broom Exercises in R: 20 Tidy Model Practice Problems
Twenty practice problems on the broom package: convert messy model objects from lm(), glm(), k-means, and hypothesis tests into tidy data frames using tidy(), glance(), and augment(). Each exercise shows the expected printed output; solutions are hidden until you click.
Section 1. tidy() for model coefficients (4 problems)
Exercise 1.1: Get a coefficient table from a simple lm fit
Task: Fit a simple linear regression of mpg on wt using the built-in mtcars dataset, then apply tidy() from broom to convert the model summary into a one-row-per-coefficient tibble. Save the tidy tibble to ex_1_1.
Expected result:
#> # A tibble: 2 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 37.3 1.88 19.9 8.24e-19
#> 2 wt -5.34 0.559 -9.56 1.29e-10Difficulty: Beginner
A fitted model object hides its coefficient table inside the summary; you want it reshaped into a flat data frame with one row per term.
Fit the model with lm(mpg ~ wt, data = mtcars), then pass the fitted object straight to tidy().
Click to reveal solution
Explanation: summary(fit)$coefficients returns a matrix with row names instead of a column, which is awkward to filter or pipe. tidy() flips row names into a term column and renames Pr(>|t|) to p.value, giving you a rectangular tibble. That shape is what every downstream verb expects: filter(p.value < 0.05), arrange(estimate), or bind_rows() across multiple models all just work.
Exercise 1.2: Attach 95 percent confidence intervals to coefficients
Task: Fit a multiple regression of mpg on wt, hp, and cyl from the mtcars dataset. Use tidy() with conf.int = TRUE to produce a coefficient table that includes the lower and upper bounds of a 95 percent confidence interval. Save the result to ex_1_2.
Expected result:
#> # A tibble: 4 x 7
#> term estimate std.error statistic p.value conf.low conf.high
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 38.8 1.79 21.7 4.80e-19 35.1 42.4
#> 2 wt -3.17 0.741 -4.28 2.06e-4 -4.69 -1.65
#> 3 hp -0.0180 0.0119 -1.52 1.40e-1 -0.0423 0.00626
#> 4 cyl -0.942 0.551 -1.71 9.85e-2 -2.07 0.189Difficulty: Intermediate
Confidence intervals are not part of the default coefficient table; an argument switches them on so lower and upper bounds become columns.
Pass the lm(mpg ~ wt + hp + cyl, ...) fit to tidy() with conf.int = TRUE.
Click to reveal solution
Explanation: confint() returns a matrix the same way coef(summary(fit)) does, so combining estimates with CIs by hand requires cbind and renaming. Passing conf.int = TRUE to tidy() does that join internally and exposes conf.low and conf.high columns ready for ggplot2::geom_pointrange(). Use conf.level = 0.99 for a 99 percent interval; the default is 0.95.
Exercise 1.3: Tidy a logistic regression on the iris dataset
Task: A biostatistician is screening flower-shape predictors for a virginica-versus-other classifier. Using the built-in iris dataset, fit a logistic regression for Species == "virginica" against Sepal.Length and Petal.Length. Use tidy() with exponentiate = TRUE to convert coefficients into odds ratios. Save to ex_1_3.
Expected result:
#> # A tibble: 3 x 5
#> term estimate std.error statistic p.value
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 (Intercept) 4.20e-19 8.81 -4.86 1.18e-6
#> 2 Sepal.Length 2.45e- 1 1.32 -1.06 2.88e-1
#> 3 Petal.Length 2.85e+ 4 2.62 3.91 9.16e-5Difficulty: Intermediate
Logistic coefficients arrive on the log-odds scale; one flag rescales them into odds ratios a non-statistician can read.
Build a 0/1 outcome, fit with glm(..., family = binomial()), then call tidy(fit, exponentiate = TRUE).
