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P-values and conclusions for proportions · Tutorial 488 of 1000

Writing a Complete Conclusion With Evidence

Practice combining a p-value, the preselected significance level, the test decision, and the research question into a complete conclusion.

Intermediate 9 min read

What You'll Learn

  • Organize a test conclusion into three parts that AP readers can identify.
  • Compare a p-value with the preselected significance level and state the correct decision.
  • Translate a one-proportion alternative hypothesis into a contextual evidence claim.
  • Write appropriate conclusions after both rejecting and failing to reject the null hypothesis.
  • Explain why a conclusion does not prove a population proportion equals a particular value.
  • Keep a conclusion within the population represented by the study.

From a Test Result to a Complete Conclusion

A significance test produces a p-value, but a p-value by itself is not a conclusion. In Comparing P-Value to the Significance Level and Rejecting the Null Hypothesis Correctly, we learned how to make the decision. Now we will connect that decision to the question being investigated.

A complete conclusion should make three things clear: how the p-value compares with the preselected significance level \(\alpha\), what decision follows from that comparison, and what the decision says about the population in context. These parts work together. A decision without a contextual claim is incomplete; a contextual claim without the decision’s evidence language can overstate what the test shows.

Three-part conclusion: (1) Report the p-value and compare it with \(\alpha\). (2) State whether you reject or fail to reject \(H_0\). (3) Say whether the data provide convincing evidence for the alternative hypothesis, translated into the setting.

The comparison rule is \(p\text{-value}\leq\alpha\): reject \(H_0\). If the p-value is greater than \(\alpha\), fail to reject \(H_0\). As emphasized in Why We Never Accept the Null Hypothesis, failing to reject is not accepting or proving the null hypothesis. In either case, the final claim should match the alternative hypothesis: “higher,” “lower,” or “different,” as appropriate.

1
Compare.
Give the p-value and the significance level, and say which is larger or whether they are equal.
2
Decide.
Use the comparison to state “reject \(H_0\)” or “fail to reject \(H_0\).”
3
Conclude in context.
State whether there is convincing evidence for the alternative, and describe that alternative in the setting and for the population represented by the data.

A useful sentence frame is: “Because the p-value of [value] is [less than, equal to, or greater than] \(\alpha=[value]\), we [decision]. At the [significance level] level, the data [do/do not] provide convincing evidence that [contextual version of \(H_a\)].” Fill in each part from the test, not from what you hope the result will show.

Worked Examples: Writing the Three Parts

Worked Example: Support for Weekend Library Hours

A fictional town library wants to know whether the proportion of adult residents who support adding weekend hours differs from 40%. A random sample of 200 adults from a list of 5,000 residents includes 94 who support the change. The library sets \(\alpha=0.05\) before collecting data. Write a complete conclusion.

State: Let \(p\) be the true proportion of adult residents on this town’s list who support adding weekend library hours. The hypotheses are \(H_0:p=0.40\) and \(H_a:p\ne0.40\).

Plan: The adults were randomly sampled, so the Random condition is met. The sample was drawn without replacement from a list of 5,000, and \(200\leq0.10(5{,}000)=500\), so the 10% condition is met. Under \(H_0\), the expected success and failure counts are \(200(0.40)=80\) and \(200(0.60)=120\). Both are at least 10, so the Large Counts condition is met. A one-proportion \(z\)-test is appropriate.

Do: The sample proportion is:

$$ \hat{p}=\frac{x}{n}=\frac{94}{200}=0.47 $$

For the test, use the null standard error and the one-proportion \(z\)-test statistic:

$$ SE_0=\sqrt{\frac{0.40(0.60)}{200}} =\sqrt{0.0012} \approx0.0346410 \qquad z=\frac{0.47-0.40}{\sqrt{0.40(0.60)/200}} \approx2.0207 $$

Because the alternative is two-sided, the p-value is twice the standard Normal area above \(\lvert2.0207\rvert\):

$$ p\text{-value}=2P(Z\geq2.0207)\approx2(0.02166)=0.0433 $$

Conclude: First, \(0.0433<0.05\). Therefore, we reject \(H_0\). At the 0.05 significance level, the sample provides convincing evidence that the proportion of adult residents on this town’s list who support adding weekend library hours differs from 40%.

The conclusion says “differs,” not “is greater,” because the alternative is two-sided. Although the sample proportion is above 40%, the direction of the claim in the conclusion must follow the alternative chosen for the test.

Worked Example: Interest in a Community Compost Program

A fictional town is considering a community compost program. Officials want to know whether more than 20% of households on a town list would subscribe. A random sample of 100 households from a list of 3,000 includes 26 that say they would subscribe. The significance level is \(\alpha=0.05\). Write a complete test conclusion.

State: Let \(p\) be the true proportion of households on this town’s list that would subscribe to the community compost program. The hypotheses are \(H_0:p=0.20\) and \(H_a:p>0.20\).

Plan: The households were randomly sampled, meeting the Random condition. Because \(100\leq0.10(3{,}000)=300\), the 10% condition is met. Under \(H_0\), the expected counts are \(100(0.20)=20\) households that would subscribe and \(100(0.80)=80\) that would not. Both counts are at least 10, so the Large Counts condition is met. A one-proportion \(z\)-test is appropriate.

