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Lesson 106 — Multiple regression

Model with p predictors, OLS matrix solution, adjusted R², multicollinearity, variable selection and assumption diagnostics.

Used in: Stochastik LK alemão (Klasse 12) · H2 Mathematics Singapura (§15) · econometria introdutória

\hat{\boldsymbol\beta} = (X^TX)^{-1}X^T\mathbf{y}

In multiple regression with $p$ predictors, the OLS estimator is the matrix solution to the normal equations. Each coefficient $\hat\beta_j$ measures the effect of $X_j$ on $Y$ holding the other predictors fixed — the so-called partial effect.

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Rigorous notation, full derivation, hypotheses

Rigorous definition

Multiple linear regression model

"The multiple regression model is $y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \cdots + \beta_k x_k + \varepsilon$ . The coefficient $\beta_i$ measures the expected change in $y$ per unit change in $x_i$ when all other predictors are held constant." — OpenIntro Statistics, §8.1, p. 362

Fit metrics

Inference

Matrix representation of the model: $\mathbf{Y} = X\boldsymbol\beta + \boldsymbol\varepsilon$ . The first column of 1s in $X$ generates the intercept $\beta_0$ .

Worked examples

Example— 106.1· Fit and interpretation with two predictors

Problem. Regression of annual salary $Y$ (R$ thousands) on experience $X_1$ (years) and education $X_2$ (years of study) produced: $\hat Y = -12 + 2.8 X_1 + 5.2 X_2$ , with $n=50$ , $R^2=0.74$ , $\bar R^2=0.73$ .

Interpret each coefficient and calculate the prediction for a professional with 10 years of experience and 16 years of education.

Strategy. Interpret each $\hat\beta_j$ as a partial effect (ceteris paribus). Substitute values into the fitted equation.

Solution.

$\hat\beta_0 = -12$ : intercept (no meaningful substantive interpretation alone — no professional has zero years of experience and education).
$\hat\beta_1 = 2.8$ : holding education constant, each additional year of experience is associated with R$ 2,800/year more, on average.
$\hat\beta_2 = 5.2$ : holding experience constant, each additional year of education is associated with R$ 5,200/year more.

Prediction: $\hat Y = -12 + 2.8 \times 10 + 5.2 \times 16 = -12+28+83.2 = 99.2$ thousand R$/year.

Verification. $\bar R^2 = 0.73$ close to $R^2 = 0.74$ : only 2 predictors, little penalization — parsimonious model.

Source. OpenIntro Statistics, §8.1, Example 8.1 — CC-BY-SA

Example— 106.2· Adjusted R² in model selection

Problem. Three models to predict household electricity consumption ( $n=80$ ):

Model	Predictors	$R^2$	$\bar R^2$
M1	area	0.61	0.61
M2	area + number of residents	0.71	0.70
M3	area + number of residents + income + temperature + day of week	0.74	0.70

Which model to prefer?

Strategy. Compare $\bar R^2$ — penalizes for number of predictors.

Solution.

M2 and M3 have the same $\bar R^2 = 0.70$ , but M3 uses 5 predictors against 2 in M2. By the principle of parsimony (Occam's razor), M2 is preferred: adding 3 predictors does not improve $\bar R^2$ — they do not contribute net information.

Note that $R^2$ of M3 (0.74) exceeds M2 (0.71) — but $R^2$ always grows with additional predictors, even irrelevant ones.

Verification. Could confirm with AIC/BIC: $AIC = n\ln(SSE/n) + 2(p+1)$ — lower is better.

Source. OpenIntro Statistics, §8.3, Table 8.7 — CC-BY-SA

Example— 106.3· Global F-test and ANOVA table

Problem. Regression with $p=3$ predictors, $n=40$ : $SST=2400$ , $SSE=720$ . Build the ANOVA table and test $H_0: \beta_1=\beta_2=\beta_3=0$ at 5% level.

Strategy. Calculate SSR, df, MS, F; compare with critical value $F_{3,36;\,0.05}$ .

Solution.

$SSR = 2400-720 = 1680$ ; $R^2 = 1680/2400 = 0.70$ .

Source	SS	df	MS	F
Regression	1680	3	560	28.0
Error	720	36	20	—
Total	2400	39	—	—

$F_{3,36;\,0.05} \approx 2.87$ . Since $F = 28.0 \gg 2.87$ , we reject $H_0$ : the model as a whole is highly significant.

Verification. $MSE = 20 \implies \hat\sigma = \sqrt{20} \approx 4.47$ — residual standard deviation in units of $Y$ .

Source. OpenStax Statistics, §13.4 — CC-BY

Example— 106.4· Detect multicollinearity via VIF

Problem. Regression of vehicle price ( $Y$ ) on power ( $X_1$ ), displacement ( $X_2$ ) and weight ( $X_3$ ). Auxiliary regressions: $R_1^2=0.89$ ( $X_1$ on $X_2, X_3$ ), $R_2^2=0.91$ , $R_3^2=0.85$ . Calculate the VIFs and evaluate.

Strategy. $VIF_j = 1/(1-R_j^2)$ .

Solution.

$VIF_1 = 1/(1-0.89) = 9.09$ ; $VIF_2 = 1/(1-0.91) = 11.1$ ; $VIF_3 = 1/(1-0.85) = 6.67$ .

$VIF_2 = 11.1 > 10$ : displacement exhibits severe multicollinearity with the other variables.

Action: remove $X_2$ from the model or combine $X_1$ and $X_2$ into a composite index (principal component analysis).

Verification. After removing $X_2$ , recalculate VIFs for remaining predictors.

Source. OpenIntro Statistics, §8.4 — CC-BY-SA

Example— 106.5· Dummy variable and interaction

Problem. Predict salary $Y$ (R$ thousands/year) using experience $X_1$ (years) and gender $D$ (1=female, 0=male). Result: $\hat Y = 45 + 2.5X_1 - 8D$ .

Interpret the coefficient of $D$ and calculate predicted salaries for: (a) man with 10 years of experience; (b) woman with 10 years of experience.

Strategy. The dummy variable creates two different intercepts (one per gender), maintaining equal slope.

