Multiple Regression Exercises in R: 12 Model-Building Problems, Solved Step-by-Step

These 12 multiple regression exercises in R walk you through fitting models, interpreting partial coefficients, testing predictor significance with t-tests and anova(), diagnosing residuals, spotting multicollinearity with VIF, adding interactions, and predicting on new data. Every problem ships with a runnable scaffold, a hint, and a collapsible solution so you can self-check the instant you finish coding.

How do you fit and interpret a multiple regression model?

One line of R fits a multiple regression: pass y ~ x1 + x2 + ... to lm() and wrap the result with summary(). The slopes you read off the output are partial effects, meaning each coefficient shows the change in the outcome for a 1-unit change in that predictor while the other predictors stay constant. We'll fit a two-predictor model on mtcars right now, then Problems 1 to 3 ask you to pull a coefficient, interpret it, and build a 95% confidence interval for it.

RFit a multiple regression on mtcars

# Predict miles per gallon from weight and horsepower fit1 <- lm(mpg ~ wt + hp, data = mtcars) summary(fit1) #> Call: #> lm(formula = mpg ~ wt + hp, data = mtcars) #> #> Coefficients: #> Estimate Std. Error t value Pr(>|t|) #> (Intercept) 37.22727 1.59879 23.285 < 2e-16 *** #> wt -3.87783 0.63273 -6.129 1.12e-06 *** #> hp -0.03177 0.00903 -3.519 0.00145 ** #> #> Residual standard error: 2.593 on 29 degrees of freedom #> Multiple R-squared: 0.8268, Adjusted R-squared: 0.8148 #> F-statistic: 69.21 on 2 and 29 DF, p-value: 9.109e-12

The weight coefficient (−3.88) says mpg drops by roughly 3.88 miles per gallon for each extra 1000 lb of weight, after holding horsepower constant. The horsepower coefficient (−0.032) says mpg drops 0.032 per extra hp, holding weight constant. Both p-values are tiny, so both predictors earn their place. Adjusted R² of 0.81 means the two predictors together explain about 81% of the variation in mpg, which is very strong for a two-variable model.

Key Insight

Every multiple-regression slope is a partial slope. The coefficient on wt is not "the effect of weight" in general, it is "the effect of weight once hp is already in the model." Change the set of other predictors and the partial slope changes too. That one idea is what separates multiple regression from a stack of simple regressions.

Problem 1: Fit a new model and extract one coefficient

Try it: Fit a model predicting mpg from wt and cyl on mtcars, then print only the cyl coefficient (a single number).

RYour turn: extract cyl coefficient

# Problem 1: fit lm(mpg ~ wt + cyl) and pull the cyl slope ex1_fit <- lm(mpg ~ wt + cyl, data = mtcars) # Print just the cyl slope here: # your code here #> Expected: approximately -1.5

Click to reveal solution

RProblem 1 solution

ex1_fit <- lm(mpg ~ wt + cyl, data = mtcars) coef(ex1_fit)["cyl"] #> cyl #> -1.507795

Explanation: coef() returns a named numeric vector of all slopes; indexing by name pulls out just the one you want. Each extra cylinder costs about 1.51 mpg after controlling for weight.

Problem 2: Write a one-sentence interpretation of a coefficient

Try it: Using fit1 from the teaching block above, write one sentence that correctly interprets the hp coefficient (−0.032). Store your interpretation as a character string in ex2_ans. The grader below checks that the sentence mentions "holding" or "constant".

RYour turn: interpret hp coefficient

# Problem 2: write a correct partial-effect interpretation ex2_ans <- "your one-sentence interpretation here" grepl("holding|constant", ex2_ans, ignore.case = TRUE) #> Expected: TRUE

Click to reveal solution

RProblem 2 solution

ex2_ans <- "Each extra unit of horsepower lowers mpg by about 0.032, holding weight constant." grepl("holding|constant", ex2_ans, ignore.case = TRUE) #> [1] TRUE

Explanation: The sentence nails all three requirements: the direction (lowers), the magnitude (0.032), and the partial qualifier (holding weight constant). Any interpretation of a multiple-regression slope that skips the partial phrase is technically wrong.

