How do you test an average and conclude in context?
You will be able to: Calculate t and p, then link a qualified conclusion to α.
How do you test an average and conclude in context?
For 25 randomly selected delivery times, x̄=22 minutes and s=5 minutes. A service claims a population mean of 20 minutes; the investigation asks whether it is higher.
A useful starting point: Which evidence makes a t procedure defensible? →
Words and symbols before equations
- Test statistic t
- Observed difference from the null, measured in estimated standard errors.
- p-value
- Null-model probability of a statistic at least as extreme in the alternative’s direction.
- Significance level α
- A prespecified rejection threshold.
- Fail to reject
- Insufficient evidence for the alternative, not proof of the null.
What this picture assumes
Synthetic summaries, with positive sample SDs. Random samples without replacement use source populations of N=100000 each, satisfying 10% for these sizes. A randomized experiment does not require that sampling-fraction check. Design and shape are assumptions chosen here, not conclusions that summary statistics can verify. Strong skewness/outlier selection conservatively withholds inference at all sizes; real data require examination.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- t=2, df=24, p=0.05694. At α=0.05, fail to reject H₀. Chosen design, sampling fraction where applicable, and shape/size assumptions support the procedure.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Assume a suitable sample shape from a much larger population. H₀:μ=20 and Hₐ:μ>20. SE=5/√25=1, so t=(22−20)/1=2 with df=24.
The upper-tail t probability is about 0.0285. Under a 20-minute population mean and the model assumptions, statistics at least this large occur about 2.85% of the time.
At α=0.05 reject H₀ because 0.0285<0.05. The data provide evidence that the population mean delivery time exceeds 20 minutes. This does not mean the probability H₀ is true is 2.85%.
A worked example, step by step
Use the same t=2,df=24 but ask whether the mean differs from 20.
- The alternative is now two-sided, prespecified as μ≠20.
- The statistic stays t=2.
- The two-sided p-value is about 2×0.0285=0.0570.
- At α=.05 fail to reject; there is not convincing evidence of a difference for that two-sided question.
A positive t does not automatically imply rejection. Tail direction, df and α matter.
Is p the probability the alternative is false?
Compare with an explanation
No. It is a probability calculated under H₀.
Predict. Change one thing. Explain.
Keep the data and alternative fixed, then change α. Explain why the p-value stays fixed while the decision may change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t=2, df=24, p=0.05694. At α=0.05, fail to reject H₀. Chosen design, sampling fraction where applicable, and shape/size assumptions support the procedure.
| Quantity | Value |
|---|---|
| Estimate (minutes) | 22 |
| SE (minutes) | 1 |
| Degrees of freedom | 24 |
Synthetic summaries, with positive sample SDs. Random samples without replacement use source populations of N=100000 each, satisfying 10% for these sizes. A randomized experiment does not require that sampling-fraction check. Design and shape are assumptions chosen here, not conclusions that summary statistics can verify. Strong skewness/outlier selection conservatively withholds inference at all sizes; real data require examination.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the means, standard errors, pairing, graph scales or model assumptions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionWith conditions verified, n=16,x̄=14,s=4, test H₀:μ=12 versus μ>12. A t calculator gives p≈.0320.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: SE=4/√16=1; df=15.
- 1 point: t=(14−12)/1=2.
- 1 point: Under μ=12 the upper-tail probability is about .0320.
- 1 point: At α=.05 reject, supporting a population mean above 12 in the response units.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What determines the tail?
The prespecified alternative.
RECALL 2Can α change p for fixed data and hypotheses?
No.
RECALL 3What must a conclusion name?
The parameter, population, context and strength of evidence.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you test an average and conclude in context?
- t=(x̄−μ₀)/(s/√n).
- df=n−1.
- Reject when p≤α; otherwise fail to reject.
Remember: A positive t does not automatically imply rejection. Tail direction, df and α matter.
Conditions: Synthetic summaries, with positive sample SDs. Random samples without replacement use source populations of N=100000 each, satisfying 10% for these sizes. A randomized experiment does not require that sampling-fraction check. Design and shape are assumptions chosen here, not conclusions that summary statistics can verify. Strong skewness/outlier selection conservatively withholds inference at all sizes; real data require examination.
Refresh Kid · AP Statistics Unit 4 · Objectives 4.5.A, 4.5.B, 4.5.C · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 4.5, objectives 4.5.A, 4.5.B, 4.5.C. Framework effective Fall 2026, checked September 17, 2026. Unit 4 includes sampling distributions of means, one-sample and paired t inference, and independent two-sample t inference; it is part of the revised five-unit course.
Examples and datasets are synthetic, independently authored teaching material. Mean inference requires a justified design and suitable shape or sample size. Paired analysis uses one sample of differences. Independent two-sample inference uses separate variance estimates and technology-computed Welch degrees of freedom. Extreme skewness and influential observations need attention even in larger samples. This model conservatively withholds inference when those warnings are selected. Conclusions are limited by random sampling and/or assignment as appropriate.
The Organic Chemistry Tutor companion title and destination were located; the full video was not reviewed. Khan Academy’s destination was checked, but its lesson content was not fully readable by the research tool. OpenStax provides optional reference reading. No provider scripts, questions or graphics were copied. Refresh Kid is not affiliated with these providers.
GitHub’s 3D website collection informed optional spatial inspection. Our original paired-data display uses self-hosted Three.js with its MIT license. Two measurement columns are connected within each labeled student lane; horizontal position is before/after, vertical position is time, and depth separates identities rather than representing a numerical variable. Rotation can separate overlapping connectors. Exact values and differences always remain in the 2D table and labeled plot. The broken-matching option is an explicit counterexample, not a legitimate alternative analysis. No autoplay; complete teaching remains available without 3D.
Independent teacher review and observation of students remain pending. Technical checks do not certify statistical accuracy, accessibility or learning effectiveness. This is a review edition.
Released AP Statistics questions and scoring guides are optional. Older exams use the earlier framework, so check alignment before selecting parts. All practice on this page is original, not official AP material.
Learn → Explore → Practice → Review is informed by the IES learning guide. This implementation has not yet been evaluated with learners.
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