Solution.

$\hat\beta_D = -8$ : controlling for experience, women earn on average R$ 8,000/year less than men — difference net (not causal, since we do not control for sector, position, hours).

(a) Man ( $D=0$ ), $X_1=10$ : $\hat Y = 45 + 25 - 0 = 70$ thousand R$/year.

(b) Woman ( $D=1$ ), $X_1=10$ : $\hat Y = 45 + 25 - 8 = 62$ thousand R$/year.

Verification. The difference is constant ( $-8$ ) for any experience level — non-interaction model assumes additive effect. To allow different slopes, add $X_1 \times D$ .

Source. OpenIntro Statistics, §8.2, Example 8.4 — CC-BY-SA

Exercise list

42 exercises · 10 with worked solution (25%)

Application 18Understanding 7Modeling 8Challenge 6Proof 3

Ex. 106.1Application
Regression: $\hat Y = 50 + 3.2X_1 + 28X_2 + 5.5X_3$ (price in R$ thousands, area in m², bedrooms, floor). Interpret each coefficient.
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hat beta_1 = 3.2 measures the total effect of area on price, ignoring bedrooms and floorhat beta_1 = 3.2 indicates that each additional m² is associated with R$ 3,200 more in price, keeping bedrooms and floor fixedhat beta_2 = 28 is the average price of an apartment with zero bedroomshat beta_0 = 50 measures the average effect of floor on price
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Ex. 106.2ApplicationAnswer key
Using $\hat Y = 50 + 3.2X_1 + 28X_2 + 5.5X_3$ , calculate the prediction and residual for an apartment of 80 m², 3 bedrooms, 5th floor with observed price of R$ 450 thousand.
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hat Y = 450 and residual = 0hat Y = 459 and residual = -9hat Y = 50 + 256 + 84 + 27.5 = 417.5 and residual = 32.5hat Y = 50 + 3.2 + 28 + 5.5 = 86.7 and residual = 363.3
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Solve online OpenIntro Statistics · §8.1
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Ex. 106.3Application
$n=40$ , $p=3$ predictors, $SST=2000$ , $SSE=800$ . Calculate $R^2$ and $\bar R^2$ .
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R² = 0.60 and adjusted R² = 0.60 (equal since degrees of freedom cancel)R² = 0.60 and adjusted R² = 0.567R² = 0.567 and adjusted R² = 0.60 (adjusted is always larger)R² = 0.40 and adjusted R² = 0.433 (complement of SSE/SST)
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Solve online OpenIntro Statistics · §8.3
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$\bar R^2 = 1 - \frac{SSE/(n-p-1)}{SST/(n-1)} = 1 - \frac{800/36}{2000/39} = 1 - \frac{22.2}{51.3} = 1 - 0.433 = 0.567$ . Compare with $R^2 = 1-800/2000 = 0.60$ : the penalized fit drops slightly.
Ex. 106.4ApplicationAnswer key
$n=50$ . Three models with $p=1,3,6$ predictors and $SSE/SST = 0.40; 0.35; 0.32$ . Calculate adjusted $\bar R^2$ for each and identify the preferable one.
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M1 is preferable since it has fewer predictors and therefore higher adjusted R²M3 is preferable since it has the lowest SSE/SST and therefore higher adjusted R²M2 is preferable since it has the best balance between R² and number of predictorsAll three models have the same adjusted R² since n is the same
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Solve online OpenIntro Statistics · §8.3
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M1: $\bar R^2_1 = 1-(1-0.60)\times(n-1)/(n-2)$ . With $n=50$ : $\bar R^2_1 = 1-0.40\times49/48 = 1-0.408 = 0.592$ . M2: $\bar R^2_2 = 1-0.35\times49/46 = 1-0.373 = 0.627$ . M3: $\bar R^2_3 = 1-0.32\times49/43 = 1-0.365 = 0.635$ . M3 has higher $\bar R^2$ : preferable among the three.
Ex. 106.5ApplicationAnswer key
$n=60$ , $p=4$ , $SST=3600$ , $R^2=0.72$ . Build the ANOVA table and test the model at 5% level.
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F = 18.3 with df1 = 55; does not reject H0F = 35.4 with df1 = 4 and df2 = 55; rejects H0F = 35.4 with df1 = 4 and df2 = 59; rejects H0 (df2 error)F = 648 with df1 = 1; the model is not globally significant
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Solve online OpenStax Statistics · §13.4
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$SSR = 0.72 \times 3600 = 2592$ ; $SSE = 3600 - 2592 = 1008$ ; $MSR = 2592/4 = 648$ ; $MSE = 1008/55 = 18.3$ ; $F = 648/18.3 = 35.4$ . Critical value $F_{4,55;\,0.05} \approx 2.54$ : reject $H_0$ .
Ex. 106.6ApplicationAnswer key
Auxiliary regressions for 3 predictors: $R_1^2=0.30$ , $R_2^2=0.70$ , $R_3^2=0.92$ . Calculate the VIFs and identify severe multicollinearity.
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VIF_1 = 1.43; VIF_2 = 3.33; VIF_3 = 12.5 — predictor 3 has severe multicollinearityVIF_1 = 0.70; VIF_2 = 0.30; VIF_3 = 0.08 — all below 1, no problemVIF_1 = 1.43; VIF_2 = 3.33; VIF_3 = 12.5 — predictor 2 has severe multicollinearity since VIF > 3VIF_3 = 1/(1-0.92) = 12.5 but the threshold is VIF > 5, so predictors 2 and 3 have problems
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Solve online OpenIntro Statistics · §8.4
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$VIF_j = 1/(1-R_j^2)$ . $VIF_1 = 1/(1-0.30) = 1.43$ ; $VIF_2 = 1/(1-0.70) = 3.33$ ; $VIF_3 = 1/(1-0.92) = 12.5$ . $VIF_3 > 10$ : predictor 3 exhibits severe multicollinearity.
Ex. 106.7Application
Regression of ENEM score on family income ( $X_1$ ) and participation in tutoring program ( $D$ : 1=yes, 0=no): $\hat Y = 480 + 8.2X_1 + 12.4D$ , $n=200$ . Interpret $\hat\beta_D = 12.4$ .
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The program raises the score by 12.4 points for any student (causal effect confirmed)Controlling for family income, participating in the program is associated with 12.4 more points on the score, on averageThe coefficient 12.4 is the total effect of the program, without controlling for incomeThe coefficient 12.4 represents the average score of program participants
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Solve online OpenIntro Statistics · §8.2
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The dummy coefficient indicates that participating in the program is associated with 12.4 more points in the score, controlling for family income. Causal effect requires appropriate experimental design.