Problem 3: Build 95% confidence intervals for every coefficient

Try it: Use confint() on fit1 to get 95% confidence intervals for all three coefficients, then read the interval for wt.

RYour turn: confidence intervals via confint()

# Problem 3: get 95% CIs for every coefficient ex3_ci <- # your code here print(ex3_ci) #> Expected: a 3-row matrix with 2.5 % and 97.5 % columns

Click to reveal solution

RProblem 3 solution

ex3_ci <- confint(fit1, level = 0.95) print(ex3_ci) #> 2.5 % 97.5 % #> (Intercept) 33.95738283 40.49715778 #> wt -5.17191604 -2.58374544 #> hp -0.05025119 -0.01328079

Explanation: The wt interval [−5.17, −2.58] does not cross zero, so the partial effect of weight is confidently negative. confint() uses the t-distribution with residual degrees of freedom, which matches the t-test in summary() exactly.

How do you test whether a predictor truly matters?

Two tests decide whether a predictor earns its place. The t-test row next to each coefficient in summary() answers "does this one predictor add value, given all the others?" The partial F-test from anova(fit_small, fit_big) answers the same question for a group of predictors at once, which is what you need when adding or dropping several variables together. Problems 4 to 6 make you read a t-row, compare two nested models, and run drop1() to test every predictor one at a time.

RCompare nested models with anova()

# Small model: weight + horsepower fit_small <- lm(mpg ~ wt + hp, data = mtcars) # Big model: small + cylinders fit_big <- lm(mpg ~ wt + hp + cyl, data = mtcars) anova(fit_small, fit_big) #> Analysis of Variance Table #> #> Model 1: mpg ~ wt + hp #> Model 2: mpg ~ wt + hp + cyl #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 29 195.05 #> 2 28 170.44 1 24.603 4.0418 0.05403 .

The F statistic compares how much extra residual sum of squares the big model reduces against the noise you would see by chance. Here F = 4.04 with p = 0.054, right at the edge of significance. Adding cyl nudges the fit but does not clearly beat chance at α = 0.05. That borderline result is a healthy reminder: real data rarely gives you a clean yes/no, and a p-value hovering near α deserves a skeptical second look (sample size, confounding, a better specification).

The partial F-test follows the same F formula every nested model comparison uses:

$$F = \frac{(RSS_{small} - RSS_{big})/q}{RSS_{big}/(n - p - 1)}$$

Where:

$RSS_{small}$, $RSS_{big}$ = residual sums of squares of the two models
$q$ = number of parameters the big model adds
$n$ = sample size
$p$ = number of predictors in the big model

Tip

Report the F statistic, its degrees of freedom, the p-value, and the change in adjusted R² together. A p-value alone is the weakest of those four numbers. The ΔR² tells the reader how much predictive power you gained, the F and df let them reproduce the test, and the p-value puts a probability on the comparison.

Problem 4: Read a t-test row and decide

Try it: Using fit_big from the teaching block, print the coefficient table and state whether hp is significant at α = 0.05. Save "yes" or "no" to ex4_ans.

RYour turn: read a t-test row

# Problem 4: inspect fit_big's summary and decide on hp at alpha=0.05 # your code here ex4_ans <- # "yes" or "no"

Click to reveal solution

RProblem 4 solution

summary(fit_big)$coefficients #> Estimate Std. Error t value Pr(>|t|) #> (Intercept) 38.75178840 1.786999400 21.6854668 4.867545e-19 #> wt -3.16697281 0.740557190 -4.2765351 2.008846e-04 #> hp -0.01804156 0.011985755 -1.5052507 1.435138e-01 #> cyl -0.94161630 0.550640701 -1.7098844 9.847503e-02 ex4_ans <- "no" #> Explanation: hp p-value is 0.144, above 0.05.

Explanation: Once cyl enters the model, hp no longer clears α = 0.05, its p-value jumps from 0.0015 in fit1 to 0.144 here. That shift is a classic multicollinearity symptom: hp and cyl carry overlapping information, so each partial slope gets diluted.

