Before and After Is Not Two Independent Groups

Five people each take one assessment before training and another after training. We have ten scores. Do we have ten independent observations?

No. We have five people, each observed twice. That relationship is the important part of the data.

Subtract first, infer second

Suppose the fictional scores are:

Person	Before	After	Change
A	60	64	4
B	65	70	5
C	70	71	1
D	75	81	6
E	80	85	5

The differences are $[4,5,1,6,5]$, with mean $4.2$. A paired t test is simply a one-sample t procedure on those five differences:

$$t=\frac{\bar D-\delta_0}{s_D/\sqrt n},$$

where $\delta_0$ is the mean change hypothesized under the null.

The variability between people at baseline cancels from each individual’s difference. That can improve precision substantially.

Not everything can be paired

Two unrelated groups of five people cannot be made into matched pairs by sorting the data and placing the rows side by side. Pairing is a property of the design: repeated measures on the same unit, or legitimately matched units chosen under a specified method.

The differences also need their own assumptions. In a small paired t test, we care about the distribution of the differences, not separately about the distributions of before and after scores.

Causality is still hiding in the hallway

Even if every student improves, we cannot immediately attribute improvement to the training. Practice on the test, aging, concurrent instruction, or ordinary fluctuation could explain some of the change.

An appropriately randomized comparative design can support stronger attribution. A before-and-after contrast on its own mainly describes change.

In R

before <- c(60,65,70,75,80)
after  <- c(64,70,71,81,85)
t.test(after, before, paired = TRUE)

Compare this with a one-sample test on after - before; they are algebraically equivalent.

Check your understanding

1. What is the observational unit and sample size in the paired analysis?
2. What is the sample mean change?
3. Why can a valid paired comparison still fail to identify a causal training effect?

Worked responses

1. One person contributes one difference; $n=5$.

2. 4.2 points.

3. Other changes through time may influence the results without a credible comparative design.

The covariance that pairing preserves

For paired measurements $A_i,B_i$, define $D_i=B_i-A_i$. Then $\operatorname{Var}(D)=\operatorname{Var}(B)+\operatorname{Var}(A)-2\operatorname{Cov}(A,B)$. Positive within-person covariance can make differences much less variable than the raw measurements. Negative covariance can have the opposite effect. Pairing does not guarantee a smaller standard error; its justification is the actual matching structure.

For synthetic differences $[4,5,1,6,5]$, the mean is $4.2$. Squared deviations sum to $14.8$, so $s_D^2=14.8/4=3.7$ and $SE=\sqrt{3.7/5}\approx0.86023$. A 95% t interval with four degrees of freedom is

$$4.2\pm2.776445\sqrt{0.74}\approx[1.8116,6.5884].$$

The target is the mean within-pair change in the population represented by the pairs. The normality condition for an exact small-sample paired t procedure concerns the distribution of differences, not separate normality of before and after measurements. Missing one member of a pair removes that difference from a complete-pair analysis and can introduce selection bias if completeness depends on outcomes.

A before-and-after increase is still not a randomized treatment effect. Practice effects, time trends, regression to the mean, and cointerventions may explain change. A randomized comparison of changes can address a different question than testing change in one treated group alone.

Additional practice

1. For differences $[1,2,3,4]$, calculate their mean, sample variance, and estimated standard error.
2. If all differences are zero, what happens to the usual t statistic for a zero null?
3. Why must changing the direction of subtraction reverse the estimate and interval signs consistently?

Worked additional solutions

1. Mean $2.5$, squared deviations sum $5$, variance $5/3$, and standard error $\sqrt{5/12}\approx0.64550$.
2. Both numerator and estimated standard error are zero; the displayed t ratio is undefined. Do not report a computed ratio or claim the finite sample proves a population equality.
3. Replacing $D$ by $-D$ changes the mean sign and preserves its standard error. An interval $[L,U]$ becomes $[-U,-L]$; the estimand’s direction has changed.