Click to reveal solution
Explanation: Logistic regression coefficients on the log-odds scale are hard to communicate to non-statisticians. exponentiate = TRUE applies exp() to estimate, conf.low, and conf.high so the column reads as an odds ratio: a one-unit rise in Petal.Length multiplies the odds of being virginica by roughly 28,500. The std.error and statistic columns stay on the original scale because they describe the log-odds estimator, not the exponentiated one.
Exercise 1.4: Convert a paired t.test htest object into a tibble
Task: Pharmacology researchers want to compare patient weights before and after a six-week diet trial. Build a small inline tibble of paired pre and post weights for 10 subjects, then run a paired t.test(). Apply tidy() to the resulting htest object to extract estimate, statistic, and p.value into one tibble row. Save to ex_1_4.
Expected result:
#> # A tibble: 1 x 8
#> estimate statistic p.value parameter conf.low conf.high method alternative
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
#> 1 2.10 4.83 0.000935 9 1.12 3.08 Paired t-test two.sidedDifficulty: Intermediate
A hypothesis-test result is a list of loose components; tidying lifts estimate, statistic, and p-value into one rectangular row.
Run t.test(pre, post, paired = TRUE) and pass the resulting object to tidy().
Click to reveal solution
Explanation: t.test() returns an htest list with named components you have to dig out (test$estimate, test$p.value) and which print awkwardly. tidy.htest() knows the standard slots and lifts them into one tidy row, which is exactly what you need to bind_rows() results from many subgroups or to feed a downstream report. tidy() also handles wilcox.test, cor.test, prop.test, and chisq.test.
Section 2. glance() for one-row model summaries (3 problems)
Exercise 2.1: Read R-squared and AIC from a regression with glance
Task: A junior analyst wants a quick fit-quality summary for a regression of mpg on wt and hp using the mtcars dataset. Use glance() from broom to extract the one-row tibble of model-level statistics (r.squared, adj.r.squared, sigma, statistic, p.value, AIC, BIC, deviance, df.residual, nobs). Save the full tibble to ex_2_1.
Expected result:
#> # A tibble: 1 x 12
#> r.squared adj.r.squared sigma statistic p.value df logLik AIC BIC deviance df.residual nobs
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int> <int>
#> 1 0.827 0.815 2.59 69.2 9.11e-12 2 -74.3 157. 163. 195. 29 32Difficulty: Beginner
Model-level fit measures like R-squared and AIC live apart from the coefficient table; you want them collapsed into a single summary row.
Fit lm(mpg ~ wt + hp, data = mtcars) and pass the fit to glance().
Click to reveal solution
Explanation: Reading summary(fit) and scraping out R-squared and AIC by hand is fragile. glance() returns a one-row tibble with named columns so you can pull(r.squared) or select(AIC, BIC) directly. The single-row shape is key: you can bind_rows() glances from many fits and end up with one tidy comparison table (see exercise 2.3).
Exercise 2.2: Pull deviance and null deviance from a glm fit
Task: A reporting analyst is comparing model fit between a fitted logistic regression and its intercept-only null. Using the inline binary outcome data below, fit glm(y ~ x1 + x2, family = binomial()), then call glance() and save the resulting one-row tibble to ex_2_2. The columns null.deviance and deviance together let you compute pseudo R-squared.
Expected result:
#> # A tibble: 1 x 8
#> null.deviance df.null logLik AIC BIC deviance df.residual nobs
#> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <int> <int>
#> 1 138. 99 -55.9 118. 126. 112. 97 100Difficulty: Intermediate
A fitted GLM stores both its own deviance and the deviance of the intercept-only null; one summary call surfaces both at once.
Fit glm(y ~ x1 + x2, family = binomial()), then call glance() on the fit.
Click to reveal solution
Explanation: Logistic regression has no native R-squared, so analysts often report a McFadden pseudo R-squared computed as 1 - deviance/null.deviance. glance.glm() exposes both quantities as columns, so you can write mutate(pseudo_r2 = 1 - deviance/null.deviance) in a pipeline rather than hunting through summary(fit). AIC and BIC are right there too, which is what you actually use when ranking competing GLMs.
