4.2 - Paired-sample t-tests

4.2 - Paired-sample t-tests

Before-and-after measurements belong together when they come from the same individual. You will turn matched observations into a single sample of differences, use a one-sample t-test on those differences, and keep the sign consistent from hypotheses to conclusion.

Working with differences

If four people are measured before and after training, do you have eight independent people or four independent changes? The second description preserves the information about who improved.

In paired data, each observation in one list has a meaningful partner in the other: the same individual at two times, or two individuals deliberately matched on relevant features. Work with one difference per pair. Independence is needed between pairs; the two measurements within a pair need not be independent.

Paired difference

Choose and state an order, such as d=before−afterd=\text{before}-\text{after}, then calculate that same difference for every pair. If there are nn pairs, the resulting sample contains nn differences. Its population mean is denoted by μD\mu_D.

Subtracting removes a person's baseline from the comparison. The mean of the differences equals the difference of the two sample means, but its variance depends on the actual pairings. Sorting the lists separately can preserve their means while destroying that variance information.

Worked example 1: one change per person

Question. Four people have before measurements 12,15,11,1812,15,11,18 and corresponding after measurements 10,14,13,1510,14,13,15. With d=before−afterd=\text{before}-\text{after}, find the differences and their mean. Interpret its sign.

Plan. Subtract within each person's pair, keeping the original alignment.

d:2,1,−2,3,dˉ=2+1−2+34=1.d: 2,1,-2,3, \qquad \bar d=\frac{2+1-2+3}{4}=1.

A positive difference means the after measurement is lower.

Answer. The differences are 2,1,−2,32,1,-2,3, with mean 11: the sample shows an average reduction of 11 unit.

Check. The before and after sample means are 1414 and 1313. Their difference is 11, agreeing with the mean of the paired differences.

Change only the subtraction order. A correct reversal changes the sign but cannot change the physical evidence about the people.

Worked example 2: reversing the definition

Question. For the same paired observations, define e=after−beforee=\text{after}-\text{before}. Find the differences and their mean. How is a population reduction expressed using μE\mu_E?

Plan. Negate each original difference and translate the direction of the claim.

e:−2,−1,2,−3,eˉ=−1.e:-2,-1,2,-3, \qquad \bar e=-1.

An after measurement below its before measurement gives a negative ee, so a population mean reduction corresponds to μE<0\mu_E<0.

Answer. The differences have mean −1-1; a reduction is tested using a negative population mean difference.

Check. Each e=−de=-d, so eˉ=−dˉ\bar e=-\bar d. The average physical reduction is still 11 unit.

Either subtraction order works. Writing it explicitly prevents an improvement from being tested in the wrong tail.

A t-test on differences

Once the differences are formed, the paired t-test is a one-sample t-test on that single list.

For a small sample, assume the population of differences is normal and the pairs are independently selected. It is the distribution of differences that matters; normality of the two separate lists alone does not guarantee the required model. Calculate the unbiased variance of the differences, not a pooled variance of the original measurements.

Paired t statistic

For n≥2n\ge2 independent pairs with normally distributed differences,

dˉ=∑dn,sd2=∑d2−(∑d)2/nn−1.\bar d=\frac{\sum d}{n},\qquad s_d^2=\frac{\sum d^2-(\sum d)^2/n}{n-1}.

Under H0:μD=δ0H_0:\mu_D=\delta_0,

T=Dˉ−δ0SD/n∼tn−1.T=\frac{\bar D-\delta_0}{S_D/\sqrt n}\sim t_{n-1}.

The unbiased variance formula and t tables are in the formula booklet; the paired t statistic is not printed separately.

The usual null of no mean change has δ0=0\delta_0=0. A zero difference remains an observation in a t-test: it affects the sample mean, variance and number of pairs.

Worked example 3: six differences

Question. Six independent differences from a normal population are 1,2,3,4,5,61,2,3,4,5,6. Calculate the t statistic for a zero population mean difference and state its degrees of freedom.

Plan. Estimate the variance of this list and standardise its mean.

∑d=21,∑d2=91,dˉ=3.5.\sum d=21,\qquad \sum d^2=91,\qquad \bar d=3.5. sd2=91−212/65=3.5t=3.53.5/6=21=4.582575…\begin{aligned} s_d^2&=\frac{91-21^2/6}{5}=3.5 \\ t&=\frac{3.5}{\sqrt{3.5/6}}=\sqrt{21}=4.582575\ldots \end{aligned}

Answer. t=4.58t=4.58 to 3 significant figures, with 6−1=56-1=5 degrees of freedom.

Check. Centring at 3.53.5 gives squared deviations 6.25,2.25,0.25,0.25,2.25,6.256.25,2.25,0.25,0.25,2.25,6.25, totalling 17.517.5. Dividing by five again gives 3.53.5.

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