Choosing Between Mean-Response and Prediction Intervals

Two Questions, Two Different Intervals

A fitted regression line gives an estimated response at a predictor value, but the uncertainty depends on what is being estimated. One question may concern the average response among all cases at a particular predictor value. Another may concern the outcome for one individual case at that value. The first calls for a confidence interval for the mean response; the second calls for a prediction interval.

The distinction matters in applications such as a hypothetical insurer studying claim-processing cost by claim size. An interval for the average processing cost among claims of a specified size does not describe the likely range of costs for one new claim. Individual outcomes vary around their average, so their prediction interval must account for both uncertainty in the estimated average and that individual variation. Choosing the appropriate interval is a basic skill aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: At predictor value $x_0$, the mean response is $E[Y\mid X=x_0]$, the average outcome for cases with that predictor value. A prediction interval estimates a range for one new response $Y_{\text{new}}$ at $x_0$. A confidence interval targets the mean; a prediction interval targets an individual outcome.

Model, Assumptions, and the Leverage Term

For the formulas below, use the simple linear regression model $Y_i=\beta_0+\beta_1x_i+\varepsilon_i$, for $i=1,\ldots,n$. Treat the observed predictor values as fixed. For exact finite-sample $t$-intervals, assume the errors are independent and normally distributed with mean zero and common variance $\sigma^2$. The fitted value at $x_0$ is $\hat{y}_0=\hat{\beta}_0+\hat{\beta}_1x_0$. Let $\bar{x}$ be the sample mean of the predictor, $S_{xx}=\sum_{i=1}^n(x_i-\bar{x})^2$, and $s=\sqrt{\mathrm{SSE}/(n-2)}$, the residual standard error.

The quantity $h_0$ measures how much uncertainty in the estimated mean is associated with the location $x_0$. It is small near $\bar{x}$ and increases as $x_0$ moves away from $\bar{x}$, relative to the spread of the observed predictor values.

h_0=\frac{1}{n}+\frac{(x_0-\bar{x})^2}{S_{xx}}

Theorem: Under the stated model assumptions, a $100(1-\alpha)\%$ confidence interval for the mean response at $x_0$ is $\hat{y}_0\pm t_{1-\alpha/2,n-2}s\sqrt{h_0}$. A $100(1-\alpha)\%$ prediction interval for one new response at $x_0$ is $\hat{y}_0\pm t_{1-\alpha/2,n-2}s\sqrt{1+h_0}$. Here $t_{1-\alpha/2,n-2}$ is the indicated quantile of the $t$-distribution with $n-2$ degrees of freedom.

Both intervals use the same center, $\hat{y}_0$, and the same critical value when their confidence level is the same. Their standard errors differ. The confidence interval uses $s\sqrt{h_0}$, while the prediction interval uses $s\sqrt{1+h_0}$. The extra $1$ represents the variability of a new individual response around its conditional mean.

Why the Prediction Interval Has an Extra Term

The fitted value estimates a mean response, so its sampling variance is $\sigma^2h_0$. This follows from the variance of the least-squares estimated coefficients: at $x_0$, the fitted mean is a linear combination of $\hat{\beta}_0$ and $\hat{\beta}_1$, and its variance is $\sigma^2[1/n+(x_0-\bar{x})^2/S_{xx}]$. A new response has an additional error term. Specifically, $Y_{\text{new}}=\beta_0+\beta_1x_0+\varepsilon_{\text{new}}$, where the new error is independent of the errors used to fit the line and has variance $\sigma^2$. Therefore,

\operatorname{Var}(Y_{\text{new}}-\hat{y}_0) =\operatorname{Var}\bigl((\beta_0+\beta_1x_0-\hat{y}_0)+\varepsilon_{\text{new}}\bigr) =\sigma^2h_0+\sigma^2 =\sigma^2(1+h_0).

Replacing $\sigma$ with $s$ and using the $t$ critical value gives the prediction standard error $s\sqrt{1+h_0}$. Since $h_0>0$, $1+h_0>h_0$, so the prediction interval is wider than the confidence interval at the same $x_0$ and confidence level. This is not a small-sample correction; it reflects the different target.

