Lesson 5 of 6 6–9% of the exam

You are reading the 2026 curriculum (exams up to November 2026). Sitting in February 2027 or later? Go to the 2027 Quantitative Methods guide.

By Daniil Tarakanov Updated Oct 16, 20262026 curriculum

Predicting Y and building a prediction interval

Plug the new X into the fitted line for the point prediction Ŷ = b₀ + b₁X. Then build the interval Ŷ ± t_c × s_f, where the standard error of the forecast s_f grows with the standard error of estimate (SEE), with smaller samples and with the distance of X from its mean.

6 min cram, 30 min full read. Based on the 2026 CFA® Program Level I curriculum, Quantitative Methods, Learning Module 10. This is the 2026 version of the lesson. Sitting in February 2027 or later? Read the updated 2027 version.

Tailor this to you: take the self-assessment

Before this: The ANOVA table and the standard error of estimate

On the exam

Calculate and interpret the predicted value for the dependent variable, and a prediction interval for it, given an estimated linear regression model and a value for the independent variable

Calculate and interpret a predicted value of Y and a prediction interval for it. Expect the standard error of the forecast to be given, so the job is Ŷ ± t_c × s_f with n − 2 degrees of freedom. Wrong options use z instead of t, or the SEE in place of s_f.

How deeply to learn each part

Know the idea: Understand it in general. It frames other topics, but you will not calculate with it.
Use it: Understand it well enough to solve a question with it. Recognize when it applies.
Memorize: Learn it by heart. The exam will not give it to you.

You know this one. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still do.

You tested strong here. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still know it.

How do you make a point prediction?

Plug the new value of X into the fitted line. The result is the predicted value of Y.

Predicted value Memorize

\hat{Y}_f = b_0 + b_1 X_f

That is a single best guess. The line is estimated, and any one retailer will miss it, so we want a range.

How do you build the prediction interval?

A prediction interval is the range expected to contain a single new observation of Y at a given X, with a stated confidence.

Prediction interval Memorize

\hat{Y}_f \pm t_c \times s_f, \qquad df = n-2

The critical t-value comes from the t-table with $n-2$ degrees of freedom. The standard error of the forecast measures the uncertainty in predicting one new Y.

Standard error of the forecast Use it

s_f = SEE\sqrt{1 + \frac{1}{n} + \frac{(X_f-\bar{X})^2}{\sum (X_i-\bar{X})^2}}

The three terms under the root are three sources of uncertainty. The 1 is the new observation’s own error. The $1/n$ reflects that the line was estimated from a finite sample. The last term grows as $X_f$ moves away from $\bar{X}$ . Since the root is always above 1, $s_f$ is always larger than the SEE.

What makes a prediction interval wider

Width of Ŷ ± t_c × s_fa bigger s_f or t_c means a wider interval

Larger SEE Use it noisier data around the line
Smaller sample, n Use it the 1/n term and the larger t_c
X_f farther from X̄ Memorize the distance term grows
Higher confidence level Use it a larger t_c

Worked example 95% interval at 7% marketing spend

For the retailers, SEE = 0.8327, $n = 6$ , $\bar{X} = 5.5$ , $\sum(X-\bar{X})^2 = 37.5$ , $\hat{Y}_f = 10.44$ and the critical t-value for 4 degrees of freedom is 2.776.

$s_f = 0.8327\sqrt{1 + 1/6 + 1.5^2 \div 37.5} = 0.8327\sqrt{1.2267} = 0.8327 \times 1.1075 = 0.9222$ .
Margin: $2.776 \times 0.9222 = 2.56$ .
Interval: $10.44 \pm 2.56$ , so 7.88% to 13.00%.

We are 95% confident that a retailer spending 7% of sales on marketing will grow between about 7.9% and 13.0%, if the model’s assumptions hold.