Problem 5: Compare two nested models with anova()

Try it: Fit lm(mpg ~ wt + hp + cyl + disp) on mtcars and compare it against fit_big using anova(). Does adding disp improve the fit at α = 0.05?

RYour turn: nested model comparison

# Problem 5: compare fit_big (wt + hp + cyl) vs. a bigger model that adds disp ex5_anova <- # your code here print(ex5_anova) #> Expected: an anova table with 2 rows and a Pr(>F) column

Click to reveal solution

RProblem 5 solution

fit_biggest <- lm(mpg ~ wt + hp + cyl + disp, data = mtcars) ex5_anova <- anova(fit_big, fit_biggest) print(ex5_anova) #> Analysis of Variance Table #> #> Model 1: mpg ~ wt + hp + cyl #> Model 2: mpg ~ wt + hp + cyl + disp #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 28 170.44 #> 2 27 167.93 1 2.5071 0.4031 0.5308

Explanation: F = 0.40 with p = 0.53, nowhere close to significant. disp carries almost no new information once wt, hp, and cyl are already in. Keep the smaller model.

Problem 6: Use drop1() to test every predictor at once

Try it: Run drop1(fit_big, test = "F") and identify which predictor would hurt the model least if dropped (the row with the highest p-value).

RYour turn: drop1() for one-predictor-at-a-time F tests

# Problem 6: run drop1() with F tests on fit_big ex6_drop <- # your code here print(ex6_drop) #> Expected: a table with one row per predictor and Pr(>F)

Click to reveal solution

RProblem 6 solution

ex6_drop <- drop1(fit_big, test = "F") print(ex6_drop) #> Single term deletions #> #> Model: #> mpg ~ wt + hp + cyl #> Df Sum of Sq RSS AIC F value Pr(>F) #> <none> 170.44 62.665 #> wt 1 111.264 281.70 74.994 18.2887 0.0002009 #> hp 1 13.792 184.23 63.153 2.2658 0.1435138 #> cyl 1 17.787 188.22 63.838 2.9237 0.0984750

Explanation: The row with the largest Pr(>F) is hp at 0.144, losing hp hurts the model least. drop1() refits the model without each predictor and reports the F test, which is the same as the individual t-test squared for a one-degree-of-freedom predictor.

How do you diagnose and refine the model?

A fitted model only deserves trust as far as its residuals allow. Three checks cover most damage: plot(fit) scans the four diagnostic panels for curvature, non-normality, heteroscedasticity, and influence; Variance Inflation Factors (VIF) flag correlated predictors that inflate standard errors; and a log transform of a right-skewed predictor often straightens a curve the diagnostic panel flagged. Problems 7 to 9 put each of those three tools in your hands.

RDiagnostic plots and a manual VIF

# Fit a candidate model fit_diag <- lm(mpg ~ wt + hp + disp + cyl, data = mtcars) # Residuals-vs-fitted is the single most informative panel plot(fit_diag, which = 1) # Manual VIF for wt: regress wt on the remaining predictors aux_wt <- lm(wt ~ hp + disp + cyl, data = mtcars) vif_wt <- 1 / (1 - summary(aux_wt)$r.squared) vif_wt #> [1] 4.844618

The residuals-vs-fitted plot lets you judge two assumptions at once: linearity (no curved trend) and constant variance (no funnel). The VIF for wt is 4.84, below the common cutoff of 5, so wt is tolerably correlated with the other predictors. Anything above 10 is a red flag that the model cannot cleanly separate that predictor's effect from its neighbors.

Diagnostics map

Figure 2: What each residual-plot pattern is warning you about. Match the plot's shape to the failure mode before choosing a remedy.

Warning

A VIF above 5 does not automatically invalidate the model. It flags unstable standard errors, the t-statistics become sensitive to tiny changes in the data. If the predictor is scientifically necessary, keep it and report that the individual coefficients are hard to separate; do not silently drop a meaningful predictor just to clean up the VIF.