Identify the target.
If the question asks for an average or expected response at $x_0$, use a confidence interval for the mean response. If it asks for one future case, use a prediction interval.

Calculate $h_0$.
Use the sample size, predictor mean, predictor sum of squares, and the stated $x_0$.

Choose the standard error and critical value.
Use $s\sqrt{h_0}$ for the mean or $s\sqrt{1+h_0}$ for an individual outcome, with $n-2$ degrees of freedom.

Form and interpret the interval.
Use $\hat{y}_0$ plus or minus the critical value times the selected standard error. State whether the interval concerns a mean or an individual response.

Worked Comparisons

Worked Example: An Average Processing Cost or One New Claim?

A hypothetical regression relates processing cost $Y$, in dollars, to claim size $x$, measured in thousands of dollars. There are $n=10$ observations, $\bar{x}=5$, $S_{xx}=50$, and $s=\$4$. At $x_0=5$, the fitted cost is $\hat{y}_0=\$40$. Find both 95% intervals, then state which one applies to the average cost and which to one new claim. Use $t_{0.975,8}=2.306$.

Method. Calculate $h_0$, then use the mean-response formula and the prediction formula separately. Because $x_0=\bar{x}$, the distance term in $h_0$ is zero:

h_0=\frac{1}{10}+\frac{(5-5)^2}{50}=0.1+0=0.1.

For the average cost, the standard error is $s\sqrt{h_0}=4\sqrt{0.1}=4(0.3162278)=1.264911$ dollars. The margin of error is $2.306(1.264911)=2.9169$, approximately $\$2.92$. Thus the 95% confidence interval for the mean cost is $\$40\pm\$2.9169$, or $\$37.08$ to $\$42.92$, rounded to the nearest cent.

For one new claim, the prediction standard error is $s\sqrt{1+h_0}=4\sqrt{1.1}=4(1.0488088)=4.195235$ dollars. The margin is $2.306(4.195235)=9.6742$, approximately $\$9.67$. The 95% prediction interval is $\$40\pm\$9.6742$, or $\$30.33$ to $\$49.67$, rounded to the nearest cent.

Check and interpretation. The prediction interval is wider: its width is about $\$19.35$, compared with about $\$5.83$ for the confidence interval. The narrower interval describes uncertainty in the average cost, not the range of costs expected for one individual claim. A common mistake is to use the confidence interval when asked about a new claim; that omits the new claim’s own response variability.

Worked Example: A Predictor Value Away From the Sample Mean

A hypothetical model predicts repair time $Y$, in hours, from equipment age $x$, in years. Here $n=8$, $\bar{x}=5$, $S_{xx}=32$, and $s=2$ hours. At $x_0=7$, the fitted repair time is 18 hours. Calculate 95% intervals for the mean repair time and for one new repair, using $t_{0.975,6}=2.447$.

Method. First account for the fact that $x_0$ is two years above $\bar{x}$:

h_0=\frac{1}{8}+\frac{(7-5)^2}{32} =0.125+\frac{4}{32} =0.125+0.125 =0.25.

For the mean, the standard error is $2\sqrt{0.25}=2(0.5)=1$ hour. The margin is $2.447(1)=2.447$ hours, so the interval is $18\pm2.447$, or 15.55 to 20.45 hours, rounded to two decimal places.

For one new repair, the standard error is $2\sqrt{1.25}=2(1.118034)=2.236068$ hours. The margin is $2.447(2.236068)=5.4717$ hours. The prediction interval is $18\pm5.4717$, or 12.53 to 23.47 hours, rounded to two decimal places.

Check and interpretation. The prediction interval is wider, as expected, and both intervals are centered at the same fitted value. The location adjustment matters for both intervals: omitting the $(x_0-\bar{x})^2/S_{xx}$ term would understate uncertainty because this predictor value is not at the sample mean.

Worked Example: Correcting the Wrong Interval for an Individual

A hypothetical study relates inspection duration $Y$, in minutes, to a machine setting $x$. There are $n=6$ observations, $\bar{x}=3$, $S_{xx}=12$, and $s=1$ minute. At $x_0=5$, the fitted duration is 10 minutes. Find 95% intervals for the mean and for one new inspection, using $t_{0.975,4}=2.776$.