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1. Model with dummy: $\hat Y = \hat\beta_0 + \hat\beta_1 X_1 + \hat\beta_2 D$ where $D=1$ if treated, $D=0$ if control.
2. $\hat\beta_2$ is the average difference in $Y$ between treated and control, controlling for $X_1$ .
3. With $\hat\beta_2 = 12.4$ : the program is associated with an average increase of 12.4 points in the score, given the same level of family income.
4. Observation: this is still not causal effect — requires randomization or quasi-experimental design.
Ex. 106.8ApplicationAnswer key
$n=50$ , $p=3$ , $\hat\beta_1=2.8$ , $SE(\hat\beta_1)=0.7$ . Test $H_0:\beta_1=0$ at 5% level (two-tailed).
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T = 2.8/0.7 = 4.0; df = n-p = 47; rejects H0T = 2.8/0.7 = 4.0; df = n-p-1 = 46; rejects H0T = 0.7/2.8 = 0.25; df = 46; does not reject H0T = 4.0; df = n-1 = 49; F-test should be used instead of t-test
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t-test: $T = 2.8/0.7 = 4.0$ . With $df = n-p-1 = 50-3-1 = 46$ , $t_{46;\,0.025} \approx 2.013$ . Since $|T| = 4.0 > 2.013$ , we reject $H_0: \beta_1 = 0$ . The predictor $X_1$ is significant at 5% level.
Ex. 106.9Application
$n=50$ , $p=3$ , $\hat\beta_2=5.2$ , $SE(\hat\beta_2)=1.8$ . Build 95% CI for $\beta_2$ . Use $t_{46;\,0.025}\approx 2.013$ .
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CI = (5.2 - 1.8; 5.2 + 1.8) = (3.4; 7.0) — uses SE directly without multiplying by tCI = (1.58; 8.82) — beta_2 is significant since the CI does not contain 0CI = (5.2 - 1.96 x 1.8; 5.2 + 1.96 x 1.8) = (1.67; 8.73) — uses z instead of tCI = (5.2 - 2.013 x 1.8; 5.2 + 2.013 x 1.8) but contains 0, so not significant
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Solve online OpenIntro Statistics · §8.1
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95% CI for $\beta_2$ : $\hat\beta_2 \pm t_{46;\,0.025} \times SE(\hat\beta_2) = 5.2 \pm 2.013 \times 1.8 = 5.2 \pm 3.62 = (1.58;\; 8.82)$ . The CI does not contain 0: $\beta_2$ is significant at 5% level.
Ex. 106.10Application
Four of the five residuals from a regression are: $3.2$ ; $-1.5$ ; $-0.8$ ; $2.1$ . What is the fifth residual?
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-3.0 (the sum of four residuals is 3.0; so the fifth is -3.0 so the total sum is zero)-5.0 (error: 3.2 - 1.5 - 0.8 + 2.1 = 3.0, not 5.0)3.0 (sum of residuals can be any value)0.0 (the fifth residual is always zero in regression with intercept)
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Solve online OpenIntro Statistics · §8.1
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Sum of residuals is always zero in any regression with intercept: $\sum e_i = 0$ . The sum of four known residuals is $3.2 + (-1.5) + (-0.8) + 2.1 = 3.0$ . Therefore the fifth residual is $e_5 = -3.0$ . Verification: $3.2 - 1.5 - 0.8 + 2.1 - 3.0 = 0$ . Correct.
Ex. 106.11Understanding
Which statement about $R^2$ and adjusted $R^2$ is CORRECT?
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R² always falls when we add an irrelevant predictorAdjusted R² can fall when we add an irrelevant predictorR² and adjusted R² are always equalAdjusted R² is always greater than R²
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Solve online OpenIntro Statistics · §8.3
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$R^2$ never falls when adding predictors (mathematical property of OLS). $\bar R^2$ penalizes for number of predictors and can fall when the added predictor is weak. That is why $\bar R^2$ is the appropriate criterion for comparing models with different numbers of predictors.
Ex. 106.12Understanding
What is the main practical effect of multicollinearity in multiple regression?
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Multicollinearity makes estimators biasedMulticollinearity inflates the standard errors of coefficientsMulticollinearity invalidates the global F-testMulticollinearity only occurs when VIF = 0
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Multicollinearity does not bias estimators (they remain BLUE), but makes $X^TX$ nearly singular, inflating $(X^TX)^{-1}$ and therefore the $SE$ . Individual $t$ -tests lose power, but the global $F$ -test and predictions remain reliable.
Ex. 106.13Understanding
Which statement about partial coefficients in multiple regression is CORRECT?
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The coefficient of X₁ in multiple regression equals the coefficient in simple regression of Y on X₁Adding X₂ to the model can change the sign of X₁Partial coefficients are always smaller than simple coefficientsIn multiple regression, β₁ measures the marginal correlation between X₁ and Y
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The phenomenon is called "sign reversal" or suppression effect. It occurs when $X_2$ is correlated with $X_1$ such that controlling for $X_2$ reveals the true relationship of $X_1$ with $Y$ . Classic example: positive correlation between fire trucks and damage — controlling for fire size, the effect can change.
Ex. 106.14Modeling
Regression of monthly household spending (R$ thousands) on 4 socioeconomic predictors: $n=80$ , $SST=18000$ , $SSE=5400$ . Calculate $MSE$ , $\hat\sigma$ and $\bar R^2$ .
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MSE = 5400/80 = 67.5; hat sigma = 8.22; bar R^2 = 0.70MSE = 5400/75 = 72.0; hat sigma = 8.49; bar R^2 = 0.684MSE = 5400/76 = 71.1; hat sigma = 8.43; bar R^2 = 0.70 (confuses df with n-p)MSE = 18000/79 = 227.8; hat sigma = 15.1 (uses SST instead of SSE)
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$MSE = SSE/(n-p-1) = 5400/75 = 72$ ; $\hat\sigma = \sqrt{72} \approx 8.49$ . $\bar R^2 = 1 - \frac{5400/75}{18000/79} = 1 - \frac{72}{227.85} \approx 1 - 0.316 = 0.684$ .
Ex. 106.15Modeling
Model: $\hat Y = 45 + 2.5X_1 - 8D$ (salary in R$ thousands, experience in years, $D$ =1 if woman). Calculate salaries for (a) man, 10 years; (b) woman, 10 years. How to add interaction to check if the gap varies with experience?
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Man: hat Y = 70 thousand; Woman: hat Y = 62 thousand; gap of R$ 8 thousand constant for any experienceMan: hat Y = 70 thousand; Woman: hat Y = 70 - 8 x 10 = -10 thousand (multiplies D by experience)Man: hat Y = 62 thousand; Woman: hat Y = 70 thousand (reverses dummy sign)Man: hat Y = 45 thousand; Woman: hat Y = 37 thousand (ignores effect of X1)