Do: The sample proportion is \(\hat{p}=26/100=0.26\). The null standard error and test statistic are:

$$ SE_0=\sqrt{\frac{0.20(0.80)}{100}} =\sqrt{0.0016} =0.04 \qquad z=\frac{0.26-0.20}{0.04} =1.50 $$

The alternative is right-tailed, so the p-value is the area to the right of \(z=1.50\):

$$ p\text{-value}=P(Z\geq1.50)\approx0.0668 $$

Conclude: The p-value \(0.0668\) is greater than \(\alpha=0.05\), so we fail to reject \(H_0\). At the 0.05 significance level, the sample does not provide convincing evidence that more than 20% of the households on this town’s list would subscribe to the community compost program.

The conclusion does not say that 20% would subscribe. It also does not say there is no interest in the program. It says that this sample did not provide convincing evidence for the specific claim that the proportion is greater than 20%.

Worked Example: The Significance Level Changes the Decision

A fictional school district takes a random sample of 400 families from a list of 8,000 and finds that 218 support a proposed change to the school calendar. The question is whether support differs from 50%. The test output gives a two-sided p-value of 0.0719. Compare the conclusions if the significance level had been set in advance at 0.05 or at 0.01.

State: Let \(p\) be the true proportion of families on this district’s list who support the proposed calendar change. The hypotheses are \(H_0:p=0.50\) and \(H_a:p\ne0.50\).

Plan: The families were randomly sampled, meeting the Random condition. The sample is no more than 10% of the list because \(400\leq0.10(8{,}000)=800\). Under \(H_0\), the expected counts are \(400(0.50)=200\) supporters and \(400(0.50)=200\) nonsupporters, both at least 10. The conditions for a one-proportion \(z\)-test are met.

Do: The sample proportion is \(218/400=0.545\). The test statistic is:

$$ SE_0=\sqrt{\frac{0.50(0.50)}{400}} =\sqrt{0.000625} =0.025 \qquad z=\frac{0.545-0.50}{0.025} =1.80 $$

For the two-sided alternative, the p-value is:

$$ p\text{-value}=2P(Z\geq1.80) \approx2(0.03593) =0.07186 \approx0.0719 $$

Conclude: If \(\alpha=0.05\) was chosen before the test, then \(0.0719>0.05\), so we fail to reject \(H_0\). At the 0.05 significance level, the data do not provide convincing evidence that the proportion of families on this district’s list who support the proposed calendar change differs from 50%. If instead \(\alpha=0.01\) was chosen before the test, then \(0.0719>0.01\), so we also fail to reject \(H_0\); at that level, the data do not provide convincing evidence that the proportion differs from 50%.

The hypotheses and sample result are the same in both comparisons; only the preselected threshold differs. In an actual test, choose \(\alpha\) before examining the data. Do not select a significance level afterward just to obtain the decision you prefer.

Common Mistakes and AP Exam Tips

  • Giving only the decision: “Reject \(H_0\)” does not explain what the result says about the research question. Add an evidence statement and translate \(H_a\) into context.
  • Reporting the p-value without comparing it with \(\alpha\): A p-value such as 0.0433 does not determine a decision by itself. State the chosen \(\alpha\) and the comparison.
  • Reversing the comparison rule: A p-value less than or equal to \(\alpha\) leads to rejecting \(H_0\). A p-value greater than \(\alpha\) leads to failing to reject \(H_0\). Equality belongs in the rejection rule.
  • Writing “accept \(H_0\)” or “prove \(H_0\)”: For a large p-value, say “fail to reject \(H_0\)” and “the data do not provide convincing evidence for \(H_a\).” Do not claim that the null value is true.
  • Changing the alternative in the conclusion: If \(H_a:p\ne p_0\), conclude that the proportion differs from \(p_0\), not that it is greater just because \(\hat{p}>p_0\). A one-sided conclusion must match the stated direction.
  • Making the claim too broad: A random sample from a particular list supports a conclusion about the population represented by that list, not automatically about everyone in a town, state, or country.
  • Leaving out “convincing evidence”: Use the standard evidence language. A significant result provides convincing evidence for the alternative; a nonsignificant result does not provide convincing evidence for it at the selected level.
AP Exam Tip: Make the logic visible: “Because the p-value is [comparison] \(\alpha\), we [decision]. At the [level] significance level, the data [do/do not] provide convincing evidence that [alternative in context].” Include the population and characteristic defined by \(p\), and keep the claim aligned with \(H_a\).

Key Takeaway

A strong test conclusion is a short, connected argument—not just a decision or a p-value. Compare the p-value with the significance level, state the resulting decision, and explain what the data do or do not show about the population claim. Use “convincing evidence” language, and never turn a failure to reject into proof that the null hypothesis is true.

Key takeaway: Compare the p-value with \(\alpha\), state “reject” or “fail to reject,” and translate the evidence decision into a careful claim about the population and characteristic in the alternative hypothesis.

Check Your Understanding

For each question, include the decision and a contextual evidence statement where requested.

  1. A test has p-value \(0.018\) and \(\alpha=0.05\). What is the decision? What does the comparison say about the evidence for \(H_a\)?
  2. A right-tailed test has p-value \(0.12\) and \(\alpha=0.10\). Write the decision and explain what the result does not prove.
  3. A two-sided test asks whether the proportion of households that use a neighborhood garden differs from 30%. The test rejects \(H_0\). Write the direction-neutral evidence claim that matches the alternative.
  4. A student writes, “Since \(p\text{-value}=0.03\), there is a 3% chance that the null hypothesis is true.” Identify the error and describe what the p-value means instead.
  5. A random sample of patients from one clinic is used to test whether more than 25% prefer appointment reminders by text. The test fails to reject \(H_0\). What should the conclusion say, and which population should it refer to?

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