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Exact t Intervals For Paired Mean Differences

Learn how to calculate an exact confidence interval for a population mean paired difference and check whether the model assumptions support it.

Intermediate 9 min read

What You'll Learn

  • Define paired differences consistently and identify the number of independent pairs.
  • Calculate the sample standard deviation and standard error of the differences.
  • Choose the Student t critical value using the correct degrees of freedom.
  • Construct and report an exact confidence interval under the normal-differences model.
  • Check when the exact procedure is not defined and avoid using the wrong critical value.

Why an Interval for Paired Differences?

When each unit contributes two related measurements, the quantity of interest is often the average change within a unit, not the difference between two unrelated group averages. Examples include paired laboratory readings on the same specimen or measurements on the same machine before and after a hypothetical adjustment. The analysis reduces each pair to one difference, then estimates the population mean of those differences.

This tutorial focuses on constructing the interval from the sample size, sample mean, and sample standard deviation of the differences. The exact Student t procedure depends on a specific model: the differences must be an independent sample from a normal population. Pairing by itself does not guarantee that model.

Definition: Let \(D_i\) be the difference for pair \(i\), defined in a stated order, such as first measurement minus second measurement. For \(n\) pairs, the sample mean difference and sample standard deviation are
$$ \bar{D}=\frac{1}{n}\sum_{i=1}^{n}D_i, \qquad s_D=\sqrt{\frac{\sum_{i=1}^{n}(D_i-\bar{D})^2}{n-1}}. $$
The population mean difference is denoted by \(\mu_D\).

The Exact Paired t Interval

Assume \(D_1,\ldots,D_n\) are independent and identically distributed as \(N(\mu_D,\sigma_D^2)\), where both population parameters are unknown. For \(n\ge 2\), calculate \(\bar{D}\) and \(s_D\) from the \(n\) differences. The two-sided confidence interval with confidence level \(1-\alpha\) is

Theorem: Under the independent normal-differences model, an exact \(100(1-\alpha)\%\) confidence interval for \(\mu_D\), for every sample size \(n\ge 2\), is
$$ \bar{D}\ \pm\ t_{1-\alpha/2,\,n-1}\frac{s_D}{\sqrt{n}}, $$
where \(t_{1-\alpha/2,\,n-1}\) is the \(1-\alpha/2\) quantile of a Student t distribution with \(n-1\) degrees of freedom. The interval is not defined for \(n=1\): there are zero degrees of freedom, and \(s_D\) cannot be calculated.

The term \(s_D/\sqrt{n}\) estimates the standard deviation of the sample mean difference. The critical value depends on the confidence level and on \(n-1\), not on the number of individual measurements considered separately. The paired data contribute \(n\) differences, so the sample size for this interval is the number of pairs.

Why the Student t Critical Value Works

The normality assumption applies to the differences \(D_i\). Under that assumption, the sample mean is normal, and the sample variance provides an independent estimate of the unknown variance. These facts produce a Student t pivot.

1
Standardize the sample mean.
For independent normal differences, \((\bar{D}-\mu_D)/(\sigma_D/\sqrt{n})\) has a standard normal distribution.
2
Account for the unknown standard deviation.
The sample variance satisfies \((n-1)s_D^2/\sigma_D^2\sim\chi^2_{n-1}\), and this quantity is independent of \(\bar{D}\).
3
Form the pivot.
Dividing the standardized mean by the square root of the independent scaled chi-square quantity divided by its degrees of freedom gives a Student t variable:
$$ T=\frac{\bar{D}-\mu_D}{s_D/\sqrt{n}}\sim t_{n-1}. $$

By the definition of the two-sided t quantile,

$$ P\!\left(-t_{1-\alpha/2,n-1}\le T\le t_{1-\alpha/2,n-1}\right)=1-\alpha. $$

Substituting the expression for \(T\), then rearranging the inequalities to isolate \(\mu_D\), gives

$$ P\!\left( \bar{D}-t_{1-\alpha/2,n-1}\frac{s_D}{\sqrt{n}} \le\mu_D\le \bar{D}+t_{1-\alpha/2,n-1}\frac{s_D}{\sqrt{n}} \right)=1-\alpha. $$

This establishes exact coverage under the stated model for every \(n\ge 2\). “Exact” does not mean that the interval has this coverage for arbitrary difference distributions; the normality and independence assumptions matter.