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Solve online OpenIntro Statistics · §8.2
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1. Non-interaction model: one line with equal slope for men and women.
2. For men ( $D=0$ ): $\hat Y = 45 + 2.5X_1$ . For women ( $D=1$ ): $\hat Y = 45 + 2.5X_1 - 8 = 37 + 2.5X_1$ .
3. Salary difference between genders is constant at R\$ 8 thousand for any experience level.
4. To test if the difference varies with experience, add the interaction term $X_1 \times D$ .
5. Observation: the interaction model is $\hat Y = \beta_0 + \beta_1 X_1 + \beta_2 D + \beta_3(X_1 \cdot D)$ . If $\hat\beta_3$ is significant, the gender difference varies with experience.
Ex. 106.16Modeling
A researcher has a regression model with 2 predictors ( $R^2=0.65$ , $\bar R^2=0.64$ ) and considers adding a third predictor. Describe two criteria to decide whether to include it.
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Include X3 if the R² of the new model is larger than 0.65 (compare raw R²)Include X3 if adjusted R² rises and/or the partial t-test of hat beta_3 has p-value < 0.05Include X3 if VIF > 10, since that indicates X3 contributes new informationInclude X3 if SSE of the new model is larger than SSE of the original model
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Compare models with and without $X_3$ : if $\bar R^2$ rises when including $X_3$ , keep it. Alternative: do partial $t$ -test for $\hat\beta_3$ . If p-value less than 0.05, $X_3$ is significant given the others. AIC criterion: $AIC = n\ln(SSE/n) + 2(p+1)$ , lower is better. Use VIF to check if $X_3$ introduces severe multicollinearity.
Ex. 106.17Challenge
Prove that the hat matrix $H = X(X^TX)^{-1}X^T$ is idempotent: $H^2 = H$ .
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H^2 = H because H is diagonalH^2 = H because X^TX(X^TX)^{-1} = I, canceling the central factorH^2 = 2H because each projection doubles the normH^2 = H^T since the matrix is symmetric but not idempotent
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OpenIntro Statistics · §8.1
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The hat matrix $H = X(X^TX)^{-1}X^T$ is symmetric ( $H^T=H$ ) and idempotent ( $H^2=H$ ). Idempotence: $H^2 = X(X^TX)^{-1}X^TX(X^TX)^{-1}X^T = X(X^TX)^{-1}X^T = H$ . This confirms that $H$ is an orthogonal projection: applying it twice gives the same result.
Ex. 106.18Challenge
Data: $n=6$ observations with $Y=(10,14,18,12,16,20)$ , $X_1=(1,2,3,2,3,4)$ , $X_2=(5,4,3,6,5,4)$ . Write the design matrix $X$ and state the procedure to calculate $\hat{\boldsymbol\beta} = (X^TX)^{-1}X^T\mathbf{y}$ (matrix inversion by hand not required — describe the steps).
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X is a 6x2 matrix (n=6, p=2 columns for X1 and X2)X is a 6x3 matrix with first column of 1s, second X1 and third X2X is a 3x6 matrix since p+1=3 parameters and n=6 observationsX is a 6x6 matrix since the design matrix is always square
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Solve online OpenIntro Statistics · §8.1
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Build the matrices with the given data, calculate the product and obtain $\hat{\boldsymbol\beta}$ . This exercise is best solved with a matrix calculator or R/Python.
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1. With $n=6$ , $p=2$ : build $X$ (column of 1s, column $X_1$ , column $X_2$ ).
2. Calculate $X^TX$ (3×3 matrix).
3. Calculate $(X^TX)^{-1}$ .
4. Calculate $X^T\mathbf{y}$ (3×1 vector).
5. Multiply: $\hat{\boldsymbol\beta} = (X^TX)^{-1}X^T\mathbf{y}$ .
6. Fun fact: in real applications, $(X^TX)^{-1}$ is computed numerically via QR decomposition, not direct inversion (more numerically stable).
Ex. 106.19Proof
Prove that in any regression with intercept, $\sum_{i=1}^n e_i = 0$ , using the orthogonality $X^T\mathbf{e} = \mathbf{0}$ .
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The proof uses that X^T e = 0 implies that the first row (vector of 1s) multiplied by e is zeroThe proof uses that H is diagonal, so X^T H = X^Tsum e_i = 0 only holds for simple regression, not for multipleX^T e = 0 implies sum e_i = 0 only when p = 1
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OpenIntro Statistics · §8.1
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We must show that $\sum e_i = 0$ . Since $\mathbf{e} = (I-H)\mathbf{Y}$ and the first column of $X$ is a vector of 1s ( $\mathbf{1}$ ), we have $X^T\mathbf{e} = X^T(I-H)\mathbf{Y} = X^T\mathbf{Y} - X^TX(X^TX)^{-1}X^T\mathbf{Y} = \mathbf{0}$ . The first row of $X^T$ is $\mathbf{1}^T$ , therefore $\mathbf{1}^T\mathbf{e} = \sum e_i = 0$ .
Ex. 106.20Challenge
Show that adding a predictor to the model increases $\bar R^2$ if and only if the test statistic $|T|$ of the new predictor is greater than 1.
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bar R^2 increases if and only if |T_j| > t_{n-p-1; 0.025} (threshold of t-test at 5%)bar R^2 increases if and only if |T_j| > 1 (less conservative threshold than t-test at 5%)bar R^2 increases if and only if T_j^2 > F_{1,n-p-1; 0.05} (threshold of F-test)bar R^2 always increases when adding any predictor, since SSE cannot rise
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Solve online OpenIntro Statistics · §8.3
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Formula for change in $\bar R^2$ when adding a predictor: $\bar R^2$ increases if and only if the partial $F$ of the new predictor is greater than 1. Equivalently: $\bar R^2$ of augmented model $>$ $\bar R^2$ of reduced model $\Leftrightarrow$ $T_j^2 > 1$ . This shows that the $\bar R^2$ criterion is less conservative than the $t$ -test at 5% level (which requires $|T_j| > t_{n-p-1;\,0.025} \approx 2$ ).
Ex. 106.21Application
A multiple model predicts baby weight ( $Y$ ) with predictors: smoking cigarettes ( $X_1$ ) and parity ( $X_2$ , number of previous children). The coefficient of $X_1$ measures which effect?
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Smoking cigarettes: coefficient measures the effect of smoking ignoring paritySmoking cigarettes: coefficient measures the effect of smoking controlling for parityParity: coefficient measures the total effect of number of children on weightBoth coefficients measure marginal effects without mutual control