Steps 1 to 3 on your calculator

0.8327×((7−5.5)x²÷37.5+1+61/x)√x= 0.922
2.776×0.9222=STO1 2.56
10.44−RCL1= 7.88
10.44+RCL1= 13.00

x² square · 1/x reciprocal · √x square root · STO store · RCL recall a stored number

7ENTER5.5−ENTER×37.5÷1+61/x+g√x0.8327× 0.922
2.776ENTER0.9222×STO1 2.56
10.44RCL1− 7.88
10.44RCL1+ 13.00

ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · g blue shift key · √x square root · STO store · RCL recall a stored number

Rearrange itStandard error of the forecast: solve for any one

The question asks for

A simple regression has SEE = 0.8327, n = 6, X̄ = 5.5 and Σ(X − X̄)² = 37.5. For X_f = 7, the standard error of the forecast is closest to:

Standard error of estimate, SEEObservations, nForecast value of X, X_fMean of X, X̄Sum of squared deviations of X, Σ(X − X̄)²

s_{f} = S E E 1 + \frac{1}{n} + \frac{( X _{f} - X ˉ ) ^{2}}{\sum ( X _{i} - X ˉ ) ^{2}}

0.8327 × √(1 + 1/6 + 1.5² ÷ 37.5) = 0.9223

Standard error of the forecast, s_f0.9223

On your calculator

0.8327×((7−5.5)x²÷37.5+1+61/x)√x=

x² square · 1/x reciprocal · √x square root

7ENTER5.5−ENTER×37.5÷1+61/x+g√x0.8327×

ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · g blue shift key · √x square root

Rearrange itPrediction interval: solve for any one

The question asks for

A regression predicts Ŷ = 10.44 with a standard error of the forecast of 0.9222. With a critical t-value of 2.776, the upper bound of the prediction interval is closest to:

Predicted value, ŶCritical t-value, t_cStandard error of the forecast, s_f

upper = Y^+ t_{c} \times s_{f}

10.44 + 2.776 × 0.9222 = 13

Upper bound13

On your calculator

2.776×0.9222+10.44=

2.776ENTER0.9222×10.44+

ENTER enter (on the HP, puts the number on the stack)

InteractivePredict Y for a new X: where is the interval tightest?

Try this 0 of 3

Drag X_f to the point where the interval is narrowest.
Load “Far from the data (X = 12)” and compare s_f with X_f = 7.
Raise n to 30 or more and look at what the interval shrinks toward.

Drag the X_f marker, tap the chart, or focus the handle and use the arrow keys.

95% prediction band
Fitted line
Interval at X_f
Observed range of X

Prediction Ŷ10.44%

Std error of forecast s_f0.9222

Critical t (df 4, 95%)2.776

95% prediction interval7.88% to 13.00%

New observation

Marketing spend X_f7.0%

What if

Sample size nn = 6

Standard error of estimate SEE0.83

Also show the interval for the mean of YDrops the “1 +”: much narrower

Show working

Ŷ = b₀ + b₁X_f = 1.3867 + 1.2933 × 7.0 = 10.4400.

s_f = SEE × √(1 + 1/n + (X_f − X̄)² ÷ Σ(X − X̄)²) = 0.8327 × √(1 + 1/6 + 2.2500 ÷ 37.5) = 0.8327 × √1.2267 = 0.9222.

Critical t, df = n − 2 = 4, 95% two-tailed: 2.7764. Half-width = 2.7764 × 0.9222 = 2.5605.

Interval: 10.4400 ± 2.5605 = 7.88 to 13.00.

s_f is always above SEE (0.9222 > 0.8327): the "1 +" is the new observation's own error, and the other two terms are the error in the estimated line.

Prediction interval on your calculator

1.2933×7+1.3867=STO1Ŷ = 10.4400
7−5.5=x²÷37.5+61/x+1=√x×0.8327=STO2s_f = 0.9222
×2.7764=STO3half-width 2.5605
RCL1−RCL3=lower 7.88
RCL1+RCL3=upper 13.00

STO store · x² square · 1/x reciprocal · √x square root · RCL recall a stored number

1.2933ENTER7×1.3867+STO1Ŷ = 10.4400
7ENTER5.5−ENTER×37.5÷61/x+1+g√x0.8327×STO2s_f = 0.9222
2.7764×STO3half-width 2.5605
RCL1RCL3−lower 7.88
RCL1RCL3+upper 13.00

ENTER enter (on the HP, puts the number on the stack) · STO store · 1/x reciprocal · g blue shift key · √x square root · RCL recall a stored number

Display: 7.88 to 13.00

Predictions are tightest near the middle of the data.

What happens far from the data?

