Confidence Intervals for Mean Paired Change

Why Estimate a Mean Paired Change?

In a paired study, each observation has a natural partner: for example, the same person measured before and after a program, or two participants matched to form a pair. The analysis focuses on the difference within each pair. A confidence interval for the population mean of those differences gives a range of plausible values for the average paired change, expressed in the original measurement units.

The interval is built from the sample mean and sample standard deviation of the differences. It therefore accounts for how much the paired changes vary from one pair to another. It is not formed by treating the two measurements in each pair as if they came from unrelated individuals.

Definition: For \(n\) pairs, let \(d_i\) be the difference for pair \(i\), using one consistent coding direction. The population mean paired difference is \(\mu_d\), and the sample mean and sample standard deviation of the differences are \(\bar{d}\) and \(s_d\). A \(100(1-\alpha)\%\) t confidence interval for \(\mu_d\) is

\bar{d}\ \pm\ t^*_{n-1}\frac{s_d}{\sqrt{n}},

where \(t^*_{n-1}\) is the positive critical value that leaves area \(\alpha/2\) in each tail of a t distribution with \(n-1\) degrees of freedom.

Here, \(n\) counts complete pairs, not the total number of individual measurements. The quantity \(s_d/\sqrt{n}\) is the standard error, \(SE\), of the sample mean difference. The margin of error is \(t^*_{n-1}SE\). The interval’s endpoints are in the same units as each \(d_i\), such as minutes, points, or millimetres of mercury.

Conditions for the Paired t Interval

The paired t interval applies a one-sample t procedure to the set of paired differences. The relevant distribution is therefore the distribution of \(d_i\), not the separate distributions of the two measurements before subtracting. The design and the differences must support inference about a population mean.

Meaningful pairing and consistent coding. Each difference must compare the two measurements that belong together, and all differences must use the same subtraction order. For example, define every change as “after minus before.”
Independent pairs or a suitable sampling design. Differences from distinct pairs should be independent, or the data should come from a design that justifies the intended inference. Dependence within a pair is expected and is handled by forming differences.
Approximately normal differences, or enough pairs for the t method to be robust. For a small sample, the differences should not show substantial skewness or influential outliers. With more pairs, the sampling distribution of \(\bar{d}\) is often closer to normal, although severe skewness, outliers, or dependence still require attention.

A graph of the differences, such as a dotplot or histogram, helps assess shape and unusual observations. No finite sample can prove that the population differences are normal. The goal is to judge whether the data and design make the t interval a reasonable approximation. If these conditions are not credible, the numerical interval should not be presented as reliable just because a formula can be evaluated.

How the Interval Is Justified

The t critical value reflects uncertainty from estimating the population standard deviation of paired differences with \(s_d\). The basic justification is the one-sample t result applied to the \(n\) differences.

Standardize the sample mean.
If the population of differences is normal, then \(\bar{d}\) has mean \(\mu_d\) and standard deviation \(\sigma_d/\sqrt{n}\), where \(\sigma_d\) is the population standard deviation of the differences.

Estimate the unknown spread.
Because \(\sigma_d\) is unknown, use \(s_d\). For independent normal differences, the resulting statistic follows a t distribution with \(n-1\) degrees of freedom.

Use the central t probability.
With \(t^*_{n-1}\) chosen to leave \(\alpha/2\) in each tail, the central area is \(1-\alpha\).

In symbols, under the normal-differences model, the statistic

T=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}

has a t distribution with \(n-1\) degrees of freedom. Thus the central t probability is

P\left(-t^*_{n-1}\leq \frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} \leq t^*_{n-1}\right)=1-\alpha.

Since \(s_d/\sqrt{n}\) is positive, multiplying through by it and rearranging the inequalities gives

P\left(\bar{d}-t^*_{n-1}\frac{s_d}{\sqrt{n}} \leq \mu_d \leq \bar{d}+t^*_{n-1}\frac{s_d}{\sqrt{n}}\right)=1-\alpha.

This rearrangement gives the interval formula. Under approximate rather than exact normality, its coverage is approximate; the conditions and diagnostics matter because they support that approximation.

Worked Examples

Worked Example: Mean Reduction in Completion Time

Suppose, in a hypothetical study, eight participants each complete a task before and after a brief practice session. Define \(d_i\) as “before time minus after time,” so a positive difference means a reduction in time. The eight differences, in minutes, are \(4, 2, 5, 1, 3, 6, 0,\) and \(3\). Construct a 95% confidence interval for the population mean paired reduction.

Check the conditions. The same participants supply both times, so the measurements are correctly paired. Assume the eight participants were sampled independently from the target population. The differences range from 0 to 6 minutes, with no isolated extreme value; for this small sample, a dotplot should be inspected and the population differences should be reasonably close to normal. These conditions are plausible for the hypothetical example, though the small sample makes the shape assessment important.

Calculate the sample summaries. The sum of the differences is 24 minutes, so

\bar{d}=\frac{24}{8}=3.00\text{ minutes}.