Calculating the Interval from Observed Differences

Worked Example: Calculate a 95% Interval from Five Paired Differences

A researcher compares a hypothetical sensor’s two readings on each of five test objects and defines \(D_i=\) reading A minus reading B. The observed differences, in measurement units, are \(2,4,6,8,10\). Find a 95% confidence interval for the population mean difference. Treat the differences as an independent sample from \(N(\mu_D,\sigma_D^2)\), with both parameters unknown. This is a measurement example, so no interest rate or mortality assumption applies.

Method and model check. Use the one-sample paired-difference t interval. The five differences, rather than the ten readings, are the observations for the calculation. The exact interval requires the differences to be independent and normally distributed.

Step 1: Calculate the mean. The sum is \(2+4+6+8+10=30\), so \(\bar{D}=30/5=6\).

Step 2: Calculate the sample standard deviation. The deviations from 6 are \(-4,-2,0,2,4\); their squared values sum to \(16+4+0+4+16=40\). Thus \(s_D^2=40/(5-1)=10\), and \(s_D=\sqrt{10}=3.1623\).

Step 3: Find the standard error and critical value. There are \(n-1=4\) degrees of freedom. For a 95% interval, \(t_{0.975,4}=2.776\). The standard error is \(s_D/\sqrt{n}=\sqrt{10}/\sqrt{5}=\sqrt{2}=1.4142\).

Step 4: Calculate the margin and endpoints. The margin is \(2.776(1.4142)=3.926\). Therefore, the interval is \(6\pm3.926\), or \((2.074,9.926)\), rounded to three decimal places.

Check. The interval is centered at 6, and its half-width is approximately 3.926, matching the calculated margin. The common mistake avoided here is using ten readings as the sample size: the sample consists of five paired differences.

Worked Example: An Interval That Includes Zero

In a hypothetical evaluation of a calibration procedure, each of ten instruments is measured under two settings. Define \(D_i=\) setting 1 reading minus setting 2 reading. The summary statistics are \(n=10\), \(\bar{D}=-3.2\) units, and \(s_D=4.5\) units. Construct a 95% confidence interval for \(\mu_D\). Assume the differences are independent and normally distributed as \(N(\mu_D,\sigma_D^2)\), with unknown parameters. No interest rate or mortality assumption applies to these measurement data.

Method. Apply the paired-difference t interval using the supplied summary statistics. The direction of the difference is already specified, so the estimate is negative when setting 1 tends to give lower readings.

The degrees of freedom are \(10-1=9\), and \(t_{0.975,9}=2.262\). The standard error is

$$ \frac{s_D}{\sqrt{n}}=\frac{4.5}{\sqrt{10}}=1.4230. $$

The margin of error is \(2.262(1.4230)=3.219\). The endpoints are \(-3.2-3.219=-6.419\) and \(-3.2+3.219=0.019\). Thus the 95% interval is \((-6.419,0.019)\) units, rounded to three decimal places.

Check. The midpoint is \((-6.419+0.019)/2=-3.200\), as required up to rounding. The interval reaches slightly above zero; that is consistent with the computed endpoints and should not be changed by rounding an intermediate value.

Confidence Level, Degrees of Freedom, and Precision

For a fixed sample, a higher confidence level requires a larger t critical value and therefore a wider interval. Holding the confidence level and \(s_D\) fixed, increasing the number of independent pairs generally reduces the standard error \(s_D/\sqrt{n}\). These facts help check whether a computed width is plausible, but they do not remove the need to use the correct degrees of freedom and assumptions.

