From a Test Decision to a Contextual Conclusion
In “Deciding a t Test by Comparing the P-Value to Alpha,” you learned how to make the decision to reject or fail to reject \(H_0\). This tutorial focuses on the next step: explaining what that decision says about the alternative hypothesis \(H_a\) in the situation being studied.
A complete conclusion does more than repeat “reject” or “fail to reject.” It names the direction or difference described by \(H_a\), uses evidence language that matches the decision, and identifies the population and measurement in context. The conclusion is about the population mean represented by \(\mu\), not just the sample mean \(\bar{x}\).
The p-value helps determine how strong the sample evidence is against \(H_0\): it is calculated assuming \(H_0\) is true. The decision at the chosen \(\alpha\) then guides the evidence wording. At \(\alpha=0.05\), a p-value at or below 0.05 leads to rejecting \(H_0\); a p-value above 0.05 leads to failing to reject it. The conclusion must use the p-value for the alternative that was actually stated.
Let the Alternative Hypothesis Shape the Sentence
As in “One-Sided Versus Two-Sided Alternatives for a Mean,” the alternative hypothesis identifies the kind of departure from the null that the test is designed to detect. Your conclusion should express that same claim in ordinary language.
- For \(H_a:\mu<\mu_0\), the claim is that the population mean is less than the reference value. A rejecting conclusion says there is convincing evidence that it is less.
- For \(H_a:\mu>\mu_0\), the claim is that the population mean is greater than the reference value. A rejecting conclusion says there is convincing evidence that it is greater.
- For \(H_a:\mu\ne\mu_0\), the claim is that the population mean differs from the reference value. A rejecting conclusion says there is convincing evidence of a difference. It does not say that the mean is specifically higher or specifically lower as the claim being tested.
A two-sided test may have a sample mean above or below \(\mu_0\), but \(H_a:\mu\ne\mu_0\) includes departures in both directions. You may describe the sample’s direction when useful, but do not replace the two-sided alternative with a one-sided conclusion.
For a decision to fail to reject \(H_0\), keep the same alternative claim in view. For example, if \(H_a:\mu>\mu_0\), write that the data do not provide convincing evidence that the population mean is greater than \(\mu_0\). Do not switch to saying there is evidence the mean equals \(\mu_0\).
A Conclusion-Focused Four-Step Process
A conclusion depends on a correctly conducted test. “Writing the Full Four-Step t Interval Solution” introduced a structured response for interval inference; for a test, use the parallel four steps below. The State, Plan, and Do steps establish the basis for the final sentence, while the Conclude step makes the evidence claim.
Define \(\mu\) in context, write \(H_0\) and \(H_a\), and state \(\alpha\).
Name the one-sample t test. Check the Random condition, the 10% condition when sampling without replacement, and the Normal/Large Sample condition.
Calculate \(t=(\bar{x}-\mu_0)/(s/\sqrt{n})\), use \(df=n-1\), find the p-value for \(H_a\), and compare it with \(\alpha\).
State “reject” or “fail to reject” \(H_0\), then describe the evidence for or against the claim in \(H_a\), in the original context.
The conditions do not change the wording rule, but they affect whether the test’s inference is justified. For a review of how to check them with evidence from a study, see “Complete Conditions Check for a Mean Inference Problem.” The examples below show the full sequence, with particular attention to the final sentence.
Worked Examples
Worked Example: Evidence for a Higher Mean
A fictional bottling workshop randomly selects 16 sealed jars from a production lot of 320. The sample mean amount of liquid is \(\bar{x}=503\) milliliters and the sample standard deviation is \(s=6\) milliliters. The workshop wants to know whether the true mean fill amount is greater than 500 milliliters. Assume the population distribution of fill amounts is approximately Normal. Use \(\alpha=0.05\).
State. Let \(\mu\) be the true mean fill amount, in milliliters, of jars in this production lot. The hypotheses are \(H_0:\mu=500\) milliliters and \(H_a:\mu>500\) milliliters. The significance level is \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test for a population mean. The jars were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(320)=32\), and \(16\leq32\), so independence is reasonable. Since \(n=16<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The degrees of freedom are \(df=16-1=15\).
Do. The standard error is \(s/\sqrt{n}=6/\sqrt{16}=1.5\) milliliters. Therefore,
The test is upper-tailed because \(H_a:\mu>500\). With \(df=15\), the p-value is \(\operatorname{tcdf}(2,1\mathrm{E}99,15)\approx0.0320\), rounded to four decimal places. Since \(0.0320<0.05\), reject \(H_0\).
Conclude. The sample provides convincing evidence that the true mean fill amount of jars in this production lot is greater than 500 milliliters.
The sentence names the population quantity, uses the direction specified by \(H_a\), and links the decision to the evidence. It does not claim that every jar contains more than 500 milliliters; the test concerns the population mean.
Worked Example: Not Convincing Evidence for a Lower Mean
A fictional school garden randomly selects 16 seedlings from a greenhouse group of 240. The sample mean time to reach a particular growth stage is \(\bar{x}=11.75\) days, and the sample standard deviation is \(s=2\) days. Gardeners want to know whether the true mean time is less than 12 days. Assume the population distribution is approximately Normal. Use \(\alpha=0.05\).