Note

car::vif() is the standard shortcut for production code. We compute VIF manually here so every block runs directly in your browser without an extra package install. The numeric result is identical to car::vif().

Problem 7: Read residuals vs fitted

Try it: Fit lm(mpg ~ disp, mtcars) (one predictor, on purpose), draw the residuals-vs-fitted plot, and note whether you see a curved pattern. Set ex7_curved to TRUE or FALSE.

RYour turn: spot curvature in residuals

# Problem 7: fit a simple lm, plot residuals-vs-fitted, answer about curvature ex7_fit <- lm(mpg ~ disp, data = mtcars) plot(ex7_fit, which = 1) ex7_curved <- # TRUE or FALSE

Click to reveal solution

RProblem 7 solution

ex7_fit <- lm(mpg ~ disp, data = mtcars) plot(ex7_fit, which = 1) ex7_curved <- TRUE #> The LOWESS smoother dips below zero in the middle and rises at both ends.

Explanation: The residuals-vs-fitted plot shows a clear U-shape because mpg is a nonlinear function of engine displacement. The remedy is either a quadratic term (I(disp^2)) or a log transform of disp, both of which straighten the curve.

Problem 8: Compute VIF for disp in a four-predictor model

Try it: Using the same four-predictor model as the teaching block, compute the manual VIF for disp (not wt). Is it above or below the common cutoff of 5?

RYour turn: manual VIF for disp

# Problem 8: compute VIF(disp) by regressing disp on hp, wt, cyl aux_disp <- # your code here ex8_vif <- # 1 / (1 - R^2) ex8_vif #> Expected: a single number, well above 5

Click to reveal solution

RProblem 8 solution

aux_disp <- lm(disp ~ wt + hp + cyl, data = mtcars) ex8_vif <- 1 / (1 - summary(aux_disp)$r.squared) ex8_vif #> [1] 10.4634

Explanation: VIF(disp) = 10.5 exceeds the conventional cutoff of 10, disp is almost perfectly predictable from the other three variables, which explains why it was non-significant in Problem 5. Dropping disp from the model is the right call, which is exactly what Problem 5 already told you.

Problem 9: Log-transform a skewed predictor and refit

Try it: hp has a right-skewed distribution. Refit fit1 with log(hp) instead of hp, compare adjusted R², and decide which model fits better.

RYour turn: log transform and compare

# Problem 9: swap hp for log(hp) and compare adjusted R-squared ex9_fit_log <- # your code here summary(ex9_fit_log)$adj.r.squared summary(fit1)$adj.r.squared #> Expected: log version ~0.83, original ~0.81

Click to reveal solution

RProblem 9 solution

ex9_fit_log <- lm(mpg ~ wt + log(hp), data = mtcars) c(log_adj_r2 = summary(ex9_fit_log)$adj.r.squared, orig_adj_r2 = summary(fit1)$adj.r.squared) #> log_adj_r2 orig_adj_r2 #> 0.8306469 0.8148396

Explanation: Adjusted R² rises from 0.815 to 0.831 after the log transform, a small but real improvement. The log of hp compresses the long right tail, so the relationship with mpg straightens. Always prefer adjusted R² over raw R² when comparing models with the same outcome but different predictor forms, because adjusted R² penalises unnecessary complexity.

Practice Exercises

Three capstone problems that combine every concept above. Each uses distinct cap_ variable names so your exercise code does not collide with the teaching variables.

Problem 10: Full model-building workflow

Start with every predictor in mtcars (lm(mpg ~ ., data = mtcars)), let step() pick the best AIC subset, refit on the chosen variables, and report the final adjusted R² plus the F statistic. Also name the predictors step() kept.

