AP Statistics Least-Squares Regression and Residuals Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Least-Squares Regression and Residuals – Maths PDF

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This regression residuals answers for AP Statistics covers least-squares regression and residuals (Common Core AP STAT 2.6, AP STAT 2.7, AP STAT 2.8, AP STAT 2.9, HSS-ID.B.6, HSS-ID.C.7), with a free PDF to print.

Each solution shows the substitution into the regression equation, the residual calculation, and the interpretation in context. Equations are rounded to the precision shown in the problem.

AP Statistics · Common Core State Standards: AP STAT 2.6, AP STAT 2.7, AP STAT 2.8, AP STAT 2.9, HSS-ID.B.6, HSS-ID.C.7

Regression residuals answers: exercises 1 to 10

Exercise 1 – Predict

$\hat{y}=40+0.12(1800)=40+216=256$.

About $256,000.

Exercise 2 – Interpret the slope

The slope 0.12 is in thousands of dollars per square foot.

For each additional square foot, the predicted price increases by about $120.

Exercise 3 – Interpret the intercept

When x = 0 classes, $\hat{y}=3.5$.

A student with no classes is predicted to do 3.5 minutes of homework. It is not meaningful because x = 0 is outside the sensible range.

Exercise 4 – Compute a residual

$\hat{y}=12+3(5)=27$. $e=30-27=3$.

Residual +3: Sofia’s score was 3 points higher than predicted.

Exercise 5 – Find the slope and intercept

$b=r\frac{s_y}{s_x}=0.60\times\frac{20}{3}=4$. $a=150-4(66)=-114$.

$\hat{y}=-114+4x$.

Exercise 6 – Point of means

$\hat{y}=-114+4(66)=-114+264=150$, which equals the mean of y.

Yes: the line goes through (66, 150).

Exercise 7 – Interpret r squared

r squared is the proportion of variation explained.

About 64% of the variation in battery life is explained by the linear relationship with screen brightness.

Exercise 8 – Get r from r squared

$r=\pm\sqrt{0.81}=\pm 0.9$. The slope is positive, so r is positive.

r = 0.9.

Exercise 9 – Residual plot: good fit

A residual plot with no pattern: a linear model is appropriate.

The residuals scatter randomly around zero with no pattern.

Yes, a linear model is appropriate.

Exercise 10 – Residual plot: curved

A curved residual pattern signals that a straight line is not the right model.

The residuals form a U shape.

No; the relationship is curved. Try a transformation of the data or a curved model.

Exercises 11 and up

Exercise 11 – Extrapolation

$\hat{y}=26.56-2.25(15)=-7.22$ thousand dollars.

-7.22 thousand dollars is below zero, which is impossible. Age 15 is outside the data range (extrapolation), so the prediction is unreliable.

Exercise 12 – LSRL for five points

Least-squares line through five points.

$b=0.775\times\frac{1.095}{1.581}\approx 0.6$. $a=4-0.6(3)=2.2$.

$\hat{y}=2.2+0.6x$.

Exercise 13 – Residuals add to zero

Predicted: 2.8, 3.4, 4.0, 4.6, 5.2. Residuals: -0.8, 0.6, 1.0, -0.6, -0.2.

Sum: $-0.8+0.6+1.0-0.6-0.2=0$.

Residuals: -0.8, 0.6, 1.0, -0.6, -0.2; their sum is 0.

Exercise 14 – Standard deviation of residuals

Squares: 0.64, 0.36, 1.00, 0.36, 0.04, sum 2.4.

$s=\sqrt{\frac{2.4}{5-2}}=\sqrt{0.8}\approx 0.894$.

s = 0.894: actual y values typically differ from the predicted values by about 0.894 units.

Exercise 15 – Read computer output

Output for sleep data.

Equation: $\hat{y}=25.6-3.2x$.

r squared = 0.78, slope negative, so $r=-\sqrt{0.78}\approx -0.883$.

For each extra hour of screen time, the predicted score drops by 3.2; r = -0.883.

Exercise 16 – Influential point

Its removal substantially changes the slope.

It is an influential point, probably with an extreme x value.

Exercise 17 – Compare two predictions

10 miles: $\hat{y}=43$, $e=39-43=-4$. 12 miles: $\hat{y}=48$, $e=52-48=4$.

Both residuals have size 4 (one low, one high), so the predictions are equally accurate; the 10-mile cost is overpredicted and the 12-mile cost underpredicted.

Exercise 18 – Why least squares?

For any candidate line, square each vertical distance from a point to the line and add them.

The least-squares line is the one that makes this sum of squared residuals as small as possible.

Exercise 19 – Describe a fit

High r squared, small s, and a patternless residual plot all support a linear model.

Linear regression fits well: 92% of the variation in mileage is explained by speed, and predictions typically miss by about 1.4 mpg.

Exercise 20 – Two predictions

$\hat{y}=15+0.8(40)=47$. $e=52-47=5$.

47 minutes predicted; residual +5 (the order took 5 minutes longer than predicted).

Key facts behind this regression residuals answers

Key facts

LSRL: y-hat = a + bx, with b = r sy/sx and a = y-bar – b x-bar; it passes through (x-bar, y-bar).
Interpret the slope as the predicted change in y per unit of x, in context.
Residual = actual – predicted; a good linear model has a residual plot with no pattern.
r squared is the percent of variation in y explained by the linear model; s is the typical prediction error.
Avoid extrapolation and check for influential points.

How to use this regression residuals answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the regression residuals answers as a PDF with the button below and keep it with your notes.

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