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In multiple regression, the coefficient of "smoking" measures the partial effect — the expected change in baby weight per unit change in smoking, holding parity constant. Each coefficient is a conditional effect given the other predictors in the model.
Ex. 106.22ApplicationAnswer key
Research with 55 Duke University students models GPA as a function of: study hours per night, number of classes missed and gender. What does the coefficient of "study hours" represent in this multiple model?
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GPA depends only on study hours; the other variables are irrelevantThe coefficient of study hours controls for number of classes missed and genderThe coefficient of gender measures the total effect of gender on GPAAdjusted R² decreases when unnecessary predictors are included
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Solve online OpenIntro Statistics · §9.1
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In a model with multiple predictors (GPA, study hours, classes missed, gender), the coefficient of each variable controls for the others. The coefficient of study hours represents the partial effect of one additional hour of study, holding the other variables (classes missed, gender) constant.
Ex. 106.23Application
Lumber mills estimate tree volume (in cubic feet) from diameter and height. A multiple regression model includes diameter ( $X_1$ ), height ( $X_2$ ) and the product diameter $\times$ height ( $X_3$ ). What is the role of the term $X_3$ in the model?
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X3 = diameter x height is a control variable to remove the effect of correlation between X1 and X2X3 is an interaction term: the effect of diameter on volume varies as height changesX3 transforms the model from linear to quadratic in X1 onlyX3 is necessary to avoid multicollinearity between X1 and X2
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Solve online OpenIntro Statistics · §9.1
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Model with 3 predictors: diameter ( $X_1$ ), height ( $X_2$ ) and interaction diameter×height ( $X_3$ ). The adjusted $\bar R^2$ should be high due to the strong geometric relationship between these dimensions and volume.
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1. Model: $\hat Y = \hat\beta_0 + \hat\beta_1 X_1 + \hat\beta_2 X_2 + \hat\beta_3 X_3$ where $X_1$ = diameter, $X_2$ = height, $X_3$ = diameter × height.
2. Wood volume is the response variable $Y$ .
3. The product $X_3 = X_1 \times X_2$ is an interaction term that allows the effect of diameter on volume to depend on height.
4. With all three predictors, the model better captures the nonlinear relationship between dimensions and volume.
Ex. 106.24ApplicationAnswer key
Comparing the simple model (smoking only) with the multiple model (smoking + parity) for baby weight: can the smoking coefficient differ in the two models? Why?
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No, since the R² of simple regression is always smallerYes, the simple R² can be smaller since it does not control for parityNo, the smoking coefficient must be equal in both modelsYes, the R² of multiple regression is always larger
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Solve online OpenIntro Statistics · §9.1
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In simple regression, the smoking coefficient absorbs also the confounding effect of parity (smoking mothers tend to have more or fewer children). In the multiple model, controlling for parity, the smoking coefficient captures only the partial effect. Coefficients can differ due to confounding.
Ex. 106.25Understanding
Why can adjusted $\bar R^2$ decrease when we add a new predictor to the multiple regression model?
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Adjusted R² cannot decrease when we add a predictorAdjusted R² can decrease when the added predictor does not contribute enoughAdjusted R² is always greater than unadjusted R²Adjusted R² penalizes for having too few predictors
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Solve online OpenStax Statistics · §12.H
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$\bar R^2 = 1 - (\mathrm{SSE}/(n-p-1))/(\mathrm{SST}/(n-1))$ . Adding a predictor reduces SSE, but also reduces the degrees of freedom $n-p-1$ . If the SSE reduction is small, the ratio $\mathrm{SSE}/(n-p-1)$ can increase, making $\bar R^2$ decrease.
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1. Formula: $\bar R^2 = 1 - \frac{\mathrm{SSE}/(n-p-1)}{\mathrm{SST}/(n-1)}$ .
2. When adding a predictor: $p \to p+1$ , so $n-p-1 \to n-p-2$ (denominator decreases).
3. SSE also decreases (more predictors reduce errors).
4. If the SSE reduction is small (uninformative predictor), the ratio $\mathrm{SSE}/(n-p-2)$ can increase.
5. In that case, $\bar R^2$ decreases — penalizing the model for including a redundant predictor.
Ex. 106.26Modeling
Researchers investigate school absenteeism ( $Y$ ) as a function of lack of discipline ( $X_1$ ) and distance from school to student's home ( $X_2$ , in km). Write the multiple regression model and interpret each coefficient.
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hat beta_1 measures the total effect of X1 on Y, summing direct and indirect effects via X2hat beta_1 measures the effect of X1 on Y holding X2 (distance) constant — partial effecthat beta_0 is the expected absenteeism for a student with X1 = 1 and X2 = 1hat beta_2 is negative by definition since greater distance reduces absenteeism
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Solve online OpenIntro Statistics · §9.1
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Absenteeism as a function of lack of discipline and distance from school. Model: $\hat Y = \hat\beta_0 + \hat\beta_1 X_1 + \hat\beta_2 X_2$ . The coefficient $\hat\beta_1$ measures the effect of lack of discipline on absenteeism controlling for distance. The coefficient $\hat\beta_2$ measures the effect of distance controlling for discipline. The intercept is the expected absenteeism when both variables are zero.