Exercise 2.3: Rank three candidate models in one glance comparison table
Task: A code reviewer wants to see whether adding cyl and hp to a baseline mpg ~ wt model is worth the extra complexity. Fit three nested models on mtcars (mpg ~ wt; mpg ~ wt + hp; mpg ~ wt + hp + cyl), apply glance() to each, bind them into a single tibble with a model label column, and save to ex_2_3.
Expected result:
#> # A tibble: 3 x 13
#> model r.squared adj.r.squared sigma statistic p.value df logLik AIC BIC deviance df.residual nobs
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int> <int>
#> 1 wt 0.753 0.745 3.05 91.4 1.29e-10 1 -80.0 166. 170. 279. 30 32
#> 2 wt+hp 0.827 0.815 2.59 69.2 9.11e-12 2 -74.3 157. 163. 195. 29 32
#> 3 wt+hp+cyl 0.843 0.826 2.51 50.2 2.18e-11 3 -72.7 155. 163. 177. 28 32Difficulty: Intermediate
Each model produces an identically shaped one-row summary, so stacking those rows yields a single comparison table.
Apply glance() to each fit, attach a label with mutate(model = ...), and combine the rows with bind_rows().
Click to reveal solution
Explanation: This pattern is a one-line answer to "which model wins?" because every row has identical columns. AIC drops nine points from wt to wt+hp (worth the extra term) but barely moves from wt+hp to wt+hp+cyl, so cyl probably is not pulling its weight. Without glance() you would build this table by hand with summary(), AIC(), and string concatenation; with broom it is three lines.
Section 3. augment() for fitted values and residuals (4 problems)
Exercise 3.1: Add fitted values and residuals to the training data
Task: Fit a regression of mpg on wt and hp using the mtcars dataset, then use augment() from broom to attach .fitted, .resid, .hat, .sigma, .cooksd, and .std.resid columns to the original predictors. Save the augmented tibble to ex_3_1.
Expected result:
#> # A tibble: 32 x 9
#> .rownames mpg wt hp .fitted .resid .hat .sigma .cooksd .std.resid
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 Mazda RX4 21 2.62 110 23.6 -2.57 0.0407 2.61 0.00865 -1.01
#> 2 Mazda RX4 W 21 2.88 110 22.6 -1.58 0.0352 2.62 0.00280 -0.622
#> 3 Datsun 710 22.8 2.32 93 25.3 -2.47 0.0584 2.61 0.0119 -0.985
#> ...
#> # 29 more rows hiddenDifficulty: Beginner
Fitted values, residuals, and leverage diagnostics belong next to the original observations, one diagnostic column per row.
Fit lm(mpg ~ wt + hp, data = mtcars) and pass the fit to augment().
Click to reveal solution
Explanation: Manually building this table means calling predict(fit), residuals(fit), hatvalues(fit), cooks.distance(fit), and rstandard(fit) separately, then cbind-ing them onto the data while hoping the row order matches. augment() does it all in one call and keeps the original rownames in a .rownames column so a ggplot2::geom_text(aes(label = .rownames)) annotation works out of the box.
Exercise 3.2: Generate predictions for new observations with augment newdata
Task: A code reviewer wants to score three new hypothetical cars against the regression fit. Build an inline tibble of three new rows with values for wt and hp, then call augment() on a fitted model with newdata = new_cars to get .fitted and .se.fit for each new row without touching the training data. Save to ex_3_2.
Expected result:
#> # A tibble: 3 x 4
#> wt hp .fitted .se.fit
#> <dbl> <dbl> <dbl> <dbl>
#> 1 2.5 100 23.7 0.726
#> 2 3.5 150 19.0 0.563
#> 3 4.5 200 14.4 1.16Difficulty: Intermediate
Scoring fresh observations means predicting on rows the model never saw, not on the training data it was fit to.
Call augment() on the fit with newdata = new_cars and se_fit = TRUE.