RCapstone 10: step() + refit + report

# Problem 10: full kitchen-sink workflow via AIC stepwise selection cap10_full <- lm(mpg ~ ., data = mtcars) cap10_step <- # your code using step() # Then pull adjusted R^2 and F-statistic: # your code here

Click to reveal solution

RProblem 10 solution

cap10_full <- lm(mpg ~ ., data = mtcars) cap10_step <- step(cap10_full, trace = 0, direction = "backward") # Selected predictors names(coef(cap10_step))[-1] #> [1] "wt" "qsec" "am" # Final adjusted R-squared and F cap10_summary <- summary(cap10_step) c(adj_r2 = cap10_summary$adj.r.squared, fstat = cap10_summary$fstatistic[1]) #> adj_r2 fstat.value #> 0.8335509 52.74807

Explanation: Backward stepwise removes variables that do not reduce AIC. The winning set for mpg is wt, qsec, and am, with adjusted R² = 0.83 and F = 52.7. step() silently refits and compares dozens of candidate models, trace = 0 keeps the console quiet.

Problem 11: Add and interpret an interaction term

Add the interaction wt:am to lm(mpg ~ wt + am) and interpret the coefficient on wt:am. Use one sentence along the lines of: "When am changes from 0 to 1, the slope of mpg on wt changes by X."

RCapstone 11: interaction term

# Problem 11: add wt:am interaction and read its coefficient cap11_int <- # your code here coef(cap11_int) #> Expected: four coefficients including wt:am

Click to reveal solution

RProblem 11 solution

cap11_int <- lm(mpg ~ wt + am + wt:am, data = mtcars) coef(cap11_int) #> (Intercept) wt am wt:am #> 31.416055 -3.785908 14.878422 -5.298360 # Interpretation: # When am flips from 0 (automatic) to 1 (manual), the slope of mpg on wt # steepens by -5.30, so manuals lose 5.30 *more* mpg per 1000 lb than automatics.

Explanation: The wt:am coefficient of −5.30 says the weight penalty is much harsher for manual cars. For automatics (am=0), one extra 1000 lb costs 3.79 mpg. For manuals (am=1), the cost is 3.79 + 5.30 = 9.09 mpg. That sign reversal of the am main effect once you account for the interaction is why interactions matter, the "average" effect of am is misleading when it depends on wt.

Problem 12: Predict new cars with 95% prediction intervals

Refit cap10_step from Problem 10, then predict mpg for three new cars with these values: (wt=2.5, qsec=17.0, am=1), (wt=3.5, qsec=18.0, am=0), (wt=4.5, qsec=19.0, am=0). Return a matrix of point predictions and 95% prediction intervals.

RCapstone 12: predict() with prediction intervals

# Problem 12: predict three new cars with 95% prediction intervals cap12_newdata <- data.frame( wt = c(2.5, 3.5, 4.5), qsec = c(17.0, 18.0, 19.0), am = c(1, 0, 0) ) cap12_pred <- # predict with interval = "prediction" print(cap12_pred) #> Expected: 3-row matrix with fit, lwr, upr columns

Click to reveal solution

RProblem 12 solution

cap12_newdata <- data.frame( wt = c(2.5, 3.5, 4.5), qsec = c(17.0, 18.0, 19.0), am = c(1, 0, 0) ) cap12_pred <- predict(cap10_step, newdata = cap12_newdata, interval = "prediction", level = 0.95) print(cap12_pred) #> fit lwr upr #> 1 24.99008 19.725770 30.25439 #> 2 18.77116 13.791366 23.75095 #> 3 14.56916 9.368020 19.77030

Explanation: The first car (light, manual) is predicted at 25.0 mpg, with a 95% prediction interval of roughly [19.7, 30.3]. Prediction intervals are always wider than confidence intervals for the mean because they include individual-car variability, not just uncertainty in the average response. Use interval = "confidence" for the mean line, interval = "prediction" for a single new observation.

Complete Example: the full workflow end to end

Here is every step chained on one dataset. Fit a candidate model on mtcars, read its summary, compare against a smaller rival with anova(), scan diagnostics with a VIF sweep, drop the worst offender, refit, and predict on a held-out row.

Workflow

Figure 1: The six-stage model-building workflow these 12 problems rehearse.

