Mean-Response and Prediction Interval Problems

Why the Interval Type Matters

A regression model can answer two different questions at a predictor setting. One asks for the mean response across the population of observations with that setting. The other asks where a single future response might fall. Both intervals are centered at the fitted value, but they have different standard errors and answer different questions. Choosing the wrong interval can produce a numerically correct calculation for the wrong target.

This tutorial focuses on identifying the target, selecting the corresponding formula, and calculating both intervals. The methods are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: At a predictor vector \(x_0\), the mean response is \(E[Y\mid x_0]\). A confidence interval for the mean response estimates this population mean. A prediction interval estimates a range for one future response \(Y_0\) observed at \(x_0\). The mean response is a fixed, unknown quantity; the future response also includes its own random error.

Model, Notation, and Interval Formulas

Use the multiple linear regression model \(Y=X\beta+\varepsilon\), where \(n\) observations are used to estimate \(p\) regression coefficients, including the intercept. At a target setting \(x_0\), the fitted value is \(\hat{y}_0=x_0^\mathsf{T}\hat{\beta}\). The leverage for that setting is \(h_0=x_0^\mathsf{T}(X^\mathsf{T}X)^{-1}x_0\). In simple linear regression, \(x_0\) contains the intercept component and the predictor value.

For the usual exact t intervals, assume the regression mean is correctly specified, the predictor settings are fixed (or inference is conditional on them), and the errors are independent and normally distributed with mean zero and common variance \(\sigma^2\). Estimate \(\sigma\) by the residual standard error \(s=\sqrt{\mathrm{SSE}/(n-p)}\). The residual degrees of freedom are \(n-p\). These formulas do not generally provide exact t intervals if the model assumptions fail; in particular, nonconstant error variance or dependence changes the standard-error calculation.

Key formulas: For a two-sided confidence level \(1-\alpha\), let \(t^*=t_{1-\alpha/2,n-p}\), the corresponding t critical value. The confidence interval for the mean response and prediction interval for one future response at \(x_0\) are

\hat{y}_0 \ \pm\ t^*s\sqrt{h_0} \qquad\text{and}\qquad \hat{y}_0 \ \pm\ t^*s\sqrt{1+h_0},

respectively. The center of each interval is \(\hat{y}_0\). The standard errors are \(s\sqrt{h_0}\) for the estimated mean and \(s\sqrt{1+h_0}\) for a future observation.

The predictor setting affects both standard errors through \(h_0\). A setting with higher leverage has greater uncertainty in its estimated mean, and therefore also in its prediction interval. The extra 1 in the prediction standard error accounts for the new observation’s random error; it is not another fitted coefficient or another degree of freedom.

Why the Prediction Standard Error Has an Extra 1

The difference between the formulas follows from separating uncertainty in the estimated regression mean from the random variation of a future response. Under the model, \(\hat{y}_0\) estimates \(x_0^\mathsf{T}\beta\), and its variance is \(\sigma^2h_0\). A future response can be written as \(Y_0=x_0^\mathsf{T}\beta+\varepsilon_0\), where \(\varepsilon_0\) is independent of the errors used to fit the model and has variance \(\sigma^2\).

Consequently, the prediction error is \(Y_0-\hat{y}_0=(x_0^\mathsf{T}\beta-\hat{y}_0)+\varepsilon_0\). The two terms are independent, so their variances add. Replacing \(\sigma\) with \(s\) gives the prediction standard error.

\operatorname{Var}(Y_0-\hat{y}_0) =\operatorname{Var}(\hat{y}_0-x_0^\mathsf{T}\beta) +\operatorname{Var}(\varepsilon_0) =\sigma^2h_0+\sigma^2 =\sigma^2(1+h_0).

For the estimated mean alone, there is no new-observation error term, so its variance is just \(\sigma^2h_0\). Since \(1+h_0>h_0\), the prediction interval is wider whenever the usual model applies. If the target were the average of \(m\) independent future responses at the same setting, the added variance would instead be \(\sigma^2/m\); the formula here is for one future response.

