Paired samples

Section of Hypothesis tests, part of Statistical inference · Also in Testing examples

Paired samples

Type Section
Taught in Hypothesis tests · Testing examples
Part Statistical inference · Data analysis

In Hypothesis tests

From Hypothesis tests

Did the weighing above show that a student’s weight increases between first and third year? It did not, because different students were weighed. A better design weighs the same ten students twice.

Student 1 2 3 4 5 6 7 8 9 10
First year 71.6 69.7 50.0 56.9 60.9 77.5 53.9 49.6 32.3 61.3
Third year 83.4 77.7 57.9 67.0 69.2 87.4 63.7 59.0 39.7 71.1
Difference 11.8 8.0 7.9 10.1 8.3 9.9 9.8 9.4 7.4 9.8

The two columns are not independent samples: a heavy first-year is a heavy third-year. A two-sample t test would be wrong. Instead, let D_i be the difference for student i, and model D_1,\ldots,D_{10} as a normal random sample with mean \mu_D. Testing H_0:\mu_D=0 against \mu_D>0 is a one-sample t test on the differences.

The differences have mean 9.24 and standard deviation 1.33, so t=9.24/(1.33/\sqrt{10})=21.9 on 9 degrees of freedom, with p-value about 2\times10^{-9}. The evidence that weight increases is overwhelming. Treating the columns as independent samples gives t=1.54 instead, because the large spread between students swamps the small, consistent change within each one. Pairing is the right analysis when the two measurements share a unit, a subject or a block. The dependence it removes is a covariance: \operatorname{Var}(X-Y)=\operatorname{Var} X+\operatorname{Var} Y-2\operatorname{Cov}(X,Y).

In Testing examples

From Testing examples

Example 1 (Two labs on the same patients) Two labs measure the same quantity on the same 15 patients.

lab1 <- c(22, 18, 28, 26, 13, 8, 21, 26, 27, 29, 25, 24, 22, 28, 15)
lab2 <- c(25, 21, 31, 27, 11, 10, 25, 26, 29, 28, 26, 23, 22, 25, 17)

Question. Do the labs differ on average?

Solution. The two columns are not independent: a patient with a low value gets it from both labs, as patient 6 does with 8 and 10. So analyse the differences D_i=X_i-Y_i as one sample, D_i\sim\operatorname{N}(\delta,\sigma^2), and test H_0:\delta=0 against \delta\neq0:

t.test(lab1, lab2, paired = TRUE)
# t = -1.7618, df = 14, p-value = 0.0999

The mean difference is -0.93 with standard deviation 2.05, so t=-0.93/(2.05/\sqrt{15})=-1.76 and p=0.10. There is weak evidence, at most, that lab 2 reads higher. Ignoring the pairing and running a two-sample test gives t=-0.42 and p=0.68: the large differences between patients swamp the small, consistent difference between labs. Pairing removes the patient effect.

Used in

See also

Null and alternative hypotheses · Critical region · p-value · Type I and type II errors · Neyman–Pearson approach · Power function · One-sample t test · Two-sample t test

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