The interval widens as X moves away from $\bar{X}$ . At $X_f = 12$ (well beyond the largest observed X of 9) the prediction is $\hat{Y} = 16.91$ and the standard error of the forecast rises to 1.261, so the 95% interval is 13.41% to 20.41%, with a margin of 3.50 against 2.56 at 7%, roughly 37% wider.

Worked example Extrapolating to 12% marketing spend

$\hat{Y}_f = 1.3867 + 1.2933 \times 12 = 16.91$ .
$s_f = 0.8327\sqrt{1 + 1/6 + 6.5^2 \div 37.5} = 0.8327 \times 1.5144 = 1.261$ .
Margin: $2.776 \times 1.261 = 3.50$ , so the interval is 13.41% to 20.41%.

The wider band is a warning. It also assumes the straight-line relationship still holds out there, which the data cannot confirm.

Step 2 on your calculator

0.8327×((12−5.5)x²÷37.5+1+61/x)√x= 1.261

x² square · 1/x reciprocal · √x square root

12ENTER5.5−ENTER×37.5÷1+61/x+g√x0.8327× 1.261

ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · g blue shift key · √x square root

Exam tip On most exam items the standard error of the forecast is given. Your job is then only Ŷ, then Ŷ ± t_c × s_f. Do not substitute the SEE.

Go deeper Prediction interval versus confidence interval for the mean

A prediction interval covers one new observation. A confidence interval for the mean of Y at that X covers the average of many observations, which is a more certain target. Its standard error drops the “1 +” under the root, so it is narrower. The exam asks for the prediction interval of a single outcome.

Go deeper The danger of extrapolation

A regression describes the relationship in the range of X it has seen. Beyond that range the true relationship could flatten or curve, and no formula can warn you. The widening interval captures only the statistical part of the risk, not that structural risk.

Memorize this

MemorizePrediction recipe

Step	Rule
Point forecast	$\hat{Y} = b_0 + b_1X_f$
Interval	$\hat{Y} \pm t_c \times s_f$ , $df = n-2$
$s_f$ vs SEE	$s_f > SEE$ always; $s_f^2 = SEE^2[1 + 1/n + (X_f-\bar{X})^2/\sum(X_i-\bar{X})^2]$
Narrowest at	$X_f = \bar{X}$ ; wider as $X_f$ moves away

Memory hook Three sources of doubt, three terms under the root: the new error itself (the 1), the line being estimated from a finite sample (1/n), and the slope error magnified away from the middle (the distance term).

Know it cold Use the t critical value with n − 2 degrees of freedom, not z, and s_f, not SEE.

What makes a prediction interval wider

Width of Ŷ ± t_c × s_fa bigger s_f or t_c means a wider interval

Larger SEE Use it noisier data around the line
Smaller sample, n Use it the 1/n term and the larger t_c
X_f farther from X̄ Memorize the distance term grows
Higher confidence level Use it a larger t_c

Predicted value

\hat{Y}_f = b_0 + b_1 X_f

Quick example

Intercept 2.0, slope 0.75, forecast X = 6. Ŷ = 2.0 + 0.75 × 6 = 6.5.

On your calculator

0.75×6+2= 6.5

0.75ENTER6×2+ 6.5

ENTER enter (on the HP, puts the number on the stack)

Standard error of the forecast

s_f = SEE\sqrt{1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{\sum (X_i-\bar{X})^2}}

Quick example

SEE = 2.0, n = 20 and X_f = X̄. s_f = 2.0 × √(1 + 1/20) = 2.05, just above the SEE.

On your calculator

2×(1+201/x)√x= 2.0494

1/x reciprocal · √x square root

201/x1+g√x2× 2.0494

1/x reciprocal · g blue shift key · √x square root

Prediction interval

\hat{Y}_f \pm t_c \times s_f,\quad df = n-2

Quick example

Ŷ = 5.5, t_c = 2.05, s_f = 1.2. Interval = 5.5 ± 2.05 × 1.2 = 5.5 ± 2.46 = 3.04 to 7.96.

On your calculator

2.05×1.2=STO1 2.46
5.5−RCL1= 3.04
5.5+RCL1= 7.96

STO store · RCL recall a stored number

2.05ENTER1.2×STO1 2.46
5.5RCL1− 3.04
5.5RCL1+ 7.96

ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number

How it is asked

Build the interval

b₀ = 1.5, b₁ = 0.8, X = 5, s_f = 1.2, t_c = 2.05. Interval?