Choosing and Calculating the Interval

Identify the target.
If the question asks for an average, expected value, or population mean at \(x_0\), use a confidence interval for the mean response. If it asks about one new case or observation, use a prediction interval.

Find the degrees of freedom and critical value.
Count \(p\) as all fitted coefficients, including the intercept. Use \(n-p\) degrees of freedom and the critical value for the stated two-sided confidence level.

Calculate the standard error and margin.
Use \(s\sqrt{h_0}\) for the mean or \(s\sqrt{1+h_0}\) for one future response. Multiply by \(t^*\) to obtain the margin of error.

Calculate and interpret the endpoints.
Subtract and add the margin to \(\hat{y}_0\). State whether the interval concerns a mean response or one future observation.

Worked Example: Comparing Both Intervals at One Setting

A hypothetical insurer models annual claim cost using one predictor and an intercept. The fitted model uses \(n=16\) observations, has residual standard error \(s=4\) thousand dollars, and gives \(\hat{y}_0=50\) thousand dollars with \(h_0=0.10\) at a specified setting. Calculate the 95% confidence interval for mean annual claim cost and the 95% prediction interval for one future policyholder. Use \(t_{0.975,14}=2.145\). Report endpoints to two decimals in thousands of dollars.

Method. First identify the targets: the population mean requires the mean-response interval; one future policyholder requires the prediction interval. There are \(p=2\) coefficients (intercept and slope), so the residual degrees of freedom are \(16-2=14\), consistent with the supplied critical value.

For the mean response, the standard error is \(4\sqrt{0.10}=4(0.316228)=1.26491\) thousand dollars. Its margin is \(2.145(1.26491)=2.71323\). Therefore, the interval is \(50\pm2.71323\), or \((47.28677,52.71323)\), which rounds to \((47.29,52.71)\) thousand dollars.

For one future policyholder, the standard error is \(4\sqrt{1+0.10}=4\sqrt{1.10}=4(1.048809)=4.19524\) thousand dollars. Its margin is \(2.145(4.19524)=8.99878\). The interval is \(50\pm8.99878\), or \((41.00122,58.99878)\), which rounds to \((41.00,59.00)\) thousand dollars.

Check. Both intervals have center 50, and the prediction interval is wider. The common mistake avoided here is using \(\sqrt{h_0}\) for a single future policyholder, which would omit that policyholder’s individual random error.

Worked Example: A Mean Response in a Multiple Regression

A hypothetical benefits analyst fits a multiple regression with three predictors and an intercept to \(n=25\) observations. At a specified combination of predictor values, the fitted mean benefit is \(\hat{y}_0=120\), the residual standard error is \(s=3\), and the leverage is \(h_0=0.08\). Find the 90% confidence interval for the population mean benefit at that setting. Use \(t_{0.95,21}=1.721\), and round endpoints to two decimals.

The question asks for a population mean, not a new individual benefit. There are \(p=4\) coefficients, so \(n-p=25-4=21\) residual degrees of freedom. The supplied critical value matches those degrees of freedom.

The mean-response standard error is \(3\sqrt{0.08}=3(0.282843)=0.84853\). The margin is \(1.721(0.84853)=1.46032\). Thus the interval is \(120\pm1.46032\), with endpoints \(118.53968\) and \(121.46032\). Rounded to two decimals, the 90% confidence interval is \((118.54,121.46)\).

Check. The interval is centered at 120: the lower and upper distances are both 1.46032 before rounding. The mistake avoided is counting only the three predictors when finding the residual degrees of freedom; the intercept is also an estimated coefficient.

Worked Example: Predicting One New Observation

A hypothetical equipment-repair model uses one predictor and an intercept, with \(n=12\) repair records. At a new equipment setting, the fitted repair time is 36 hours, \(s=2.5\) hours, and \(h_0=0.25\). Find a 99% prediction interval for one new repair time. Use \(t_{0.995,10}=3.169\), and round endpoints to two decimals.