The squared deviations from 3 are \(1,1,4,4,0,9,9,0\), which sum to 28. Therefore,

s_d=\sqrt{\frac{28}{8-1}}=2.00\text{ minutes}, \qquad SE=\frac{s_d}{\sqrt{n}}=\frac{2.00}{\sqrt{8}}=0.7071\text{ minutes}.

For a 95% interval, \(df=n-1=7\), and the t critical value is approximately \(t^*_7=2.365\). The margin of error is

2.365(0.7071)=1.672\text{ minutes}.

The interval is

3.00\pm1.672=(1.328,\ 4.672)\text{ minutes}.

We are 95% confident that the population mean reduction in completion time is between about 1.33 and 4.67 minutes. The positive endpoints correspond to reductions because of the chosen “before minus after” coding.

Worked Example: Mean Change in Quiz Score

In a hypothetical sample of 16 students, each student takes a quiz before and after a review session. Define \(d_i\) as “after score minus before score,” measured in points. The differences have sample mean \(\bar{d}=-1.80\) points and sample standard deviation \(s_d=3.20\) points. Construct a 90% confidence interval for the population mean paired change.

Check the conditions. Each after score is matched to that same student’s before score. Assume the 16 students’ differences are independent and that the sample was obtained in a way appropriate to the target population. Suppose a plot of the differences shows no severe skew or influential outlier. Those checks support using a t interval, while the inference remains limited to the population represented by the sampling design.

The number of pairs is \(n=16\), so \(df=15\). For a 90% interval, \(t^*_{15}\) is approximately 1.753. The standard error and margin of error are

SE=\frac{3.20}{\sqrt{16}}=0.800\text{ points}, \qquad ME=1.753(0.800)=1.402\text{ points}.

Thus,

-1.80\pm1.402=(-3.202,\ -0.398)\text{ points}.

We are 90% confident that the population mean change in quiz score, defined as after minus before, is between about \(-3.20\) and \(-0.398\) points. Both endpoints are negative, indicating a mean decrease under this coding. The interval estimates an average change; it does not say that every student’s score decreased by an amount in this range.

Worked Example: Mean Paired Change in Systolic Blood Pressure

A hypothetical study records systolic blood pressure before and after a program for 25 participants. Define \(d_i\) as “before minus after,” in millimetres of mercury (mm Hg). The sample differences have \(\bar{d}=12.4\) mm Hg and \(s_d=8.0\) mm Hg. Construct a 95% confidence interval for the population mean reduction.

Check the conditions. The two readings are paired within participant. Assume the participants’ differences are independent and the sampling design supports the intended population inference. With 25 pairs, inspect the difference distribution for severe skew or extreme outliers; suppose the hypothetical plot is roughly unimodal without an influential outlier. These observations make the t approximation reasonable, but do not establish that the program caused any change.

Here \(n=25\), so \(df=24\), and the 95% critical value is \(t^*_{24}\approx2.064\). Then

SE=\frac{8.0}{\sqrt{25}}=1.60\text{ mm Hg}, \qquad ME=2.064(1.60)=3.302\text{ mm Hg}.

The interval is

12.4\pm3.302=(9.098,\ 15.702)\text{ mm Hg}.

We are 95% confident that the population mean reduction in systolic blood pressure is between about 9.10 and 15.7 mm Hg. The statement concerns the mean of the paired reductions, not a prediction for an individual participant.

Interpreting Confidence and Avoiding Common Errors

A 95% confidence level describes the long-run performance of the method: if the same sampling process were repeated many times and a 95% interval constructed each time, approximately 95% of those intervals would contain the fixed population mean paired difference. Once an interval has been calculated, it either contains that parameter or it does not. The confidence level is not a probability that the parameter lies in this particular interval.

The endpoints should be read in context and with units. If differences are coded “after minus before,” a positive mean indicates an increase; if coded “before minus after,” a positive mean indicates a decrease. Changing the subtraction order changes the signs of the sample mean and both endpoints, but not the underlying information. State the coding so the direction is unambiguous.

A frequent mistake is to calculate a two-sample interval from the before measurements and the after measurements separately. That approach ignores the pairing: it does not use the within-person differences and their variability. The correction is to calculate one difference per complete pair, then use \(\bar{d}\), \(s_d\), and the number of pairs \(n\) in the paired t interval. Another mistake is to use \(n-1\) as the number of pairs; it is the degrees of freedom, while the standard error divides by \(\sqrt{n}\).

Key takeaway: A paired t confidence interval estimates \(\mu_d\) using the mean and standard deviation of the within-pair differences. Its center is \(\bar{d}\), its margin of error is \(t^*_{n-1}s_d/\sqrt{n}\), and both endpoints retain the units of the original paired measurements.

Check Your Understanding

Answer each question using the paired-difference framework.

A study has 12 complete pairs. What are the sample size used in the standard error and the degrees of freedom for the t critical value?
Differences are defined as “after minus before.” What does a positive value of \(\bar{d}\) indicate?
A paired sample has \(\bar{d}=2.5\), \(s_d=4.0\), and \(n=16\). What is the standard error of \(\bar{d}\)?
Why should a plot of the differences be examined instead of relying only on separate plots of the before and after measurements?
In repeated sampling, what does a 95% confidence level mean?