AP Statistics / Unit 4: Inference for Quantitative Data: Means / Topic 4.10
NUM8ERS study notes · Topic 4.10

Carrying Out a Test for the Difference Between Two Population Means

Your hypotheses are ready. Now measure the evidence: calculate a two-sample tt statistic, find the correct pp-value and explain what the results say about the population means.

2026–27 curriculum4 worked examples8 practice questionsVisuals + calculator guidance

By the end of this lesson, you should be able to:

  • Calculate the standard error and unpooled two-sample tt statistic.
  • Use the appropriate degrees of freedom and alternative to find a pp-value.
  • Interpret that pp-value assuming equal population means.
  • Compare pp with α\alpha and justify a conclusion in context.

Before you start: Review Topic 4.9: setting up a two-sample tt-test. You need 22 independent groups with a quantitative response and justified conditions.

First time learning this? Follow the delivery-time calculation, then compare a lower-tail test, a two-sided test and a randomized experiment.

Here to revise? Use the complete test checklist, then attempt the practice questions before opening the solutions.

The concept in 60 seconds

The null says the 22 population means are equal. The test asks how unusual your observed sample mean difference would be under that assumption.

Difference → standardized evidence → probability → conclusion. Divide the observed difference by its standard error to get tt. Use the appropriate tt distribution to find the pp-value. Compare that pp-value with the planned significance level α\alpha.

A small pp-value makes the observed result hard to explain by sampling variation under equal population means. A large pp-value means the evidence is not strong enough to reject equality at the chosen α\alpha; it does not prove the means are equal.

The direction of the alternative comes from the original question. You cannot switch to the favorable tail after seeing the data.

Finish the delivery-time test

Continue the study from Topic 4.9. Independently selected SRSs compare service A’s 2,0002{,}000 deliveries with service B’s 1,8001{,}800 deliveries during a defined period. The question is whether A has a longer population mean delivery time. Use α=0.05\alpha =0.05.

Delivery sample summaries
ServicennSample meanSample SD⁡\operatorname{SD}
A404032 min32 \,\mathrm{min}8 min8 \,\mathrm{min}
B363628 min28 \,\mathrm{min}6 min6 \,\mathrm{min}

Define μA\mu_{A} and μB\mu_{B} as the population mean delivery times for these services during that period. Test H0:μA−μB=0H_{0}: \mu_{A}-\mu_{B} = 0 against Ha:μA−μB>0H_{a}: \mu_{A}-\mu_{B} \gt 0.

The independent SRSs support random sampling and group independence. The 10%10\% checks are 40≤20040\le 200 and 36≤18036\le 180. Both sample sizes are at least 3030. We can use an unpooled two-sample tt-test.

SE⁡=8240+6236=2.6≈1.612 min\operatorname{SE}=\sqrt{\frac{8^2}{40}+\frac{6^2}{36}}=\sqrt{2.6}\approx1.612\,\mathrm{min}
t=(32−28)−02.6≈2.481t=\frac{(32-28)-0}{\sqrt{2.6}}\approx2.481

Technology gives df≈71.753\mathrm{df}\approx 71.753. The greater-than alternative uses the right-tail probability: p≈0.00773p\approx 0.00773.

Locate the observed result under the null model
Right-tail delivery comparison under equal population meansDelivery comparison under equal population mean times: t distribution with approximately 71.75 degrees of freedom. The right tail beyond observed t=2.481 is shaded, with p approximately 0.00773. −4 −2 0 2 4 Standardized t statistic (df ≈ 71.75) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Probability density Observed t ≈ 2.481
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