Worked Example: Construct a 90% Interval from Summary Statistics

A hypothetical quality study records paired differences in the output of two testing methods for eight specimens. The difference is method A minus method B. The reported summaries are \(n=8\), \(\bar{D}=1.75\) units, and \(s_D=2.00\) units. Find a 90% confidence interval for \(\mu_D\). Assume independent differences from \(N(\mu_D,\sigma_D^2)\), with unknown mean and variance. There is no interest rate or mortality assumption because the observations are test measurements.

Method. Use a two-sided paired-difference t interval. The degrees of freedom are \(8-1=7\). For a 90% interval, \(\alpha=0.10\), so each tail has probability \(0.05\), and \(t_{0.95,7}=1.895\).

The standard error is \(2.00/\sqrt{8}=0.7071\) units. The margin is \(1.895(0.7071)=1.340\) units. Therefore, the endpoints are \(1.75-1.340=0.410\) and \(1.75+1.340=3.090\). The interval is \((0.410,3.090)\) units, rounded to three decimal places.

Check. The midpoint is 1.750 and the half-width is 1.340. The margin is positive and smaller than the estimate, giving a positive lower endpoint.

Common Exam Trap: Using the Wrong Degrees of Freedom

The degrees of freedom are \(n-1\) because \(s_D\) is calculated after estimating the mean difference from the same \(n\) observations. For example, in the first worked example \(n=5\), so the correct critical value is \(t_{0.975,4}=2.776\), not \(t_{0.975,5}=2.571\). Using 2.571 would give a margin of \(2.571(1.4142)=3.636\), which is too small for the stated exact procedure. The corrected margin is \(2.776(1.4142)=3.926\).

Other checks help prevent avoidable errors:

  • Count pairs, not measurements. If there are \(n\) pairs, use \(n\) differences and \(n-1\) degrees of freedom.
  • Use the sample standard deviation of the differences. The interval requires \(s_D\), not separate standard deviations from the two sets of measurements.
  • Split the tail probability correctly. For a two-sided \(100(1-\alpha)\%\) interval, use \(t_{1-\alpha/2,n-1}\), not \(t_{1-\alpha,n-1}\).
  • Do not claim exactness without the model. The t procedure is exact when the differences are an independent normal sample; pairing alone does not establish normality.
  • Check that \(n\ge2\). With one difference, no sample standard deviation or t interval can be calculated.
Key takeaway: For \(n\ge2\) independent normal paired differences, calculate \(\bar{D}\) and \(s_D\), then use \(\bar{D}\pm t_{1-\alpha/2,n-1}s_D/\sqrt{n}\). The exact coverage claim belongs to this stated model, and the degrees of freedom are based on the number of pairs.

Check Your Understanding

For each problem, use the stated paired-difference model and round interval endpoints to three decimal places unless the question asks for a degrees-of-freedom check.

  1. A hypothetical instrument study has \(n=6\), \(\bar{D}=2.5\), and \(s_D=3.0\). Find a 90% interval for \(\mu_D\), assuming independent normal differences. (Answer: \((0.032,4.968)\))
  2. For 12 hypothetical paired measurements, \(\bar{D}=-1.8\) and \(s_D=2.4\). Find a 95% interval under the independent normal-differences model. (Answer: \((-3.325,-0.275)\))
  3. A study of nine matched components gives \(\bar{D}=4\) and \(s_D=3\). Find a 99% interval under the independent normal-differences model. (Answer: \((0.645,7.355)\))
  4. For 15 hypothetical paired readings, \(\bar{D}=0.6\) and \(s_D=1.5\). Find a 95% interval under the independent normal-differences model. (Answer: \((-0.231,1.431)\))
  5. A researcher has five independent normal differences with \(\bar{D}=8\) and \(s_D=4\). A calculation uses \(t_{0.95,5}\) for a two-sided 90% interval. Identify the degrees-of-freedom error and give the corrected interval. (Answer: Use \(4\) degrees of freedom, not \(5\); \(8\pm2.132(4/\sqrt{5})=(4.186,11.814)\).)

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