• Use lm() with two-group and three-group categorical predictors.
• Interpret the reference group and dummy-coded coefficients.
• Count dummy coefficients, beta coefficients, and total model parameters.
• Calculate group means from coefficients, then use predict() as a check.
• Read a coefficient t-test and an overall model F-test.

Complete the online lecture before the workshop and the Week 7 problem set afterwards.

In Weeks 5–6, each continuous predictor entered as a single linear term added one slope coefficient. A categorical predictor can add more than one coefficient:

One categorical predictor with G levels adds G − 1 coefficients. We still fit one intercept as well.

These are general linear models, whether predictors are continuous, categorical, or both.

Two population means can be represented using one regression equation:

\[ \mu_i = \beta_0 + \beta_1 X_i. \]

• For the reference group, \(X=0\): its mean is \(\beta_0\).
• For the other group, \(X=1\): its mean is \(\beta_0+\beta_1\).
• Therefore, \(\beta_1\) (by itself) is the other group’s mean minus the reference mean.

lm() estimates these population parameters from our sample.

Like the equal variance independent samples t-test, the linear model assumes a common residual SD, sigma, around the group means.

Open your module’s RStudio Project and check that workshop_data is inside it.

1. Download week_07_workshop.R from NOW.
2. Move it into your project folder using your file explorer.
3. Open it in RStudio.

Wait for each exercise checkpoint. Run one line at a time and keep your coefficient output visible for the worked examples.

blomkvist <- read_csv("workshop_data/analysis_blomkvist.csv")
blomkvist <- drop_na(blomkvist, rt, meds_group)
count(blomkvist, meds_group)

meds_group already contains “medicated” and “not medicated”. We do not need to create a new grouping variable.

Do the medication groups differ in mean hand reaction time?

The prepared data are adapted from Blomkvist et al. (2017).

Complete the data-loading steps and Part A1 with me.

m_lm <- lm(rt ~ meds_group, data = blomkvist)

The formula has the same structure as Week 5; the predictor now contains group labels.

Identify which coefficient is the reference mean and which is a difference.

coef(m_lm)

            (Intercept) meds_groupnot medicated 
               738.8143               -157.4875 

A continuous coefficient usually uses its variable name, such as age. Here R joins the predictor name and the comparison level:

Here the reference is medicated. This coefficient gives not medicated minus medicated, not the not-medicated mean on its own.

coef(m_lm)

            (Intercept) meds_groupnot medicated 
               738.8143               -157.4875 

Read the coefficient names to identify the reference mean and the other group’s difference.

With these text labels, R orders the levels alphabetically. The first level is the reference and the remaining level gets a binary indicator.

“medicated” comes before “not medicated”, so it is coded 0; “not medicated” is coded 1.

Two levels give 2 - 1 = 1 dummy variable. In general, G levels give G - 1 dummy variables. R creates these codes inside lm().

A model predicts quiz score from study_mode, with two levels: classroom and online.

     (Intercept) study_modeonline 
              70               -8 

• Which level is the reference group, whose predicted mean is the intercept?
• How does the coefficient name help you identify it?
• What does the slope coefficient -8 represent?

Discuss: What is the predicted mean quiz score for each group? No calculator needed.

• classroom is the reference level: its dummy code is 0, so its predicted mean is 70.
• study_modeonline joins the predictor study_mode and the comparison level online. It represents online minus classroom.
• The dummy-coded slope is -8: the predicted online mean is 8 points lower.

The intercept gives the reference mean; the other coefficient gives a difference, rather than the other group’s mean.

Use rt ~ meds_group, our model with two medication levels. Answer the counting questions in Part A1; no new R commands are needed.

1. How many predictor variables are in the model?
2. How many dummy slope coefficients represent the two levels?
3. Add the intercept: how many beta coefficients are there altogether?
4. Add the shared residual SD, sigma: how many model parameters are there altogether?

Check the beta count against the names printed by coef(m_lm). Sigma is a parameter, but is reported separately from the beta coefficients.

coef(m_lm)

            (Intercept) meds_groupnot medicated 
               738.8143               -157.4875 

Here X = 0 for medicated and X = 1 for not medicated.

\[ \widehat{\mu}_i = \widehat{\beta}_0 + \widehat{\beta}_1 X_i. \]

new_meds <- tibble(
  meds_group = c("medicated", "not medicated")
)
predict(m_lm, newdata = new_meds)

Now that we have calculated the means by hand, use predict() to check them. tibble() supplies one row per group.

predict() uses the fitted model and returns one predicted mean per row. The column name and group labels must match the model.

predict() uses the fitted model directly. For the manual calculation, read and add the printed coefficient values.

Use the coefficient row for the group difference, not the intercept row.

m_lm_summary <- summary(m_lm)
m_lm_summary$coefficients

                         Estimate Std. Error   t value     Pr(>|t|)
(Intercept)              738.8143   21.95403 33.652780 4.444748e-77
meds_groupnot medicated -157.4875   28.50877 -5.524177 1.222335e-07

Find the meds_groupnot medicated row. Read its t value and Pr(>|t|) columns, as in Week 6.

The null hypothesis is that the population group difference is 0.

Compare with equal-variance t-test.