State. Let \(\mu\) be the true mean number of days for seedlings in this greenhouse group to reach the specified growth stage. The hypotheses are \(H_0:\mu=12\) days and \(H_a:\mu<12\) days, with \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The seedlings were randomly selected, supporting the Random condition. For the 10% condition, \(0.10(240)=24\), and \(16\leq24\), so independence is reasonable. Since \(n=16<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. Use \(df=16-1=15\).
Do. The standard error is \(2/\sqrt{16}=0.5\) days. The test statistic is
Because the alternative is lower-tailed, the p-value is the area to the left of \(-0.50\). With \(df=15\), \(\operatorname{tcdf}(-1\mathrm{E}99,-0.5,15)\approx0.3121\), rounded to four decimal places. Since \(0.3121>0.05\), fail to reject \(H_0\).
Conclude. The sample does not provide convincing evidence that the true mean time for seedlings in this greenhouse group to reach the specified growth stage is less than 12 days.
This conclusion does not say that the mean time is 12 days, nor that the data support a mean greater than 12 days. It says that this lower-tailed test did not find convincing evidence for the specific claim that the mean is less than 12 days.
Worked Example: A Two-Sided Conclusion
A fictional community center randomly selects 16 lockers from a building containing 280 lockers. The sample mean interior width is \(\bar{x}=32\) centimeters, and the sample standard deviation is \(s=4\) centimeters. The center wants to know whether the true mean locker width differs from 30 centimeters. Assume the population distribution is approximately Normal. Use \(\alpha=0.05\).
State. Let \(\mu\) be the true mean interior width, in centimeters, of lockers in this building. The hypotheses are \(H_0:\mu=30\) centimeters and \(H_a:\mu\ne30\) centimeters. Use \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. Random selection supports the Random condition. For the 10% condition, \(0.10(280)=28\), and \(16\leq28\), so independence is reasonable. Since \(n=16<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The degrees of freedom are \(df=16-1=15\).
Do. The standard error is \(4/\sqrt{16}=1\) centimeter, so
The alternative is two-sided, so the p-value includes results at least as far from 0 as \(2.00\) in either direction. With \(df=15\), \(2\operatorname{tcdf}(2,1\mathrm{E}99,15)\approx0.0639\), rounded to four decimal places. Since \(0.0639>0.05\), fail to reject \(H_0\).
Conclude. The sample does not provide convincing evidence that the true mean interior width of lockers in this building differs from 30 centimeters.
Although the sample mean is above 30 centimeters, the conclusion must match the two-sided alternative: it addresses whether the mean differs, not whether it is greater. The observed direction alone is not enough to change the claim tested.
Common Mistakes and Full-Credit Wording
A conclusion can lose clarity even when the t statistic and p-value are correct. Before finishing, check that the evidence statement follows from the decision and describes the alternative in the problem’s context.
- Saying “accept \(H_0\).” A p-value above alpha does not establish that the null is true. Write “fail to reject \(H_0\)” and explain that the data do not provide convincing evidence for the alternative.
- Leaving out the claim in \(H_a\). “Reject \(H_0\)” alone is not a contextual conclusion. Complete the response by stating what the sample provides convincing evidence about, including the population and measured quantity.
- Using the wrong direction. For a lower-tailed alternative, the conclusion should say “less than”; for an upper-tailed alternative, “greater than.” For a two-sided alternative, say “differs from.”
- Claiming evidence for equality after failing to reject. Failure to reject does not show that \(\mu=\mu_0\). It only means the sample did not provide convincing evidence for the stated alternative at the chosen significance level.
- Calling a result proof. Even when \(p\leq\alpha\), the test does not prove \(H_a\). Say “provides convincing evidence,” not “proves.”
- Confusing a mean claim with an individual claim. If \(\mu\) represents a population mean, a conclusion about \(\mu\) is not a claim about every individual measurement or a specified percentage of them.
- Overstating what the p-value means. The p-value is not the probability that \(H_0\) is true, and it is not the probability that chance alone caused the result. It is calculated assuming \(H_0\) is true and measures how unusual results at least as extreme as the observed one would be.
- Equating statistical evidence with importance. A small p-value supports the alternative under the test, but it does not by itself show that the difference is large or practically important.
Check Your Understanding
For each prompt, focus on the conclusion language as well as the test decision. Use \(\alpha=0.05\) where a significance level is needed.
- A one-sample t test of \(H_a:\mu>18\) has \(p=0.021\). Write a one-sentence conclusion for a setting where \(\mu\) is the population mean waiting time, in minutes, at a clinic.
- A test of \(H_a:\mu<6\) has \(p=0.084\). What decision should be made, and what should the contextual conclusion say?
- A two-sided test of \(H_a:\mu\ne42\) rejects \(H_0\). The sample mean is 40.5. What direction should the evidence claim use: “less than,” “greater than,” or “differs from”?
- Explain why “There is a 91% chance that \(H_0\) is true” is not an appropriate conclusion when a test has \(p=0.09\).
- Rewrite this statement for a test with \(p=0.18\): “We accept the null, so the population mean equals 25.”