Click to reveal solution
Explanation: Pass newdata = <tibble> and augment() predicts on the new rows rather than the training set. Setting se_fit = TRUE exposes a .se.fit column with the standard error of each prediction; multiply by 1.96 for a 95 percent CI on the mean response. To get prediction intervals for individual observations (wider, includes residual variance) use predict(fit, newdata, interval = "prediction") directly instead.
Exercise 3.3: Attach k-means cluster labels back to the original rows
Task: A scout for a sports analytics team wants to group the 32 cars in mtcars by horsepower and weight to identify clusters of similar vehicles. Run kmeans() with three centres on the scaled wt and hp columns, then use augment() from broom to attach the .cluster column to the original mtcars rows. Save the result to ex_3_3.
Expected result:
#> # A tibble: 32 x 12
#> .rownames mpg cyl disp hp drat wt qsec vs am gear .cluster
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
#> 1 Mazda RX4 21 6 160 110 3.9 2.62 16.5 0 1 4 1
#> 2 Mazda RX4 W 21 6 160 110 3.9 2.88 17.0 0 1 4 1
#> 3 Datsun 710 22.8 4 108 93 3.85 2.32 18.6 1 1 4 2
#> ...
#> # 29 more rows hiddenDifficulty: Advanced
A clustering result returns bare assignments with no tie back to the source rows; you need them joined onto the original data.
Pass the kmeans object and mtcars to augment() to get a .cluster column appended to each row.
Click to reveal solution
Explanation: kmeans() returns cluster assignments as an integer vector inside km$cluster with no link back to the source data. augment(km, data) joins the assignments back to the data by row index and converts them to a factor in a .cluster column. broom also provides tidy.kmeans() for cluster centres and glance.kmeans() for total within-cluster sum-of-squares, completing the triple. The same pattern works for hclust() cut trees via tidy() and augment().
Exercise 3.4: Flag high-leverage observations using Cook's distance
Task: An audit team wants to spot any cars in mtcars that exert disproportionate influence on a fitted regression. Fit lm(mpg ~ wt + hp, data = mtcars), augment it, then filter the augmented tibble to keep only rows whose .cooksd exceeds 4 / n (the common rule-of-thumb threshold). Save the filtered tibble to ex_3_4, ordered by .cooksd descending.
Expected result:
#> # A tibble: 3 x 10
#> .rownames mpg wt hp .fitted .resid .hat .sigma .cooksd .std.resid
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 Maserati Bora 15 3.57 335 12.0 3.04 0.299 2.62 0.236 1.42
#> 2 Chrysler Imperial 14.7 5.34 230 9.74 4.96 0.121 2.55 0.187 2.04
#> 3 Toyota Corolla 33.9 1.84 65 28.4 5.49 0.0581 2.51 0.0975 2.18Difficulty: Intermediate
Influential observations are flagged by comparing a per-row diagnostic against a threshold that depends on the sample size.
After augment(), use filter(.cooksd > 4 / n) and arrange(desc(.cooksd)) on the result.
Click to reveal solution
Explanation: 4 / n (here 0.125) is one widely used cut-off for Cook's distance; observations above it are worth investigating but are not automatically wrong. The Maserati Bora has the largest leverage because its 335 hp sits far from the rest of the design, so even a moderate residual produces a large Cook value. Augment plus filter() plus arrange() is the canonical broom workflow for any diagnostic that lives on the row of the fit.
Section 4. Tidy many models with nest, map, and broom (3 problems)
Exercise 4.1: Fit one regression per cylinder group with nest and map
Task: A statistician wants to know whether the mpg versus wt slope differs by cylinder count. Using the mtcars dataset, group by cyl, nest the rows, fit a separate lm(mpg ~ wt) per group, tidy each fit, and unnest into one long tibble with columns cyl, term, estimate, std.error, statistic, p.value. Save to ex_4_1.
