AP Statistics · Lesson 3

Paired and two-sample means

Know Know it

First decide how the data were made. Paired data come in matched pairs: the same individual measured twice, or two similar individuals split between treatments. Two independent samples are separate groups with no link between a value in one group and a value in the other.

Paired t: compute each pair's difference, in a stated order such as after minus before. Then run a one-sample t procedure on the differences: xˉd±t∗sdn\bar{x}_d \pm t^* \frac{s_d}{\sqrt{n}} or t=xˉd−0sd/nt = \frac{\bar{x}_d - 0}{s_d / \sqrt{n}}, with n = number of pairs and df = n - 1. The parameter is μd\mu_d, the true mean difference. Check the conditions on the differences.

Two-sample t: for μ1−μ2\mu_1 - \mu_2, use (xˉ1−xˉ2)±t∗s12n1+s22n2(\bar{x}_1 - \bar{x}_2) \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}, or t=xˉ1−xˉ2s12/n1+s22/n2t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} for H0:μ1−μ2=0H_0: \mu_1 - \mu_2 = 0. Take df from technology, or use the smaller of n1−1n_1 - 1 and n2−1n_2 - 1 as a cautious choice. Check Random and Normal/Large Sample for each group, and 10% for each sample.

Paired or two-sample?
DesignProceduredf
Each of 20 runners runs in shoe A and in shoe BPaired t on 20 differences19
Twins, one twin in each programPaired t on the twin differencespairs - 1
30 phones of model A and 35 different phones of model BTwo-sample tfrom technology

Trap Do not treat paired data as two samples

If the same people are measured twice, the two sets of values are not independent, so a two-sample procedure does not fit. It also ignores the pairing: when people differ a lot from one another, it leaves that variation in the standard error, which usually makes the interval wider and the test weaker than it should be. Work with the differences.

Worked example Easy

Twenty runners from a fictional club each run 400 meters once in shoe A and once in shoe B, in a random order, on the same track.

Which procedure fits a comparison of the mean times in the two shoes?

  1. A paired t procedure on the 20 differences, shoe A minus shoe BAnswer
  2. A two-sample t procedure on the 20 shoe A times and the 20 shoe B times
  3. A one-sample t procedure on the 20 shoe A times alone
  4. A two-sample z procedure for proportions comparing the shoes

How to solve it

  1. The same runner is measured in both shoes, so the data are paired.
  2. Compute 20 differences, shoe A minus shoe B.
  3. Run one-sample t procedures on the differences, with df = 19.

Why each choice is right or wrong

  • A. Correct. Each runner gives one time in each shoe, so the times come in pairs, and the analysis uses each runner's difference.
  • B. Incorrect. A two-sample procedure treats the shoe groups as unrelated, but the same 20 runners are in both, so the data are paired.
  • C. Incorrect. Using the shoe A times alone ignores shoe B, so it cannot compare the two shoes at all.
  • D. Incorrect. Run times are quantitative, so a procedure for proportions does not fit; the data call for t procedures on means.

Explain Explain it

Task verbs: justify, calculate

Justify the choice of procedure from the design, not from the numbers: say what makes the values pairs (same subject, matched subjects) or what makes the groups independent (separate random samples, or random assignment to separate groups).

Why pairing helps: people differ a lot from one another. A difference within one person cancels out that person's own level, so the differences vary much less than the raw values, and the standard error of the mean difference is smaller.

Worked example Medium

Twelve typists chosen at random from a large office each type one passage on a new keyboard and one on their old keyboard. For each typist, the difference is new minus old, in words per minute. The 12 differences have mean 1.8 and standard deviation 2.4, with no strong skew or outliers.

For a paired t test of H0:μd=0H_0: \mu_d = 0, what are the test statistic and its degrees of freedom?

  1. t≈0.75t \approx 0.75, df = 11
  2. t≈2.60t \approx 2.60, df = 22
  3. t≈2.60t \approx 2.60, df = 11Answer
  4. t≈9.00t \approx 9.00, df = 11

How to solve it

  1. The data are paired, so work with the 12 differences.
  2. Standard error: 2.412≈0.693\frac{2.4}{\sqrt{12}} \approx 0.693.
  3. t=1.80.693≈2.60t = \frac{1.8}{0.693} \approx 2.60 with df = 11.

Why each choice is right or wrong

  • A. Incorrect. 0.75 divides the mean difference by sd=2.4s_d = 2.4 itself instead of by the standard error 2.412\frac{2.4}{\sqrt{12}}.
  • B. Incorrect. The statistic 2.60 is right, but df = 22 treats the data as two samples; paired data have 12 differences, so df = 11.
  • C. Correct. t=1.8−02.4/12≈2.60t = \frac{1.8 - 0}{2.4/\sqrt{12}} \approx 2.60, and 12 pairs give df = 12 - 1 = 11.
  • D. Incorrect. 9.00 divides sds_d by 12 instead of by 12\sqrt{12}, which makes the standard error too small.

Justify Your turn

A researcher measures the reaction time of each of 15 volunteers once with caffeine and once without, on two days in a random order. Justify using paired t procedures here, and say why pairing gives a more precise comparison than two separate groups would.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says each volunteer is measured under both conditions, so the two sets of times are paired, and the analysis uses each volunteer's difference
  • Explains that pairing removes the variation between volunteers, so the standard error of the mean difference is smaller

Model answer

Paired t procedures fit because the same 15 volunteers are timed both with and without caffeine, so the two times for a volunteer belong together. We compute each volunteer's difference, with caffeine minus without, and analyze those 15 differences. Pairing is more precise because some people are simply faster than others; taking a difference within each person cancels that person-to-person variation, so the differences vary less and the standard error of the mean difference is smaller.

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