The target is one future repair time, so use the prediction standard error. With \(p=2\), the residual degrees of freedom are \(12-2=10\). The prediction standard error is \(2.5\sqrt{1+0.25}=2.5\sqrt{1.25}=2.5(1.118034)=2.79508\) hours.

The margin is \(3.169(2.79508)=8.85761\) hours. The interval is \(36\pm8.85761\), giving endpoints \(27.14239\) and \(44.85761\). To two decimals, the 99% prediction interval is \((27.14,44.86)\) hours.

Check. The fitted value lies midway between the endpoints, and the prediction interval includes the added uncertainty for a new repair. The mistake avoided is treating a 99% prediction interval as though it were a confidence interval for the mean; they have different standard errors even when their centers and confidence levels match.

Common Mistakes and Interpretation

The interval’s endpoints alone do not say what has been estimated. Interpretation must name the target and the predictor setting. A confidence interval for a mean response describes uncertainty about the population average at that setting; it does not claim that a specified proportion of individual responses must lie inside it. A prediction interval concerns one future response and is wider because individuals vary around the mean.

For example, if a 95% prediction interval is calculated under the stated model, its repeated-sampling interpretation is that the procedure will cover the future response about 95% of the time. It is not a statement that 95% of all possible future responses are guaranteed to fall within the particular interval calculated from this sample.

Wrong target: using the mean-response formula for a request about one new observation. Correction: look for words such as “one future,” “new case,” or “individual,” and include the added 1 under the square root.
Wrong degrees of freedom: using \(n-1\) or forgetting the intercept. Correction: subtract the total number \(p\) of fitted regression coefficients from \(n\).
Wrong critical value: using a one-sided quantile for a two-sided interval. Correction: for a two-sided level \(1-\alpha\), use \(t_{1-\alpha/2,n-p}\).
Wrong center: using a sample mean or an individual observed response rather than the fitted value at \(x_0\). Correction: both intervals are centered at \(\hat{y}_0\).
Wrong interpretation: describing a mean-response interval as a range expected to contain individual outcomes. Correction: say explicitly whether the endpoints estimate the population mean or predict a future response.

Key takeaway: At a specified predictor setting, both intervals are centered at the fitted value. Use \(s\sqrt{h_0}\) for uncertainty in the population mean and \(s\sqrt{1+h_0}\) for one future response. Then multiply by the two-sided t critical value with \(n-p\) degrees of freedom.

Check Your Understanding

For each problem, identify the interval type before calculating. Unless stated otherwise, use the usual regression assumptions described above.

A hypothetical model with one predictor and an intercept is fit to \(n=18\) records. At \(x_0\), \(\hat{y}_0=40\), \(s=5\), and \(h_0=0.04\). Find the 95% prediction interval using \(t_{0.975,16}=2.120\). Round to two decimals. (Answer: \((29.19,50.81)\))
A multiple regression with two predictors and an intercept uses \(n=30\) observations. At a specified setting, \(\hat{y}_0=15\), \(s=2\), and \(h_0=0.25\). Find the 90% confidence interval for the mean response using \(t_{0.95,27}=1.703\). Round to three decimals. (Answer: \((13.297,16.703)\))
A regression with two predictors and an intercept uses \(n=20\) observations. At \(x_0\), \(\hat{y}_0=72\), \(s=6\), and \(h_0=0.09\). Find a 95% prediction interval for one future response using \(t_{0.975,17}=2.110\). Round to two decimals. (Answer: \((58.78,85.22)\))
At one setting, a future-observation standard error is 5, while the residual standard error is \(s=4\). Under the usual prediction formula, find \(h_0\) and the standard error for the estimated mean response. (Answer: \(h_0=0.5625\); mean-response standard error \(=3\))
A report gives a 95% interval centered at a fitted value of 80, with standard error \(s\sqrt{h_0}\). The question asks where one new observation at that setting may fall. Is the reported standard error appropriate, and what standard error should be used instead? (Answer: No; use \(s\sqrt{1+h_0}\) for one future observation.)

What You'll Learn