Why a Paired t Interval Uses Differences
When measurements are taken in matched pairs, the quantity of interest is usually the average difference within a pair. Examples include two measurements on the same person or measurements on two deliberately matched items. The pairing matters because the observations in a pair are connected. For inference about the mean difference, the data can instead be represented by one difference per pair.
The paired t interval follows directly from the one-sample t result applied to those differences. This connection explains both the standard error and why the degrees of freedom depend on the number of pairs, not on twice that number.
The sign convention must be set before calculating: here the difference is the first measurement minus the second. The interval estimates the mean under that convention. Reversing the subtraction reverses every difference and the interval's endpoints.
Model and Assumptions
The exact paired t result requires the differences—not necessarily the two measurements considered separately—to form a random sample of independent, identically distributed normal random variables:
The pair is the sampling unit, so the independence assumption applies across pairs. The two measurements within a pair may be related; that relationship is one reason to work with their difference. Exact normal-theory inference requires the differences to be normally distributed. If this condition is not reasonable, the formula below is not an exact t interval merely because there are two observations in each pair.
Once the differences have been formed, the paired problem is a one-sample problem. There are \(n\) difference observations, their sample mean is \(\bar D\), and their sample standard deviation is \(S_D\). The one-sample t result therefore applies to this sample of differences.
Deriving the Paired t Pivot
Under the stated model, the sample mean difference has distribution \(\bar D\sim N(\mu_D,\sigma_D^2/n)\). Thus, if the population standard deviation \(\sigma_D\) were known, standardizing \(\bar D\) would give a standard normal variable. In practice, \(\sigma_D\) is unknown and is estimated by \(S_D\). The one-sample t result states that the corresponding studentized statistic has a Student t distribution with \(n-1\) degrees of freedom.
For clarity, the result uses two facts for a normal sample: \((\bar D-\mu_D)/(\sigma_D/\sqrt n)\) is standard normal, and \((n-1)S_D^2/\sigma_D^2\) has a chi-square distribution with \(n-1\) degrees of freedom and is independent of the standardized sample mean. Dividing the standard normal quantity by the square root of the independent chi-square quantity divided by its degrees of freedom gives a Student t variable. Substituting those quantities gives the paired pivot:
Use the declared order, such as \(D_i=X_i-Y_i\), for every pair. This produces \(n\) observations, not \(2n\).
Find \(\bar D\) and \(S_D^2\), using \(n-1\) in the denominator of the sample variance.
Use \(S_D/\sqrt n\) as the estimated standard error and compare the resulting statistic with a t distribution on \(n-1\) degrees of freedom.
The degrees of freedom are \(n-1\) because the variance is calculated from the \(n\) differences after estimating their mean. Once the deviations from \(\bar D\) are specified, they sum to zero, leaving \(n-1\) free deviations. The number of original measurements does not replace \(n\): two measurements in each pair produce one difference and one contribution to the sample of differences.
From the Pivot to a Confidence Interval
Let \(0<\alpha<1\), and let \(t_{1-\alpha/2,n-1}\) denote the \(1-\alpha/2\) quantile of the Student t distribution with \(n-1\) degrees of freedom. By the pivot's distribution, the probability that it falls between the two central critical values is \(1-\alpha\). Rearranging that probability statement to isolate \(\mu_D\) gives the paired t confidence interval:
The interval is centered at the observed mean difference. Its margin of error is the t critical value multiplied by the estimated standard error. Both the standard error and the degrees of freedom come from the difference sample.
Worked Example: Checking a Pivot and a 95% Interval
In a hypothetical study, four independent pairs produce differences \(-2,0,1,3\), measured in a specified score unit, with each difference defined as the first score minus the second. Assume the differences are independent observations from \(N(\mu_D,\sigma_D^2)\). Calculate the observed pivot for the reference value \(\mu_D=0\), then calculate a 95% confidence interval for \(\mu_D\). This calculation illustrates the method and is not a claim about an actual study.
Method. Treat the four differences as a one-sample normal data set. First calculate the mean:
The deviations from the mean are \(-2.5,-0.5,0.5,2.5\); their squared values are \(6.25,0.25,0.25,6.25\), with sum \(13\). Therefore:
For the reference value \(\mu_D=0\), the observed pivot is:
The degrees of freedom are \(4-1=3\). For a two-sided 95% interval, use \(t_{0.975,3}=3.182\). The margin of error is \(3.182(1.0408)\approx3.3119\), so the interval is \(0.5\pm3.3119\), or \((-2.8119,3.8119)\). The endpoints are rounded to four decimal places. The interval contains zero, and the calculated pivot is correspondingly between \(-3.182\) and \(3.182\).
Common mistake avoided. The squared deviations sum to \(13\), not \(14\). Using \(14\) would give the wrong sample variance, standard error, observed pivot, and interval. Rechecking each squared deviation prevents that arithmetic error from propagating.
Worked Example: Constructing an Interval from Five Paired Differences
Suppose five hypothetical matched pairs of components are measured under two inspection procedures. Define each difference as the first procedure's reading minus the second procedure's reading. The differences are \(2,4,3,5,6\) reading units. Assume these five differences are independent and normally distributed with common mean \(\mu_D\) and variance \(\sigma_D^2\). Find a 90% confidence interval for the mean difference.