Expected result:
#> # A tibble: 6 x 6
#> cyl term estimate std.error statistic p.value
#> <dbl> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 4 (Intercept) 39.6 4.35 9.10 7.77e- 6
#> 2 4 wt -5.65 1.85 -3.05 1.37e- 2
#> 3 6 (Intercept) 28.4 4.18 6.79 1.05e- 3
#> 4 6 wt -2.78 1.33 -2.08 9.18e- 2
#> 5 8 (Intercept) 23.9 3.01 7.94 4.05e- 6
#> 6 8 wt -2.19 0.739 -2.97 1.18e- 2Difficulty: Advanced
Fitting one model per group means each group's rows must travel together in their own cell before any model touches them.
After group_by(cyl) and nest(), use map() to fit lm(mpg ~ wt) per group, map() again with tidy, then unnest().
Click to reveal solution
Explanation: Nest plus map plus tidy is the canonical "many models" pipeline. Each group lands in its own row with a list-column of data and a list-column of model objects; tidying inside mutate() and unnesting flattens the result. The slope is steepest for 4-cylinder cars (-5.65) and shallowest for 8-cylinder cars (-2.19), which would be invisible from a single pooled regression. Same shape supports glance and augment if you swap the inner function.
Exercise 4.2: Compare group-level R-squared with grouped glance
Task: A performance reviewer is asking how well a per-cylinder mpg ~ wt model fits inside each group. Using the same nest plus map approach as the previous exercise, apply glance() instead of tidy() to get one row of fit statistics per cyl level. Save the resulting tibble (cyl plus all glance columns) to ex_4_2.
Expected result:
#> # A tibble: 3 x 13
#> cyl r.squared adj.r.squared sigma statistic p.value df logLik AIC BIC deviance df.residual nobs
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int> <int>
#> 1 4 0.509 0.454 3.33 9.32 0.0137 1 -28.2 62.4 63.6 99.7 9 11
#> 2 6 0.465 0.357 1.17 4.34 0.0918 1 -10.0 26.0 25.2 6.87 5 7
#> 3 8 0.423 0.376 2.02 8.80 0.0118 1 -28.7 63.5 65.7 49.0 12 14Difficulty: Advanced
Switching what you compute per group from coefficients to fit statistics leaves the grouping scaffold completely untouched.
Reuse the group_by() plus nest() plus map() pipeline, but apply glance instead of tidy before unnest().
Click to reveal solution
Explanation: Same scaffold as exercise 4.1, only the inner map swaps tidy for glance. The 4-cylinder model explains the most variance (R² = 0.509), while the 6-cylinder model has only seven rows and its p.value (0.092) reflects that. This is the right way to spot heterogeneity: low per-group n inflates uncertainty, and grouped glance forces it into the open instead of hiding it in a global fit.
Exercise 4.3: Build a wide tibble of slopes by group with pivot_wider
Task: A product manager wants a one-line dashboard showing the wt slope for each cyl level side by side. Start from the tidy long-format tibble of the previous nested fits (cyl, term, estimate). Filter to keep only term == "wt", then pivot the estimate column to one row with three columns named cyl_4, cyl_6, cyl_8. Save the wide-format tibble to ex_4_3.
Expected result:
#> # A tibble: 1 x 3
#> cyl_4 cyl_6 cyl_8
#> <dbl> <dbl> <dbl>
#> 1 -5.65 -2.78 -2.19Difficulty: Advanced
A long table of one slope per group becomes a one-row dashboard once the group identifier moves into the column headers.
From the long tibble, filter(term == "wt") then reshape with pivot_wider(names_from = cyl, values_from = estimate).
Click to reveal solution
Explanation: Tidy output is long-by-default because that is what most downstream operations want, but stakeholders often want wide. The recipe is broom keeps your data tidy; pivot reshapes for display. Note that wide tibbles lose the std.error and p.value columns unless you pivot multiple value columns with names_from = cyl, values_from = c(estimate, std.error), which generates estimate_cyl_4, std.error_cyl_4, and so on.
Section 5. Hypothesis test results (3 problems)
Exercise 5.1: Tidy a chi-squared test of independence
Task: A marketing analyst is checking whether email opens are independent of campaign segment. Build a 3-by-2 contingency matrix of opens versus non-opens for three segments, run chisq.test() on it, then call tidy() on the htest object to extract statistic, p.value, parameter (df), and method in one tidy row. Save to ex_5_1.
