A Complete Test Answer Is a Connected Argument
A one-sample t test response is more than a correct calculator result. A strong free-response answer explains what population mean is being tested, why the procedure is appropriate, how the test statistic and p-value were obtained, and what the evidence says in context. The parts should form one clear argument from the research question to the conclusion.
Earlier in this course, “Writing the Full Four-Step t Interval Solution” introduced the general State, Plan, Do, Conclude structure. A test uses the same overall organization, but the details differ: the State includes hypotheses and a significance level, the Do includes a p-value, and the conclusion reports whether the evidence is convincing for the alternative. The tutorials on “Stating Hypotheses for a Population Mean,” “Complete Conditions Check for a Mean Inference Problem,” and “Writing a Conclusion for a One-Sample t Test” developed those components. Here, the goal is to assemble them into an effective complete response.
The Four Steps in a Free-Response Answer
Use the four steps as headings or as clearly signaled parts of your written solution. The headings are not a substitute for explaining your reasoning: each step should include the details that make the answer specific to the study.
Define \(\mu\) as the true population mean of the quantitative variable, including its units and target population. Write \(H_0:\mu=\mu_0\), the appropriate alternative \(H_a\), and the given significance level \(\alpha\).
Name the one-sample t test for a population mean. Check the Random condition, the 10% condition when sampling without replacement from a finite population, and the Normal/Large Sample condition. Use study details as evidence for each check.
Calculate the standard error, t statistic, and degrees of freedom. Use \(H_a\) to find the correct p-value, state it with suitable rounding, and compare it with \(\alpha\).
State “reject \(H_0\)” or “fail to reject \(H_0\).” Then describe whether the sample provides convincing evidence for the specific alternative, naming the population and variable in context.
A useful response keeps the four steps aligned. For example, the direction of \(H_a\) in State determines which tail is used in Do and which claim is addressed in Conclude. The population named in the definition of \(\mu\) should also be the population named in the conclusion. A response that shifts populations or directions partway through is not a complete argument, even if its arithmetic is correct.
Worked Examples
Worked Example: Testing Whether Seedlings Are Taller
A fictional greenhouse randomly selects 9 trays from the 120 trays in a particular growing cycle. For each selected tray, the staff record the average height of its seedlings after 21 days. The sample mean is 52 centimeters, and the sample standard deviation is 3 centimeters. Assume the population distribution of tray averages is approximately Normal. At \(\alpha=0.05\), test whether the true mean height is greater than 50 centimeters.
State. Let \(\mu\) be the true mean 21-day seedling height, in centimeters, across all trays in this growing cycle. Test \(H_0:\mu=50\) centimeters against \(H_a:\mu>50\) centimeters at \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test for a population mean. The trays were randomly selected, supporting the Random condition. They were sampled without replacement from 120 trays, and \(9\leq0.10(120)=12\), so the 10% condition is met and independence is supported. Since \(n=9<30\), the large-sample route is not met; however, the stated approximately Normal population supports the Normal/Large Sample condition. The conditions support using the one-sample t test.
Do. First calculate the standard error, test statistic, and degrees of freedom:
Because \(H_a:\mu>50\), the p-value is the area to the right of \(t=2.0000\) in a t distribution with 8 degrees of freedom. A calculator gives \(p=0.0403\), rounded. Since \(0.0403<0.05\), reject \(H_0\).
Conclude. At the 0.05 significance level, the sample provides convincing evidence that the true mean 21-day seedling height across trays in this growing cycle is greater than 50 centimeters.
Worked Example: Testing Whether Parcel Weights Fall Short
A fictional packing facility randomly selects 16 parcels from 300 parcels prepared during a shift. The sample mean weight is 246 grams, and the sample standard deviation is 8 grams. Assume parcel weights in this group are approximately Normally distributed. Test at \(\alpha=0.05\) whether the true mean weight is less than 250 grams.
State. Let \(\mu\) be the true mean parcel weight, in grams, for all parcels prepared during this shift. Test \(H_0:\mu=250\) grams against \(H_a:\mu<250\) grams at \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. Random selection supports the Random condition. Because the sample was drawn without replacement, check the 10% condition: \(16\leq0.10(300)=30\), so independence is supported. The sample size is \(16<30\), so the large-sample route is not met. The stated approximately Normal population supports the Normal/Large Sample condition. Therefore, the conditions support the test.