Method. Calculate the sample mean:
The deviations from \(4\) are \(-2,0,-1,1,2\), whose squares sum to \(4+0+1+1+4=10\). Thus:
There are \(5-1=4\) degrees of freedom. For a central 90% interval, \(t_{0.95,4}=2.132\). The margin of error is \(2.132(0.7071)\approx1.5076\), giving:
The endpoints are rounded to four decimal places. As a check, the interval is centered at \(4\), and its half-width is the critical value times the standard error. The calculation uses the number of pairs, five, rather than counting ten individual readings as ten independent differences.
Common mistake avoided. The sample variance denominator is \(n-1=4\), not \(n=5\). Dividing the squared-deviation sum by five would underestimate the sample variance and produce an incorrectly small standard error and margin of error.
Worked Example: Applying the Degrees of Freedom Correctly
A hypothetical program records paired changes for six independently selected participants. Each difference is the second measurement minus the first, in change units. The differences are \(1,3,2,4,0,2\). Under the model that these differences are independent and normally distributed with common mean and variance, calculate a 95% confidence interval for the mean change.
Method. The differences sum to \(12\), so \(\bar D=12/6=2\). Their deviations from \(2\) are \(-1,1,0,2,-2,0\), and the squared deviations sum to \(1+1+0+4+4+0=10\). Hence:
The degrees of freedom are \(6-1=5\), so the 95% critical value is \(t_{0.975,5}=2.571\). The margin of error is \(2.571(0.5774)\approx1.4844\). Therefore the interval is:
The endpoints are rounded to four decimal places. The calculation is internally consistent: the six differences give five degrees of freedom, and the interval's midpoint is the sample mean, \(2\).
Common mistake avoided. Using \(11\) degrees of freedom because there are twelve original measurements would not apply the one-sample t result to the correct data. The analysis uses six differences, so it has five degrees of freedom.
Assumption Checks and Exam Traps
The derivation depends on the distribution of the differences. Normality of the two measurements separately is not the statement required here; the relevant model is that the \(D_i\) are independent and normally distributed with a common mean and variance. Conversely, pairing alone does not guarantee that the differences follow this model. Always state the model before calling the interval exact.
- Count pairs, not measurements. With \(n\) pairs there are \(n\) differences, the standard error is \(S_D/\sqrt n\), and the degrees of freedom are \(n-1\).
- Use the sample-variance denominator. Calculate \(S_D^2\) by dividing the squared deviations from \(\bar D\) by \(n-1\), not \(n\).
- Keep the difference order consistent. If \(D_i=X_i-Y_i\), apply that same order to every pair before calculating the mean and variance.
- Do not confuse a t pivot with a known-variance z statistic. The paired t pivot uses the sample standard deviation of the differences because the population standard deviation is unknown.
A useful final check is to reconstruct the interval from its components: the center must be \(\bar D\), the half-width must equal \(t_{1-\alpha/2,n-1}S_D/\sqrt n\), and the degrees of freedom must be \(n-1\). These checks catch many calculation and setup errors before interpretation.
Check Your Understanding
For each problem, use the paired-difference model as stated and retain enough intermediate precision to round only the final interval endpoints.
- Four independent matched pairs yield differences \(1,2,2,3\), defined consistently as first measurement minus second. Assume the differences are normal. Calculate \(\bar D\), \(S_D^2\), the standard error, and the 95% interval for \(\mu_D\). Use \(t_{0.975,3}=3.182\). (Answer: \(\bar D=2\), \(S_D^2=2/3\), standard error \(=\sqrt{1/6}\approx0.4082\), degrees of freedom \(3\), interval approximately \((0.7010,3.2990)\).)
- Five independent paired differences are \(0,2,4,2,2\). Assume they are normal. Calculate the observed pivot for the reference value \(\mu_D=1\), and state its degrees of freedom. (Answer: \(\bar D=2\), \(S_D^2=2\), standard error \(=\sqrt{0.4}\approx0.6325\), \(t_{\mathrm{obs}}\approx1.5811\), degrees of freedom \(4\).)
- Six independent paired differences are \(2,4,3,5,4,6\), and are normally distributed. Calculate the 90% interval for \(\mu_D\), using \(t_{0.95,5}=2.015\). (Answer: \(\bar D=4\), \(S_D^2=2\), standard error \(=\sqrt{1/3}\approx0.5774\), margin of error \(\approx1.1634\), interval approximately \((2.8366,5.1634)\).)
- Seven independent paired differences are \(1,2,3,4,5,4,2\), with a normal model. Calculate the 95% interval for \(\mu_D\), using \(t_{0.975,6}=2.447\). (Answer: \(\bar D=3\), \(S_D^2=2\), standard error \(=\sqrt{2/7}\approx0.5345\), margin of error \(\approx1.3080\), degrees of freedom \(6\), interval approximately \((1.6920,4.3080)\).)
- Five independent paired differences are \(-1,1,2,2,6\). Assume they are normal. Find the observed pivot for \(\mu_D=1\) and the degrees of freedom. (Answer: \(\bar D=2\), \(S_D^2=6.5\), standard error \(=\sqrt{1.3}\approx1.1402\), \(t_{\mathrm{obs}}\approx0.8771\), degrees of freedom \(4\).)