Expected result:
#> # A tibble: 1 x 4
#> statistic p.value parameter method
#> <dbl> <dbl> <int> <chr>
#> 1 66.5 3.66e-15 2 Pearson's Chi-squared testDifficulty: Intermediate
A contingency-table test prints as a paragraph; tidying turns the same information into one row of named columns you can log.
Run chisq.test() on the matrix and wrap the returned object in tidy().
Click to reveal solution
Explanation: chisq.test() prints a paragraph; tidy.htest() turns the same information into one row with named columns so you can append it to a results log. A near-zero p.value plus df = 2 means the segments differ on open rate in a way that is unlikely under independence. To see the cell-level standardized residuals (where the difference is largest) call augment() on the chisq object instead.
Exercise 5.2: Tidy a Wilcoxon signed-rank test for non-normal data
Task: A clinical reviewer wants a paired comparison without assuming normality. Using the inline pre and post pain scores below for nine patients, run wilcox.test(post, pre, paired = TRUE) and apply tidy() to convert the htest object into a one-row tibble. Save the result to ex_5_2.
Expected result:
#> # A tibble: 1 x 4
#> statistic p.value method alternative
#> <dbl> <dbl> <chr> <chr>
#> 1 3 0.0273 Wilcoxon signed rank test with continuity correction two.sidedDifficulty: Intermediate
A rank-based paired test still produces a result object that collapses to a single fixed-column row when tidied.
Run wilcox.test(post, pre, paired = TRUE) and pass the result to tidy().
Click to reveal solution
Explanation: The wilcox.test() print output is text-heavy and changes its column shape depending on whether ties exist, which makes scripted reporting brittle. tidy() normalizes the output into a fixed-column tibble, so the same bind_rows() pipeline that consumes t-tests also consumes Wilcoxon tests. The reported statistic is V, the sum of positive signed ranks.
Exercise 5.3: Loop t.tests across measurements and bind tidy rows
Task: A reporting analyst needs a side-by-side significance table for Sepal.Length, Sepal.Width, Petal.Length, and Petal.Width comparing setosa versus versicolor from the iris dataset. For each variable, run t.test(value ~ Species), tidy the result, prepend a variable column, and row-bind the four single-row tibbles into one. Save to ex_5_3.
Expected result:
#> # A tibble: 4 x 9
#> variable estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 Sepal.Length -0.930 5.01 5.94 -10.5 3.75e-17 86.5 -1.11 -0.755
#> 2 Sepal.Width 0.658 3.43 2.77 9.45 2.48e-15 94.7 0.519 0.796
#> 3 Petal.Length -2.80 1.46 4.26 -39.5 9.93e-46 62.1 -2.94 -2.66
#> 4 Petal.Width -1.08 0.246 1.33 -34.1 2.72e-47 74.8 -1.14 -1.02Difficulty: Intermediate
Running the same test on several variables and stacking the resulting rows gives one side-by-side significance table.
Use map_dfr() over the variable names, building each formula with as.formula() and tidying every t.test().
Click to reveal solution
Explanation: map_dfr() from purrr applies a function to each element and row-binds the results into one tibble, which is exactly the shape you want for a multi-variable significance table. Because every tidy(t.test(...)) returns the same columns, the bind is safe. Without broom, each htest would have a different structure and the bind would fail or produce ragged results.
Section 6. End-to-end reporting workflows (3 problems)
Exercise 6.1: Build a publication-style coefficient table
Task: A statistician is preparing a paper draft and needs a clean coefficient table for lm(mpg ~ wt + hp + cyl, data = mtcars) containing term, estimate (rounded to 3 dp), 95 percent CI as a [lo, hi] string, and a significance marker (*, , *, ., blank) following the standard <0.001 / <0.01 / <0.05 / <0.1 thresholds. Save the formatted tibble to ex_6_1.