Opposite subtraction order: opposite signs, equal magnitudes and identical two-sided p-values.

Complete Part A2.

Stop after Part A2.

After the break, we will extend the same logic to three groups.

Alphabetical order of the text levels: former smoker → non-smoker → smoker.

Three levels give 3 - 1 = 2 dummy variables.

The first level, “former smoker”, is the reference: both its codes are 0. Each other level has a 1 in its own column and a 0 in the other.

Start Part A3 with me: fit the model and read its coefficients.

m_smoker <- lm(rt ~ smoker, data = blomkvist_smoker)
coef(m_smoker)
     (Intercept) smokernon-smoker     smokersmoker 
       652.51316        -17.95079        -19.06444 

The first smoker is the predictor name; the rest is the level. In smokersmoker, only the second “smoker” names the level.

coef(m_smoker)

     (Intercept) smokernon-smoker     smokersmoker 
       652.51316        -17.95079        -19.06444 

Match A, B and C to the groups: former smoker, non-smoker and smoker.

Which coefficient name tells you which calculation to use?

Repeat the steps in Part A3 for rt ~ smoker, using its three levels and the coefficient names you just read.

1. Count the predictor variables.
2. Count the dummy slope coefficients for the group differences.
3. Add the intercept to count all beta coefficients.
4. Add the shared sigma to count all model parameters.

Compare with the two-group model: what increased, and what stayed the same?

K counts non-intercept beta coefficients. Use n − K − 1 for residual df; sigma is excluded from K.

anova() now gives us the ESS and RSS supplied in Week 6.

anova(m_smoker)
Analysis of Variance Table

Response: rt
           Df      Sum Sq     Mean Sq F value  Pr(>F)
smoker      2   17793.332  8896.66595 0.24369 0.78391
Residuals 262 9565066.423 36507.88711                

Sum Sq: smoker = ESS (explained); Residuals = RSS (unexplained).

TSS = ESS + RSS, so R² = ESS / TSS.

Part A3: read the two Sum Sq values, calculate TSS and R-squared, then check against summary(m_smoker)$r.squared.

Analysis of Variance Table

Response: rt
           Df      Sum Sq     Mean Sq F value  Pr(>F)
smoker      2   17793.332  8896.66595 0.24369 0.78391
Residuals 262 9565066.423 36507.88711                

\[ \text{Mean Sq} = \frac{\text{Sum Sq}}{\text{Df}}. \]

Part A3: divide each row’s Sum Sq by its own Df. Do your two results match the Mean Sq column?

These are explained and unexplained variation per degree of freedom. The divisions from Week 6’s F formula are already done for us.

m_smoker_summary <- summary(m_smoker)
m_smoker_summary$fstatistic
 value  numdf  dendf 
0.2437      2    262 

anova(m_smoker)
Analysis of Variance Table

Response: rt
           Df      Sum Sq     Mean Sq F value  Pr(>F)
smoker      2   17793.332  8896.66595 0.24369 0.78391
Residuals 262 9565066.423 36507.88711                

Both outputs give the same overall F-statistic.

Sum Sq: smoker = ESS, Residuals = RSS; together they make TSS.

Null hypothesis: all three population means are equal.

Analysis of Variance Table

Response: rt
           Df      Sum Sq     Mean Sq F value  Pr(>F)
smoker      2   17793.332  8896.66595 0.24369 0.78391
Residuals 262 9565066.423 36507.88711                

\[ F = \frac{\text{explained Mean Sq}}{\text{unexplained Mean Sq}}. \]

• Part A3: calculate F from the smoker and Residuals Mean Sq values.
• Calculate its p-value with pf(f_stat, df1, df2, lower.tail = FALSE), using the two Df entries.
• Explain how the overall F-test differs from the non-smoker coefficient’s t-test.

Complete Part B with the Chinese lexical-decision data.

• Fit rt ~ accuracy_group and calculate both means from its coefficients.
• Count its dummy slopes, add the intercept, then add sigma. Read the group t-statistic and p-value.
• Then fit rt ~ list_group using all four prepared list groups.
• Repeat the counting steps for four groups; calculate all four means and read the overall F-test.

After Part B, discuss your results. Complete the Week 7 problem set after the workshop for consolidation. The separate stretch task is optional.

• How do coefficient names reveal the reference group?
• Why are there three group coefficients for four list groups?
• How does an overall F-test differ from one coefficient’s t-test?

The Week 7 problem set asks for two-group predictions and a three-group overall test. Use the same logic even when the data set and labels change.

Recommended reading: Andrews (2021), Chapter 9, Normal Linear Models.

Complete the Week 7 problem set after this workshop. Calculate group means from the intercept and group differences; use predict() when a check is helpful. No downloaded dummy-coding function or extra prediction package is needed.

The first three questions use a two-group model. The remaining questions use three groups and ask for an overall test. Apply the same workflow to the new data and group labels.

• The intercept is the reference-group mean in a group-only model.
• Each dummy coefficient is a difference from that mean.
• A predictor with G groups needs G - 1 dummy coefficients.
• A group-only normal model has G beta coefficients and one shared sigma.
• Group predictions, coefficient tests, and overall tests answer different questions.