Lesson 5 of 7 11–14% of the exam

By Daniil Tarakanov Updated Oct 22, 20262027 curriculum

Predicting Y and building a prediction interval

Plug X_f into the fitted line: Ŷ_f = b̂₀ + b̂₁X_f. The interval is Ŷ_f ± t_c × s_f. The standard error of the forecast, s_f, is larger with a larger standard error of the estimate (s_e), a smaller sample, an X_f farther from X̄ and less spread in X.

6 min cram, 30 min full read. Based on the 2027 CFA® Program Level I curriculum, Quantitative Methods, Learning Module 10. Sitting in November 2026? The 2026 version of this lesson is still available.

Tailor this to you: take the self-assessment

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

On the exam

Calculate and interpret predicted values, the standard error of the estimate, and prediction intervals for the dependent variable in a simple linear regression model and describe different functional forms

Calculate and interpret predicted values and prediction intervals. Expect a predicted value from a regression output (keep the intercept), s_f given or built from s_e, and Ŷ_f ± t_c × s_f with n − 2 degrees of freedom, plus what makes the interval wider. Wrong options use z instead of t, or s_e 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 = \hat{b}_0 + \hat{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. It is built from the standard error of the estimate ( $s_e$ , also called SEE; lesson 4).

Standard error of the forecast Use it

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

The same formula can be written in variance form, where $(n-1)s_X^2$ is $\sum(X_i-\bar{X})^2$ written with the sample variance of X (for the retailers, 5 × 7.5 = 37.5):

Standard error of the forecast (variance form) Use it

s_f^2 = s_e^2\left[1 + \frac{1}{n} + \frac{(X_f-\bar{X})^2}{(n-1)\,s_X^2}\right]

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 $\bar{X}$ . So four things set $s_f$ (the drivers of forecast uncertainty ): the standard error of the estimate (fit), the sample size, the distance of $X_f$ from $\bar{X}$ , and the spread of X (more spread pins the slope down better). Since the root is always above 1, $s_f$ is always larger than $s_e$ .

What makes a prediction interval wider

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

Larger s_e 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
Less spread in X Use it the slope is estimated less precisely
Higher confidence level Use it a larger t_c

Worked example 95% interval at 7% marketing spend

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

$s_f = 0.8327\sqrt{1 + 1/6 + 1.5^2 \div 37.5} \approx 0.8327 \times 1.1075 = 0.9222$ .
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 s_e = 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 the estimate, s_eObservations, nForecast value of X, X_fMean of X, X̄Sum of squared deviations of X, Σ(X − X̄)²

s_{f} = s_{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, Ŷ_fCritical t-value, t_cStandard error of the forecast, s_f

upper = Y^_{f} + 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.
Make the band narrower at X_f = 12 without adding data.

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 the estimate, s_e0.83

Spread out XDoubles each X's distance from X̄Also show the interval for the mean of YDrops the “1 +”: much narrower

Show working

Ŷ_f = b̂₀ + b̂₁X_f = 1.3867 + 1.2933 × 7.0 = 10.4400.

s_f = s_e × √(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 s_e (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.

Exam tip In a spreadsheet, FORECAST.LINEAR (FORECAST in Google Sheets) returns $\hat{Y}_f$ and TREND returns fitted values. On the exam you calculate them by hand.

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_f, then Ŷ_f ± t_c × s_f. Do not substitute s_e.

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	$\hat{Y}_f = \hat{b}_0 + \hat{b}_1X_f$
Interval	$\hat{Y}_f \pm t_c \times s_f$ , $df = n-2$
$s_f$ vs $s_e$	$s_f > s_e$ always; $s_f^2 = s_e^2[1 + 1/n + (X_f-\bar{X})^2/\sum(X_i-\bar{X})^2]$
Wider when	larger $s_e$ , smaller n, $X_f$ far from $\bar{X}$ , less spread in X, higher confidence

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 t with n − 2 degrees of freedom, not z, and s_f, not s_e.

What makes a prediction interval wider

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

Larger s_e 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
Less spread in X Use it the slope is estimated less precisely
Higher confidence level Use it a larger t_c

Predicted value

\hat{Y}_f = \hat{b}_0 + \hat{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 = s_e\sqrt{1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{\sum (X_i-\bar{X})^2}}

Quick example

s_e = 3.0, n = 25 and X_f = X̄. s_f = 3.0 × √(1 + 1/25) = 3.06, just above s_e.