Expected result:
#> # A tibble: 4 x 4
#> term estimate ci sig
#> <chr> <dbl> <chr> <chr>
#> 1 (Intercept) 38.8 [35.063, 42.430] ***
#> 2 wt -3.17 [-4.685, -1.647] ***
#> 3 hp -0.018 [-0.042, 0.006] ""
#> 4 cyl -0.942 [-2.071, 0.189] .Difficulty: Intermediate
Once the coefficient table is a flat data frame, rounding, CI strings, and significance stars are just ordinary column transformations.
Start from tidy(fit, conf.int = TRUE), then mutate() using sprintf() for the CI string and case_when() for the marker.
Click to reveal solution
Explanation: Once tidy() gives you a rectangular tibble, every formatting step is just dplyr. sprintf() builds the CI string with controlled precision; case_when() builds the conventional significance markers from p.value. The same pipeline scales to a for loop over many models, then a gt::gt() or knitr::kable() call to render the final table.
Exercise 6.2: Compute model lift over a baseline by tidy slope comparison
Task: A growth team measured conversion rate as a function of two predictors: a control feature x_ctrl and a new feature x_new. Fit lm(y ~ x_ctrl + x_new) on the inline data below. Use tidy() to extract the two slopes, compute the ratio new_lift = estimate(x_new) / estimate(x_ctrl), and save a one-row tibble with columns slope_ctrl, slope_new, and new_lift to ex_6_2.
Expected result:
#> # A tibble: 1 x 3
#> slope_ctrl slope_new new_lift
#> <dbl> <dbl> <dbl>
#> 1 0.499 1.51 3.02Difficulty: Advanced
Reducing a model to a single headline number means isolating the two slopes and dividing one by the other.
tidy() the fit, filter() to the two slope terms, reshape with pivot_wider(), then compute the ratio in transmute().
Click to reveal solution
Explanation: Stakeholders want a single number ("the new feature is 3x as effective") that hides the model behind it. tidy plus pivot_wider plus transmute is the bridge: tidy gives row-per-coefficient, pivot reshapes to one row, transmute computes the derived metric. Bake the simulation values into your head as a sanity check: the true ratio is 1.5 / 0.5 = 3, and the estimate (3.02) is within sampling noise.
Exercise 6.3: Detect residual patterns using augment plus group_by
Task: A code reviewer wants to know whether the regression mpg ~ wt + hp from the mtcars dataset systematically over- or under-predicts cars of different cylinder counts. Augment the fit, group by cyl (joined back from the original mtcars data on .rownames), and compute the mean and standard deviation of .resid per group. Save the per-group residual summary to ex_6_3.
Expected result:
#> # A tibble: 3 x 4
#> cyl n mean_resid sd_resid
#> <dbl> <int> <dbl> <dbl>
#> 1 4 11 1.71 2.69
#> 2 6 7 -1.85 1.46
#> 3 8 14 -0.421 2.21Difficulty: Advanced
Checking for residual bias across a category means the residuals must carry that category alongside them, then be summarised per group.
augment() the fit, left_join() the cyl column on .rownames, then group_by(cyl) and summarise() the mean and sd of .resid.
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Explanation: Group residual means should be near zero if the model is unbiased within each group. Here, 4-cylinder cars are systematically under-predicted (mean residual 1.71 mpg) and 6-cylinder cars are over-predicted (-1.85), which is a strong hint that cyl should enter the model as a predictor or interaction. The augment plus join plus group_by recipe generalizes: replace cyl with any grouping variable to audit residual bias across cohorts.
What to do next
Now that you can move from messy model objects to tidy tibbles, line these up next:
- dplyr Exercises in R: Real-World Practice Problems for the verb fluency that consumes broom output.
- Linear Regression Exercises in R to deepen the statistical side of what tidy() returns.
- Machine Learning Exercises in R for end-to-end modelling workflows where glance() and augment() shine.
- tidyverse Exercises in R for cross-package patterns combining purrr, tidyr, and broom.
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