Ex. 106.27Understanding
What is the main practical effect of multicollinearity in a multiple regression model?
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Multicollinearity violates the normality assumption of errorsMulticollinearity increases the standard errors of coefficients, making the t-test impreciseMulticollinearity makes OLS coefficients biasedMulticollinearity reduces the R² of the model
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Solve online OpenIntro Statistics · §9.2
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When predictors are highly correlated, the matrix $X^TX$ becomes nearly singular, amplifying coefficient standard errors. OLS estimators remain unbiased, but have very high variance, making individual $t$ -tests powerless to detect real effects.
Ex. 106.28ApplicationAnswer key
The VIF (Variance Inflation Factor) of a predictor is 5. What does this mean for the standard error of the estimated coefficient?
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VIF = 1 indicates perfect multicollinearityVIF = 5 indicates that the standard error of beta_j is sqrt(5) times larger than without collinearityVIF > 1 always indicates serious multicollinearity problemVIF measures the correlation between beta_j and the other coefficients
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Solve online OpenIntro Statistics · §9.2
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The Variance Inflation Factor is $\mathrm{VIF}_j = 1/(1 - R_j^2)$ , where $R_j^2$ is the $R^2$ of regressing $X_j$ on the other predictors. $\mathrm{VIF}_j = 5$ means that $\mathrm{Var}(\hat\beta_j)$ is 5 times larger than without collinearity; so the SE is $\sqrt{5} \approx 2.24$ times larger.
Ex. 106.29Modeling
A researcher wants to model ENEM score ( $Y$ ) of students as a function of: family income ( $X_1$ ), parental education ( $X_2$ ), weekly study hours ( $X_3$ ) and school type ( $X_4$ : 1=private, 0=public). Write the model, identify possible multicollinearity and describe how to verify it.
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X1 and X2 cannot be collinear since income and parental education are different types of variablesX1 (income) and X2 (parental education) tend to be highly correlated — VIF can be high for bothX4 (school dummy) never causes multicollinearity since it only has values 0 and 1Multicollinearity is diagnosed by the R² of the full model, not by individual VIFs
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Solve online OpenIntro Statistics · §9.2
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Model: $\hat Y = \hat\beta_0 + \hat\beta_1 \cdot \text{income} + \hat\beta_2 \cdot \text{education} + \hat\beta_3 \cdot \text{hours} + \hat\beta_4 \cdot \text{school}$ . Check multicollinearity between income and parental education via VIF.
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1. Proposed predictors to forecast ENEM scores: family income, parental education, study hours, school type (public/private).
2. Check multicollinearity: income and parental education may be highly correlated. Calculate $\mathrm{VIF}$ for each predictor.
3. If $\mathrm{VIF} > 10$ for any predictor, consider removing the most collinear or using principal components.
4. Check assumptions: plot residuals vs. fitted (homoscedasticity), QQ-plot (normality), plot residuals vs. order (independence).
Ex. 106.30Modeling
Explain the variable selection strategies forward selection and backward elimination. Which criterion is more recommended when you have many candidate predictors?
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Forward adds the least significant predictor first; backward removes the most significantForward includes predictors from most to least significant; backward removes from least to most significant; AIC is recommended with many candidatesForward and backward are identical when the number of predictors is smallThe AIC criterion penalizes models with few predictors, always favoring saturated models
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Solve online OpenIntro Statistics · §9.3
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In stepwise selection, start without predictors (or with all) and add (or remove) predictors one by one by AIC criterion or partial $F$ -test p-value. Forward selection includes first the most significant predictor, then the second most significant given the first, etc. Backward elimination starts with all and removes the least significant. The AIC criterion balances fit and parsimony: $\mathrm{AIC} = n\ln(\mathrm{SSE}/n) + 2(p+1)$ .
Ex. 106.31ApplicationAnswer key
In the model of the previous exercise, the variable $X_4$ is a dummy (1=private, 0=public). How to interpret the coefficient $\hat\beta_4$ ?
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The coefficient of X4 measures the effect of being a private school relative to the interceptThe coefficient of X4 measures the expected difference in Y between private and public school, controlling for the other predictorsThe coefficient of X4 is always positive since private > publicDummy variables cannot be used in multiple regression
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Solve online OpenIntro Statistics · §9.1
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When $X_4 = 1$ (private), $\hat Y$ increases by $\hat\beta_4$ compared to $X_4 = 0$ (public), holding the other predictors constant. $\hat\beta_4$ is the expected average difference between the two school categories, partially adjusted for income, parental education and study hours.
Ex. 106.32Understanding
What is the relationship between SST (total variation), SSR (variation explained by regression) and SSE (variation of residuals)?
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SSR = 0 indicates perfect model fitSST = SSR + SSE, where SSR is the explained variation and SSE is the residual variationSSE = SST + SSR since residuals contribute to total variationSSR = SSE when the model explains 50% of the variation
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Solve online OpenStax Statistics · §12.H
