistical evidence that the claim is false."
"Since the p-value is 0.147 and 0.147 > 0.05, we reject $H_0$ ."	The decision rule is backwards.	"Since $0.147 > 0.05$ , we fail to reject $H_0$ ."
"The p-value is small, so we reject $H_a$ ."	Rejects the wrong hypothesis.	"The p-value is small, so we reject $H_0$ , which supports $H_a$ ."
"There is a 1.7% chance the null is true."	Misreads the p-value as a probability about the hypothesis.	"If $H_0$ were true, results this extreme would occur about 1.7% of the time."

The pattern behind four of these seven is the same: the writer treated the test as a verdict on the truth of $H_0$ rather than as a measure of how surprising the data would be if $H_0$ held.

Chi-square, two-sample tests, and one-sided alternatives

The structure does not change across procedures, only the wording of $H_a$ .

Chi-square. For a test for independence, $H_a$ says there is an association between the two categorical variables in the population. A large p-value tempts the sentence "there is no association", which is accepting the null. The AP course suggests non-definitive wording such as: the data do not provide strong enough evidence to conclude that the variables are associated.

Two-sample tests. Name both groups and the direction of subtraction. "There is convincing statistical evidence that the mean commute time for downtown employees is longer than for suburban employees" is complete; "there is a difference" is not.

One-sided alternatives. Keep the direction all the way through. If $H_a: \mu > 12$ , the conclusion says "greater than 12", never "different from 12". And if the sample result lands on the wrong side of the null value, you fail to reject no matter how large the gap, because the evidence points away from $H_a$ .

You can confirm any p-value you use with the p-value calculator.

What the conclusion does not settle

Rejecting $H_0$ does not make the effect large or important. With a big enough sample, a difference too small to matter can produce a p-value under 0.05. Report the estimate alongside the decision so a reader can judge size, not just significance.

Every conclusion also carries a chance of being wrong, and naming which kind is a separate question type. A Type I error occurs when there is convincing statistical evidence that the alternative is true, due to a small p-value, but it is not. A Type II error occurs when there is not convincing statistical evidence that the alternative is true, due to a large p-value, but it is. So a reject conclusion risks a Type I error and a fail-to-reject conclusion risks a Type II error; see type 1 vs type 2 errors.

Finally, the conclusion inherits whatever the data collection allows. A test on observational data supports an association, not a cause. A test on a randomized experiment supports a cause-and-effect conclusion for the treatments compared.

A conclusion that rejects: one-proportion z-test

A university help desk claims that 80% of the 12,000 tickets filed last year were resolved within one day. A student group takes a random sample of 200 of those tickets and finds 148 resolved within one day. Using $\alpha = 0.05$ , test whether the true proportion is less than 0.80, and write the conclusion.

State the hypotheses. Let $p$ be the proportion of all 12,000 tickets filed last year that were resolved within one day. $H_0: p = 0.80$ and $H_a: p < 0.80$ , with $\alpha = 0.05$ .
Check conditions. The 200 tickets are a random sample. $0.10(12{,}000) = 1{,}200$ and $200 \leq 1{,}200$ . Using the null value, $np_0 = 200(0.80) = 160$ and $n(1 - p_0) = 200(0.20) = 40$ , both at least 10.
Find the sample proportion. $\hat{p}$ ("p-hat") $= \frac{148}{200} = 0.74$ .
Find the standard deviation of the sampling distribution under $H_0$ . $\sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.80(0.20)}{200}} = \sqrt{\frac{0.16}{200}} = \sqrt{0.0008} = 0.0283$ .
Compute the test statistic. $z = \frac{0.74 - 0.80}{0.0283} = \frac{-0.06}{0.0283} = -2.12$ .
Find the p-value. The alternative is one-sided to the left, so the p-value is $P(Z < -2.12)$ . From the z-table this is 0.0170 (technology gives 0.0169).
Part 1, the formal decision. Because the p-value of 0.0170 is less than the significance level of 0.05, we reject $H_0$ .
Part 2, the answer in context. There is convincing statistical evidence that the proportion of all tickets filed at this help desk last year that were resolved within one day is less than 0.80.