On your calculator

3×(1+251/x)√x= 3.0594

1/x reciprocal · √x square root

251/x1+g√x3× 3.0594

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

Standard error of the forecast (variance form)

s_f^2 = s_e^2\left[1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{(n-1)\,s_X^2}\right]

Quick example

s_e = 1.5, n = 31, s_X = 2.0 and X_f is 3 above X̄: s_f = 1.5 × √(1 + 1/31 + 9 ÷ 120) = 1.5 × 1.0523 = 1.58.

On your calculator

9÷120+311/x+1=√x×1.5= 1.5784

1/x reciprocal · √x square root

9ENTER120÷311/x+1+g√x1.5× 1.5784

ENTER enter (on the HP, puts the number on the stack) · 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 = 8.4, t_c = 2.04, s_f = 1.4. Interval = 8.4 ± 2.04 × 1.4 = 8.4 ± 2.86 = 5.54 to 11.26.

On your calculator

2.04×1.4=STO1 2.856
8.4−RCL1= 5.544
8.4+RCL1= 11.256

STO store · RCL recall a stored number

2.04ENTER1.4×STO1 2.856
8.4RCL1− 5.544
8.4RCL1+ 11.256

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

b̂₀ = 2.2, b̂₁ = 0.9, X = 6, s_f = 1.3, t_c = 2.06. Interval?

Ŷ = 7.6; 7.6 ± 2.68 = 4.92 to 10.28. Trap: z = 1.96 gives 5.05 to 10.15.
Where is it narrowest?

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

At the sample mean of X.
s_e or s_f?

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

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 s_e instead of the standard error of the forecast. s_e 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.
Dropping the intercept when computing the predicted value.

Checkpoint

4 exam-style questions. About 90 seconds each.

1A fitted model has b̂₀ = 1.5 and b̂₁ = 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 the estimate, the standard error of the forecast is:
4A regression has a standard error of the estimate of 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_f = b̂₀ + b̂₁X_f for the point prediction.
Interval: Ŷ_f ± t_c × s_f with n − 2 degrees of freedom.
s_f is larger than s_e and grows with the distance of X_f from X̄; more data and more spread in X shrink it.

Predicted value

\hat{Y}_f = \hat{b}_0 + \hat{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 = s_e\sqrt{1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{\sum (X_i-\bar{X})^2}}

Quick example

s_e = 3.0, n = 25 and X_f = X̄. s_f = 3.0 × √(1 + 1/25) = 3.06, just above s_e.

On your calculator

3×(1+251/x)√x= 3.0594

1/x reciprocal · √x square root

251/x1+g√x3× 3.0594

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

Standard error of the forecast (variance form)

s_f^2 = s_e^2\left[1 + \dfrac{1}{n} + \dfrac{(X_f-\bar{X})^2}{(n-1)\,s_X^2}\right]

Quick example

s_e = 1.5, n = 31, s_X = 2.0 and X_f is 3 above X̄: s_f = 1.5 × √(1 + 1/31 + 9 ÷ 120) = 1.5 × 1.0523 = 1.58.

On your calculator

9÷120+311/x+1=√x×1.5= 1.5784

1/x reciprocal · √x square root

9ENTER120÷311/x+1+g√x1.5× 1.5784

ENTER enter (on the HP, puts the number on the stack) · 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 = 8.4, t_c = 2.04, s_f = 1.4. Interval = 8.4 ± 2.04 × 1.4 = 8.4 ± 2.86 = 5.54 to 11.26.

On your calculator

2.04×1.4=STO1 2.856
8.4−RCL1= 5.544
8.4+RCL1= 11.256

STO store · RCL recall a stored number

2.04ENTER1.4×STO1 2.856
8.4RCL1− 5.544
8.4RCL1+ 11.256

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

All Quantitative Methods formulas, every lesson

Practice exam-style questions on Simple Linear Regression in Finance

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 in Finance
Financial Statement Forecasting Equities

Questions candidates ask

How do you calculate a prediction interval in regression?

Find the predicted value Ŷ_f = b̂₀ + b̂₁X_f, then compute Ŷ_f ± 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 standard error of the estimate (s_e, also called 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 (s_e), the uncertainty in the estimated line (1/n), the distance of the new X from the sample mean and how spread out X is. It is s_f = s_e × √(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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