The leverage term is $h_0=1/6+(5-3)^2/12=1/6+4/12=1/6+1/3=1/2$. For the mean, the standard error is $1\sqrt{1/2}=0.707107$ minute. The margin is $2.776(0.707107)=1.9629$ minutes, giving $10\pm1.9629$, or 8.037 to 11.963 minutes, rounded to three decimal places.

For an individual inspection, the standard error is $1\sqrt{1+1/2}=\sqrt{1.5}=1.224745$ minutes. The margin is $2.776(1.224745)=3.3999$ minutes. The prediction interval is $10\pm3.3999$, or 6.600 to 13.400 minutes, rounded to three decimal places.

Common mistake and correction. A wrong calculation for the individual outcome uses $s\sqrt{h_0}=\sqrt{0.5}$, producing the mean-response interval, 8.037 to 11.963. That calculation does not include the new inspection’s random error. For one inspection, use $s\sqrt{1+h_0}=\sqrt{1.5}$, which gives the wider interval, 6.600 to 13.400. The target in the question—not the fact that both intervals use the same fitted line—determines which formula is correct.

Interpretation and Limits

A 95% confidence interval for the mean response is a procedure that, over repeated samples under the model, covers the true mean response at $x_0$ about 95% of the time. A 95% prediction interval is a procedure that covers a new individual response at $x_0$ about 95% of the time. Once the data have been collected and the interval calculated, it is not correct to interpret either interval as assigning a 95% probability to a fixed parameter or to a particular fixed outcome.

The formulas rely on the stated regression model. If the errors are not independent, do not have a common variance, or are not normally distributed, the exact finite-sample $t$-interval interpretation may not hold. In particular, normal errors are needed for the exact small-sample $t$ result given here. The intervals also do not repair an inappropriate regression model or make predictions at unsupported predictor values reliable.

Question being answered	Target	Standard error
What is the average response at $x_0$?	Mean response	$s\sqrt{h_0}$
What response might one new case have at $x_0$?	Individual outcome	$s\sqrt{1+h_0}$

Key takeaway: First decide whether the question concerns an average or one individual outcome. Both intervals are centered at the fitted value and use the same $t$ critical value, but only a prediction interval includes the additional variability of a new response.

Check Your Understanding

For each item, identify the target before calculating. Round interval endpoints to two decimal places where needed.

A hypothetical analyst asks for the average monthly repair cost among all devices with a specified usage level. Should the analyst use a confidence interval for the mean response or a prediction interval? What if the question instead asks for the cost on one newly serviced device? (Answer: Use a confidence interval for the mean response for the average; use a prediction interval for one new device.)
A regression has $n=8$, $\bar{x}=2$, $S_{xx}=16$, $s=2$, and $\hat{y}_0=12$ at $x_0=2$. Using $t_{0.975,6}=2.447$, calculate the 95% mean-response interval and prediction interval. (Answer: $h_0=0.125$; mean interval 10.27 to 13.73; prediction interval 6.81 to 17.19.)
A hypothetical model has $n=10$, $\bar{x}=4$, $S_{xx}=40$, $s=2.5$, and fitted value 30 at $x_0=6$. Using $t_{0.975,8}=2.306$, calculate the two 95% intervals. (Answer: $h_0=0.2$; mean-response interval 27.42 to 32.58; prediction interval 23.68 to 36.32.)
For a 95% interval calculation, suppose $h_0=0.2$, $s=4$, and the critical value is 2.2. Find the full widths of the mean-response interval and prediction interval, and state which applies to one individual outcome. (Answer: Mean-response width $2(2.2)(4\sqrt{0.2})\approx7.87$; prediction width $2(2.2)(4\sqrt{1.2})\approx19.28$; the prediction interval applies to one individual outcome.)

Question being answered	Target	Standard error
What is the average response at \(x_0\)?	Mean response	\(s\sqrt{h_0}\)
What response might one new case have at \(x_0\)?	Individual outcome	\(s\sqrt{1+h_0}\)

What You'll Learn