$z = -2.12$ and the p-value is 0.0170. Because $0.0170 < 0.05$ , we reject $H_0$ . There is convincing statistical evidence that the proportion of all tickets filed at this help desk last year that were resolved within one day is less than 0.80. Note what the conclusion avoids: it does not say the claim has been disproved, and it does not say the true proportion equals 0.74.

A conclusion that fails to reject: one-sample t-test

A roaster fills bags labeled 340 grams. From a day's production of 4,000 bags, a random sample of 25 bags has mean $\bar{x} = 337.6$ grams and sample standard deviation $s = 8.0$ grams, and a dotplot of the 25 weights is roughly symmetric with no outliers. Using $\alpha = 0.05$ , test whether the true mean fill weight differs from 340 grams, and write the conclusion.

State the hypotheses. Let $\mu$ ("mu") be the true mean fill weight of all bags in that day's production. $H_0: \mu = 340$ and $H_a: \mu \neq 340$ , with $\alpha = 0.05$ .
Check conditions. The 25 bags are a random sample. $0.10(4{,}000) = 400$ and $25 \leq 400$ . Since $n = 25 < 30$ and no population normality is stated, the dotplot argument applies: it is roughly symmetric with no outliers, so the sample data condition is met.
Find the standard error. $\frac{s}{\sqrt{n}} = \frac{8.0}{\sqrt{25}} = \frac{8.0}{5} = 1.6$ grams.
Compute the test statistic. $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{337.6 - 340}{1.6} = \frac{-2.4}{1.6} = -1.50$ , with $df = n - 1 = 24$ .
Find the p-value. The alternative is two-sided, so the p-value is $2 \times P(T < -1.50)$ for 24 degrees of freedom. The t-table brackets it between 0.10 and 0.20; technology gives 0.1467.
Part 1, the formal decision. Because the p-value of 0.1467 is greater than the significance level of 0.05, we fail to reject $H_0$ .
Part 2, the answer in context. There is not convincing statistical evidence that the true mean fill weight of all bags in that day's production differs from 340 grams.
Check what was not written. The conclusion does not say the mean is 340 grams, does not say the bags are filled correctly, and does not say the machine is accurate. Each of those would be accepting $H_0$ .

$t = -1.50$ with 24 degrees of freedom, and the two-sided p-value is 0.1467. Because $0.1467 > 0.05$ , we fail to reject $H_0$ . There is not convincing statistical evidence that the true mean fill weight of all bags in that day's production differs from 340 grams. The sample mean of 337.6 grams is below 340, but not far enough below to rule out ordinary sampling variability.

Frequently asked questions

Can I ever write "we accept the null hypothesis"?

No. A hypothesis test can lead to rejecting or not rejecting $H_0$ , but it can never conclude or prove that $H_0$ is true. The AP course names "we can accept" and "we have proven" as errors to avoid. The correct phrase is always "fail to reject".

What if the p-value is exactly equal to alpha?

Reject $H_0$ . The AP rule is written with an inequality that includes equality: if the p-value is less than or equal to $\alpha$ , reject; if it is greater than $\alpha$ , fail to reject. A p-value of exactly 0.05 against $\alpha = 0.05$ falls on the reject side.

What do I use for alpha when the problem does not give one?

The significance level may be given or chosen by the researcher, so state the one you are using. By convention that is 0.05, and writing "using $\alpha = 0.05$ " before the decision makes the comparison in Part 1 legible. Never pick $\alpha$ after seeing the p-value.

Should the conclusion be phrased in terms of the null or the alternative?

The alternative. The AP course asks for a conclusion stated in context consistent with, and in terms of, the alternative hypothesis. Both outcomes use the same $H_a$ wording; only the word "not" moves. Phrasing a failure to reject in terms of $H_0$ is what leads students into claiming the null is true.

Does rejecting the null mean the effect is large?

No. Statistical significance and practical importance are different questions. A large sample can produce a small p-value from a difference too small to act on, so report the estimate or the confidence interval alongside the decision.