Ŷ = 5.5; 5.5 ± 2.46 = 3.04 to 7.96. Trap: z = 1.96 gives 3.15 to 7.85.
Where is it narrowest?

Other things equal, for which X_f is the interval narrowest?

At the sample mean of X.
SEE or s_f?

How does the standard error of the forecast compare with the SEE?

s_f is larger: the root is always above 1.
Interpret the interval

A 95% prediction interval for 7% marketing spend is 7.9% to 13.0%.

We are 95% confident that a single new retailer with 7% spend will grow between 7.9% and 13.0%, if the assumptions hold.

Exam traps

Using the SEE instead of the standard error of the forecast. The SEE ignores the uncertainty in the estimated line.
Using a z critical value (1.96) instead of t with n − 2 degrees of freedom.
Forgetting the interval widens as X moves away from the sample mean of X.
Extrapolating far beyond the observed X range: the linear relation may not hold there.

Checkpoint

4 exam-style questions. About 90 seconds each.

1A model has b₀ = 1.5 and b₁ = 0.8. For X = 5, the standard error of the forecast is 1.2 and the critical t-value is 2.05. The 95% prediction interval is closest to:
2Other things equal, the prediction interval is narrowest when the new X equals:
3Compared with the standard error of estimate, the standard error of the forecast is:
4A regression has SEE = 2.0 and n = 20. For a forecast at X equal to the sample mean, the standard error of the forecast is closest to:

Key takeaways

Ŷ = b₀ + b₁X_f for the point prediction.
Interval: Ŷ ± t_c × s_f with n − 2 degrees of freedom.
s_f is larger than SEE and grows with the distance of X_f from X̄.

Predicted value

\hat{Y}_f = b_0 + b_1 X_f

Quick example

Intercept 2.0, slope 0.75, forecast X = 6. Ŷ = 2.0 + 0.75 × 6 = 6.5.

On your calculator

0.75×6+2= 6.5

0.75ENTER6×2+ 6.5

ENTER enter (on the HP, puts the number on the stack)

Standard error of the forecast

s_f = SEE\sqrt{1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{\sum (X_i-\bar{X})^2}}

Quick example

SEE = 2.0, n = 20 and X_f = X̄. s_f = 2.0 × √(1 + 1/20) = 2.05, just above the SEE.

On your calculator

2×(1+201/x)√x= 2.0494

1/x reciprocal · √x square root

201/x1+g√x2× 2.0494

1/x reciprocal · g blue shift key · √x square root

Prediction interval

\hat{Y}_f \pm t_c \times s_f,\quad df = n-2

Quick example

Ŷ = 5.5, t_c = 2.05, s_f = 1.2. Interval = 5.5 ± 2.05 × 1.2 = 5.5 ± 2.46 = 3.04 to 7.96.

On your calculator

2.05×1.2=STO1 2.46
5.5−RCL1= 3.04
5.5+RCL1= 7.96

STO store · RCL recall a stored number

2.05ENTER1.2×STO1 2.46
5.5RCL1− 3.04
5.5RCL1+ 7.96

ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number

Next lesson Functional forms: log-lin, lin-log and log-log models

Questions about this lesson?

Where this comes back

Functional forms: log-lin, lin-log and log-log models Simple Linear Regression
Company Analysis: Forecasting Equity Investments

Questions candidates ask

How do you calculate a prediction interval in regression?

Find the predicted value Ŷ = b₀ + b₁X_f, then compute Ŷ ± t_c × s_f. The critical t-value has n − 2 degrees of freedom, and s_f is the standard error of the forecast, which is larger than the SEE. The result is the range expected to contain a new observation.

What is the standard error of the forecast?

It measures the uncertainty in predicting a single new Y. It combines the error scatter (SEE), the uncertainty in the estimated line (1/n) and the distance of the new X from the sample mean. It is s_f = SEE × √(1 + 1/n + (X_f − X̄)² ÷ Σ(X − X̄)²).

Why is a prediction interval wider away from the mean?

Any error in the estimated slope is magnified the farther X is from X̄, because the line pivots around the point of means. So the forecast is least certain at the extremes and beyond the data, where extrapolating is especially risky.

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