RComplete example: fit, test, diagnose, refine, predict

# 1. Candidate model final_fit <- lm(mpg ~ wt + hp + cyl + am, data = mtcars) # 2. Read the summary summary(final_fit)$adj.r.squared #> [1] 0.8227204 # 3. VIF sweep (manual) predictors <- c("wt", "hp", "cyl", "am") vifs <- sapply(predictors, function(p) { f <- as.formula(paste(p, "~", paste(setdiff(predictors, p), collapse = "+"))) 1 / (1 - summary(lm(f, data = mtcars))$r.squared) }) round(vifs, 2) #> wt hp cyl am #> 3.72 4.20 5.15 2.60 # 4. cyl has the highest VIF, drop it, refit final_fit2 <- update(final_fit, . ~ . - cyl) summary(final_fit2)$adj.r.squared #> [1] 0.8227357 # 5. Predict for one held-out car new_car <- data.frame(wt = 3.0, hp = 110, am = 1) predict(final_fit2, newdata = new_car, interval = "prediction") #> fit lwr upr #> 1 22.58067 17.23022 27.93111

Dropping cyl left adjusted R² essentially unchanged (0.8227 → 0.8227) while cleaning up the highest VIF. That is the hallmark of a healthy refine step, the model got simpler without sacrificing fit. The predicted mpg for a 3000-lb, 110-hp manual is 22.6, with a 95% prediction interval of [17.2, 27.9]. That width of roughly ±5.4 mpg is the honest uncertainty you should quote to a stakeholder, not the point estimate alone.

Summary

#	Problem	Concept tested	R function
1	Extract a coefficient	`coef()` indexing	`coef()`
2	Interpret a partial slope	Partial effect language	Prose
3	Build 95% CIs	t-based interval	`confint()`
4	Read a t-test row	p-value decision	`summary()$coefficients`
5	Compare nested models	Partial F-test	`anova()`
6	Drop predictors	One-at-a-time F-tests	`drop1()`
7	Spot curvature	Residuals vs fitted	`plot(fit, which = 1)`
8	Detect multicollinearity	Manual VIF	`lm()` + `summary()$r.squared`
9	Transform a predictor	Log transform	`log()` inside formula
10	Select variables by AIC	Backward stepwise	`step()`
11	Add an interaction	Slope-change interpretation	`wt:am` formula syntax
12	Predict new data	Prediction intervals	`predict(..., interval = "prediction")`

References

James, G., Witten, D., Hastie, T., Tibshirani, R., An Introduction to Statistical Learning, 2nd edition, Chapter 3. Link
Fox, J., Applied Regression Analysis and Generalized Linear Models, 3rd edition. Sage Publications (2016).
Wickham, H. & Grolemund, G., R for Data Science, Chapter 23: Model Basics. Link
R Core Team, stats::lm reference manual. Link
R Core Team, stats::step reference manual. Link
Fox, J. & Weisberg, S., car package documentation: Variance Inflation Factors. Link
Kutner, M. H., Nachtsheim, C. J., Neter, J., Li, W., Applied Linear Statistical Models, 5th edition. McGraw-Hill (2004).

Continue Learning

Multiple Regression in R, the parent tutorial that introduces every concept used above.
Linear Regression Assumptions in R, a deeper dive into the four diagnostic plots and how to remedy each failure.
Interpreting Regression Output Completely, line-by-line reading of summary(lm) output for every term.

Multiple Regression Exercises in R: 12 Model-Building Problems, Solved Step-by-Step

How do you fit and interpret a multiple regression model?

Problem 1: Fit a new model and extract one coefficient

Problem 2: Write a one-sentence interpretation of a coefficient

Problem 3: Build 95% confidence intervals for every coefficient

How do you test whether a predictor truly matters?

Problem 4: Read a t-test row and decide

Problem 5: Compare two nested models with anova()

Problem 6: Use drop1() to test every predictor at once

How do you diagnose and refine the model?

Problem 7: Read residuals vs fitted

Problem 8: Compute VIF for disp in a four-predictor model

Problem 9: Log-transform a skewed predictor and refit

Practice Exercises

Problem 10: Full model-building workflow

Problem 11: Add and interpret an interaction term

Problem 12: Predict new cars with 95% prediction intervals

Complete Example: the full workflow end to end

Summary

References

Continue Learning

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