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In the variance decomposition: $\mathrm{SST} = \mathrm{SSR} + \mathrm{SSE}$ , where SST = total sum of squares, SSR = regression sum of squares (explained) and SSE = sum of squared residuals (unexplained). By definition, $R^2 = \mathrm{SSR}/\mathrm{SST} = 1 - \mathrm{SSE}/\mathrm{SST}$ .
Ex. 106.33ApplicationAnswer key
A multiple regression model with $p = 3$ predictors fitted to $n = 55$ observations has $\mathrm{SSR} = 120$ and $\mathrm{SSE} = 80$ . Calculate the global $F$ statistic and conclude at 5% level (critical value $F_{0.05; 3; 51} \approx 2.79$ ).
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F = (120/3)/(80/55) = 40/1.45 = 27.6; df2 = n = 55 (df error)F = (120/3)/(80/51) = 25.5; rejects H0 at 5% levelF = (80/3)/(120/51) = 11.3; accepts H0 (reverses SSR and SSE)F = (120/51)/(80/3) = 0.088; does not reject H0 (reverses MSR and MSE)
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Solve online OpenIntro Statistics · §9.1
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$F = 25.5 > 2.79$ . We reject $H_0$ : the 3 predictors significantly explain the variation in Y. (Resp: $F = 25.5$ )
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1. Data: $p = 3$ , $n = 55$ , $\mathrm{SSR} = 120$ , $\mathrm{SSE} = 80$ .
2. Degrees of freedom: $df_1 = p = 3$ , $df_2 = n - p - 1 = 55 - 3 - 1 = 51$ .
3. F statistic: $F = (\mathrm{SSR}/df_1)/(\mathrm{SSE}/df_2) = (120/3)/(80/51) = 40/1.569 = 25.5$ .
4. Critical value: $F_{0.05;\,3;\,51} \approx 2.79$ . Since $25.5 \gg 2.79$ , we reject $H_0$ .
5. Conclusion: the set of 3 predictors significantly explains the variation in Y at 5% level.
Ex. 106.34Modeling
Tree volume in an orchard (in cubic feet) is modeled as a function of diameter ( $X_1$ , in inches) and height ( $X_2$ , in feet). Based on the multiple model, interpret the coefficient $\hat\beta_1$ . What does an $\bar R^2 = 0.944$ indicate about the fit?
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hat beta_1 measures the average increase in volume per inch of diameter, without controlling height; bar R² = 0.944 indicates poor fithat beta_1 measures the average increase in volume per inch of diameter keeping height constant; bar R² = 0.944 indicates excellent fithat beta_1 is the intercept of the model when diameter is zerobar R² = 0.944 indicates that 5.6% of volume variation is explained by the model
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Solve online OpenIntro Statistics · §9.1
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Model: $\hat{\text{vol}} = \hat\beta_0 + \hat\beta_1 \cdot \text{diam} + \hat\beta_2 \cdot \text{height}$ . Each coefficient is the partial effect holding the other dimension constant. A high $\bar R^2$ confirms that larger trees have larger volume, as geometrically expected.
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1. Base model: $\hat\beta = (X^TX)^{-1}X^T\mathbf{y}$ . With $p = 2$ , $n = 3$ : $X^TX$ is $3 \times 3$ .
2. Data: $X_1 =$ diameter, $X_2 =$ height (in feet), $Y =$ volume (cubic feet).
3. Calculate $\hat\beta_0, \hat\beta_1, \hat\beta_2$ numerically and interpret: per extra unit of diameter (height fixed), volume changes by $\hat\beta_1$ cubic feet.
4. Check $\bar R^2$ . If high (>0.90), the model fits well. Plot residuals vs. fitted to check homoscedasticity.
Ex. 106.35Challenge
In a model with three predictors $X_1, X_2, X_3$ where $X_3 = 2X_1 + X_2$ (perfect multicollinearity), explain why the OLS solution $\hat\beta = (X^TX)^{-1}X^T\mathbf{y}$ does not exist. What does statistical software do in this situation?
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X^TX has full rank even with X3 = 2X1 + X2, since linear dependence does not affect inversionX^TX is singular (det = 0) when X3 is a linear combination of X1 and X2, making (X^TX)^{-1} undefinedX^TX is singular but the OLS solution still exists as an inconsistent systemThe software resolves the problem automatically without removing any predictor
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Solve online OpenIntro Statistics · §9.2
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If $X_3 = \alpha X_1 + \gamma X_2 + c$ (exact linear combination of the others), the matrix $X^TX$ has determinant zero and is not invertible. The OLS solution $\hat\beta = (X^TX)^{-1}X^T\mathbf{y}$ does not exist. On the computer, the software detects perfect multicollinearity and automatically discards the redundant variable.
Ex. 106.36Challenge
A multiple regression model with 15 predictors and $n = 20$ observations has $R^2 = 0.97$ . Explain why this high $R^2$ can be misleading and how cross-validation provides a more honest measure of predictive power.
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R² = 0.97 with 15 predictors and n = 20 is reliable since R² penalizes for complexityWith p = 15 and n = 20, the model almost interpolates the data (overfitting); cross-validation estimates out-of-sample errorCross-validation is unnecessary when R² > 0.90R² = 0.97 guarantees that the model generalizes well to new data
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Solve online OpenIntro Statistics · §9.3
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The cross-validation RMSE (leave-one-out or k-fold) is more honest than in-sample $R^2$ because it measures error on data not used in fitting. A model with many predictors can have $R^2 \approx 1$ (overfitting) but high out-of-sample RMSE. Adjusted $\bar R^2$ and AIC penalize complexity, but only cross-validation directly estimates prediction error.
Ex. 106.37Application
In a multiple model, the intercept $\hat\beta_0$ is not significant at 5% level. Should it be removed from the model? What is the interpretation of the intercept?
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The intercept is the prediction of Y when all variables are zero, even without practical meaningThe intercept measures the effect of X1 when the other predictors are zeroThe intercept is only interpretable when X = 0 is within the sampled rangeThe intercept should be removed if not significant at 5% level
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Solve online OpenIntro Statistics · §9.1
