AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/4/4-5)
Unit 4 · Topic 4.5
4.5 Carrying Out a Test for a Population Mean or Population Mean Difference
Carrying out a one-sample or paired t-test means computing t = (x̄ − μ₀)/(s/√n) with df = n − 1, finding the p-value from a t-distribution and writing a conclusion in context about the population mean or mean difference.
Key terms
- t test statistic
- degrees of freedom
- p-value
- conclusion in context
Test statistic and p-value
t = (x̄ − μ₀) ÷ (s/√n), with df = n − 1. For paired data, use the differences: t = (x̄d − 0) ÷ (sd/√nd).
If H₀ is true, t follows a t-distribution with n − 1 degrees of freedom. The p-value is the area in the direction of Hₐ: left tail for <, right tail for >, both tails for ≠. Use tcdf, T-Test, or the t-table (which gives a range for the p-value rather than an exact value).
Interpreting the p-value
"Assuming the true mean fill volume is 500 mL, there is about a 0.021 probability of getting a sample mean of 497.8 mL or lower in a random sample of 25 bottles by chance alone." Include the null assumption, the observed statistic and "or more extreme."
Decision and conclusion
If the p-value ≤ α, reject H₀ and say there is convincing evidence for Hₐ in context. If the p-value > α, fail to reject H₀ and say there is not convincing evidence for Hₐ. Never conclude H₀ is true.
Your conclusion is the answer to the study's investigative question, stated for the population the sample came from. For a paired experiment with random assignment of treatment order, a rejection can support a cause-and-effect conclusion.
If the sample data condition isn't met (say, a small sample with a clear outlier), the t-distribution may not describe the test statistic well, and the p-value can be misleading. Say so in your answer rather than ignoring it.
Using a t-table
Table B lists t* values for common tail areas. Find the row for your df, then locate where your |t| falls between two columns; the p-value is between those columns' tail probabilities (double them for a two-sided test). If your df isn't listed, use the next smaller one.
Technology is quicker and gives an exact value, but knowing how to bracket a p-value from the table helps on multiple-choice questions.
Tests and intervals agree
A two-sided t-test at α = 0.05 rejects H₀: μ = μ₀ exactly when μ₀ is outside the 95% t-interval built from the same data. For the paired example, the 95% interval (1.60, 10.15) excludes 0, so a two-sided test at α = 0.05 would reject H₀: μd = 0. The one-sided test, with p ≈ 0.007, agrees.
Significant vs. important
In the bottle test, the evidence says the mean fill is below 500 mL, and the sample mean was about 2.2 mL low. Whether 2.2 mL matters is a business question, not a statistics question. A 95% interval for μ would show the plausible size of the shortfall, which is often more useful than the yes-or-no answer of a test.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
One-sample t-test
Continue the bottle test from 4.4: n = 25, x̄ = 497.8 mL, s = 5.1 mL. Test H₀: μ = 500 vs. Hₐ: μ < 500 at α = 0.05.
Show the solutionHide the solution
- Step 1: SE = 5.1/√25 = 1.02.
- Step 2: t = (497.8 − 500)/1.02 ≈ −2.16, df = 24.
- Step 3: p-value = P(t ≤ −2.16) ≈ 0.021.
- Step 4: 0.021 < 0.05, so reject H₀.
Answer: t ≈ −2.16, df = 24, p-value ≈ 0.021. There is convincing evidence that the true mean fill volume of today's bottles is less than 500 mL.
- Example 2Calculator allowed
Paired t-test
Finish the paired test from 4.4: differences (after − before) 12, 5, −3, 8, 10, 6, 0, 9; x̄d = 5.875, sd ≈ 5.11. Test H₀: μd = 0 vs. Hₐ: μd > 0 at α = 0.05.
Show the solutionHide the solution
- Step 1: SE = 5.11/√8 ≈ 1.807.
- Step 2: t = (5.875 − 0)/1.807 ≈ 3.25, df = 7.
- Step 3: p-value = P(t ≥ 3.25) ≈ 0.007.
- Step 4: 0.007 < 0.05, so reject H₀.
Answer: t ≈ 3.25, df = 7, p-value ≈ 0.007. There is convincing evidence that the true mean score improvement (after − before) is greater than 0 for all students at the school.
- Example 3Calculator allowed
Trap: wrong df
For the paired test, a student uses df = 14 because there are 16 scores in all. What's the correct df, and why does it matter?
Show the solutionHide the solution
- Step 1: The paired test is a one-sample test on the 8 differences, so df = 8 − 1 = 7.
- Step 2: With df = 14, the t-distribution has lighter tails, so the p-value comes out too small (about 0.003 instead of 0.007). That overstates the evidence.
Answer: df = 7. Paired data give one sample of differences.
Common mistakes
- Using df = n instead of n − 1, or counting all values instead of pairs in a paired test.
- Using a normal (z) p-value for a t statistic.
- Concluding "the mean is 500" after failing to reject H₀.
- Using a one-tailed p-value for a two-sided test.
On the exam
- Show the formula with numbers, the t value, df and the p-value. A bare calculator result is not enough for full credit.
- Link the conclusion to the p-value and α, and write it about the population mean in context.
Connected topics
Videos
Check yourself
4 questions on 4.5 Carrying Out a Test for a Population Mean or Population Mean Difference. Pick an answer to see if you got it, and why.
A one-sample t-test of H₀: μ = 50 versus Hₐ: μ ≠ 50 uses a random sample of 18 and gives t = 2.20. What is the p-value?
A test of H₀: μ = 8 versus Hₐ: μ < 8, where μ is the mean number of hours that all students at a school sleep on school nights, gives a p-value of 0.12. Which conclusion is correct at α = 0.05?
A random sample of 36 adults from a city spends a mean of 52 minutes a day exercising, with a standard deviation of 18 minutes. For a test of H₀: μ = 45, what is the test statistic?
Why is a t-distribution, rather than the standard normal distribution, used for a confidence interval for a population mean?
0 of 4 answered