Do. Show the standard error and test statistic, including the direction of the difference from the null value:
For the lower-tailed alternative \(H_a:\mu<250\), use the area to the left of \(-2.0000\) for \(df=15\). The p-value is \(0.0320\), rounded. Since \(0.0320<0.05\), reject \(H_0\).
Conclude. At the 0.05 significance level, the sample provides convincing evidence that the true mean weight of parcels prepared during this shift is less than 250 grams. The conclusion follows the lower-tailed alternative; it does not claim that every parcel weighs less than 250 grams.
Worked Example: A Two-Sided Test That Does Not Reject
A fictional library randomly selects 9 evening sessions from 160 sessions held during a month. For each session, staff calculate the average customer wait time. The sample mean is 34 minutes, and the sample standard deviation is 6 minutes. Assume the population distribution of session averages is approximately Normal. At \(\alpha=0.05\), test whether the true mean differs from 30 minutes.
State. Let \(\mu\) be the true mean customer wait time, in minutes, per evening session at this library during the month. Test \(H_0:\mu=30\) minutes against \(H_a:\mu\ne30\) minutes at \(\alpha=0.05\).
Plan and check conditions. Use a one-sample t test. The sessions were randomly selected, supporting the Random condition. The 10% condition is met because \(9\leq0.10(160)=16\), so independence is supported for this sample drawn without replacement. Since \(n=9<30\), the large-sample route is not met; the stated approximately Normal population supports the Normal/Large Sample condition. The conditions support using the test.
Do. Calculate the test statistic and degrees of freedom:
The two-sided alternative counts results at least as far from zero as \(2.0000\) in either direction. With \(df=8\), the p-value is \(0.0805\), rounded. Since \(0.0805>0.05\), fail to reject \(H_0\).
Conclude. At the 0.05 significance level, the sample does not provide convincing evidence that the true mean customer wait time per evening session at this library during the month differs from 30 minutes. Failing to reject does not establish that the population mean is exactly 30 minutes.
Common Mistakes and What a Complete Answer Says
A response can lose clarity or credit by leaving out a link in the argument. Before moving on, check that the parameter definition, hypotheses, condition checks, tail choice, decision, and conclusion all refer to the same question.
- Defining the sample mean instead of the parameter. The parameter is the population mean \(\mu\), not the observed \(\bar{x}\). Define the population and the measured variable with units.
- Writing hypotheses about \(\bar{x}\). Hypotheses for this test describe the population mean. Put the null equality and the directional or two-sided claim on \(\mu\).
- Saying only “conditions are met.” Name the Random, 10%, and Normal/Large Sample conditions and give the evidence for each. If \(n<30\), do not use the large-sample route; state what supports the shape condition instead.
- Leaving out degrees of freedom or the tail choice. Show \(df=n-1\), and make clear whether the p-value comes from a lower tail, upper tail, or both tails. The sign of \(t\) alone does not determine the p-value; \(H_a\) does.
- Reporting a p-value without a decision. Compare \(p\) with the stated \(\alpha\), then explicitly say “reject” or “fail to reject” \(H_0\).
- Giving a conclusion that is too broad. State whether there is convincing evidence for the exact claim in \(H_a\), in context. Do not say the null is true after failing to reject, or claim a causal result from a random sample alone.
For an AP free-response answer, concise writing is fine as long as the evidence is visible. A statement such as “Random sample, conditions met” does not show which design fact supports randomness or how a small sample meets the shape condition. In contrast, one complete sentence per condition makes the reasoning easy to evaluate. Likewise, “\(p=0.0403\)” is not a conclusion: include the comparison with \(\alpha\), the test decision, and the population claim.
Check Your Understanding
Use the four-step structure to identify what a complete response should include.
- A random sample of 20 bags is taken from 500 bags, and the variable is the mass of the contents in grams. What information should a context-specific definition of \(\mu\) include?
- A student writes “\(H_0:\bar{x}=12\), \(H_a:\bar{x}<12\).” What should the hypotheses refer to instead?
- A sample of 14 observations is randomly selected without replacement from a population of 200. Check the 10% condition and state what it supports.
- For \(H_a:\mu\ne\mu_0\), what tail area or areas belong in the p-value?
- A test has \(p=0.08\) and \(\alpha=0.05\). Write the decision and explain what the conclusion should—and should not—claim.