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The intercept $\hat\beta_0$ is the predicted value of $Y$ when all predictors are zero. In many practical contexts, this makes no sense (height = 0, income = 0). Still, the intercept is necessary to correctly center the regression and should be kept even when not significant, to avoid bias in the other coefficients.
Ex. 106.38Proof
Derive the normal equations $X^TX\hat\beta = X^T\mathbf{y}$ from the minimization of $\|\mathbf{y} - X\boldsymbol\beta\|^2$ with respect to $\boldsymbol\beta$ . Use the result to show that residuals are orthogonal to the columns of $X$ .
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The derivative of ||y - X beta||^2 with respect to beta is 2X(y - Xbeta), set to zero gives X^TXbeta = X^TyThe derivative of ||y - X beta||^2 is 2||y - Xbeta||, which has no closed matrix formThe normal equations are X X^T beta = X y, not X^TX beta = X^TyX^T e = 0 implies that residuals are orthogonal only to vector y, not the columns of X
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Solve online OpenIntro Statistics · §9.1
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The normal equations are $X^TX\hat\beta = X^T\mathbf{y}$ . Multiplying on the left by $(X^TX)^{-1}$ : $\hat\beta = (X^TX)^{-1}X^T\mathbf{y}$ . To verify $X^T\mathbf{e} = 0$ : $\mathbf{e} = \mathbf{y} - X\hat\beta$ , so $X^T\mathbf{e} = X^T\mathbf{y} - X^TX\hat\beta = X^T\mathbf{y} - X^T\mathbf{y} = 0$ . This implies $\sum e_i = 0$ (first row of $X$ is 1s).
Ex. 106.39Proof
Define the hat matrix $H = X(X^TX)^{-1}X^T$ . Prove that $H$ is idempotent ( $H^2 = H$ ) and symmetric. What is the trace of $H$ and what does it represent geometrically?
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tr(H) = n since H is an n x n matrixtr(H) = p+1 (number of estimated parameters), since eigenvalues of H are 0 and 1tr(H) = 0 since the sum of eigenvalues of a projection is zerotr(H) = n - p - 1 since that is the number of degrees of freedom of residuals
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Solve online OpenIntro Statistics · §9.1
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The hat matrix is $H = X(X^TX)^{-1}X^T$ . Fitted values are $\hat{\mathbf{y}} = H\mathbf{y}$ and residuals $\mathbf{e} = (I-H)\mathbf{y}$ . Properties: (1) $H$ is symmetric: $H^T=H$ ; (2) $H$ is idempotent: $H^2=H$ (orthogonal projection); (3) eigenvalues of $H$ are 0 and 1; (4) $\mathrm{tr}(H) = p+1$ (number of estimated parameters).
Ex. 106.40Understanding
Which diagnostic plot specifically verifies the normality assumption of errors in a multiple regression model? What do we observe in this plot when normality is satisfied?
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Residuals must have mean zero and constant variances across the fitted valuesThe plot of residuals vs. fitted should show a fan-shaped pattern (heteroscedasticity)The QQ-plot of residuals should follow a line, indicating normalityResiduals must be positive for the model to be valid
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Solve online OpenIntro Statistics · §9.2
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The QQ-plot (quantile-quantile plot) of residuals versus quantiles of a standard normal distribution should show points on a diagonal line if errors are normally distributed. Deviations at the tails indicate skewness or heavy tails. It is one of the standard diagnostic plots for the normality assumption (N in the LINE mnemonic).
Ex. 106.41Modeling
For the model of exercise 106.33 ( $p = 3$ , $n = 55$ , $\mathrm{SSR} = 120$ , $\mathrm{SSE} = 80$ ): calculate $R^2$ and adjusted $\bar R^2$ . Compare the two and interpret the difference.
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R² = 0.60 and bar R² = 0.576 — adjusted R² is lower since it penalizes for the number of predictorsR² = 0.576 and bar R² = 0.60 — adjusted is always greater than unadjustedR² = 0.60 and bar R² = 0.60 — they are identical since n is largeR² = 120/80 = 1.5 — SSR divided by SSE
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Solve online OpenIntro Statistics · §9.3
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For the regression with $n = 55$ , $p = 3$ , $\mathrm{SSR} = 120$ , $\mathrm{SSE} = 80$ : $\mathrm{SST} = 200$ . $R^2 = 120/200 = 0.60$ . $\bar R^2 = 1 - (80/51)/(200/54) = 1 - 1.569/3.704 = 1 - 0.424 = 0.576$ . The model explains 57.6% of Y variance after penalizing for complexity.
Ex. 106.42Challenge
In a model with $p = 10$ predictors, each coefficient is tested at level $\alpha = 0.05$ . What is the probability of at least one Type I error (false positive) under the global $H_0$ ? How does Bonferroni adjustment correct this?
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P(at least one false positive) = 10 x 0.05 = 0.50; Bonferroni uses alpha* = alpha x pP(at least one false positive) = 1 - 0.95^10 ≈ 0.40; Bonferroni uses alpha* = alpha/p = 0.005P(at least one false positive) = 0.05 since tests are independentP(at least one false positive) = 0.95^10 ≈ 0.60; Bonferroni increases alpha to compensate
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Solve online OpenIntro Statistics · §9.3
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With $p$ predictors, the individual $t$ -test of $H_0: \beta_j = 0$ at level $\alpha$ has probability $\alpha$ of false positive per predictor. With $p = 10$ independent tests and $\alpha = 0.05$ : probability of at least one false positive $= 1 - (1-0.05)^{10} \approx 0.40$ . Corrections: Bonferroni ( $\alpha^* = \alpha/p$ ), Holm (sequential procedure), or false discovery rate control (Benjamini-Hochberg FDR).

Sources

OpenIntro Statistics (4th ed.) — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA · Chapter 8 (Multiple and logistic regression). Primary source for coefficient interpretation, $\bar R^2$ , multicollinearity and dummy variables.
Statistics — OpenStax — Illowsky, Dean · CC-BY · Chapter 13 (Linear Regression and Correlation — Multiple). Source for ANOVA tables of multiple regression and global F-test.
Probabilidade e Estatística — Wikilivros — collaborative · CC-BY-SA · Multiple regression section. Portuguese-language reference with matrix notation compatible with Brazilian engineering curriculum.