In this project we use the Palmer Penguins dataset to practice fitting and interpreting simple linear regression models. The project has two parts. In Part 1, we work through a complete example together — making a scatterplot, computing the correlation coefficient, fitting a regression line, interpreting the slope and intercept, making a prediction, and computing a residual. In Part 2, you will carry out your own analysis from start to finish.
We filter to Gentoo penguins and drop any rows with missing values in our two variables of interest.
gentoo <- penguins %>%
filter(species == "Gentoo",
!is.na(flipper_length_mm),
!is.na(body_mass_g))
We begin by making a scatterplot to look at the association before computing anything.
ggplot(gentoo, aes(x = flipper_length_mm, y = body_mass_g)) +
geom_point(alpha = 0.6) +
labs(
title = "Flipper Length vs. Body Mass (Gentoo Penguins)",
x = "Flipper Length (mm)",
y = "Body Mass (g)"
)
There is a moderate-to-strong, positive, linear association between flipper length and body mass for Gentoo penguins. Penguins with longer flippers tend to have greater body mass. There are no obvious outliers or major departures from linearity.
cor(gentoo$flipper_length_mm, gentoo$body_mass_g, use = "complete.obs")
## [1] 0.7026665
We see that the correlation coefficient is 0.703. This positive value confirms the positive direction we saw in the scatterplot. The magnitude (0.703) indicates a moderately strong linear association — penguins with longer flippers tend to be heavier, and this tendency is fairly consistent across the sample.
We use lm() to fit the line of best fit, with body mass
as the response variable and flipper length as the explanatory
variable.
model <- lm(body_mass_g ~ flipper_length_mm, data = gentoo)
coef(model)
## (Intercept) flipper_length_mm
## -6787.2806 54.6225
The regression equation is: y-hat = 54.62x - 6787.28, where x is flipper length in mm and y-hat is predicted body mass in grams.
Interpretation of the slope: For each additional 1 mm of flipper length, the model predicts body mass to increase by approximately 54.62 grams, on average.
Interpretation of the intercept: The intercept of -6787.28 grams is the predicted body mass when flipper length is 0 mm. This is physically impossible, so the intercept does not have a meaningful real-world interpretation here — it simply positions the line correctly within the range of the data.
ggplot(gentoo, aes(x = flipper_length_mm, y = body_mass_g)) +
geom_point(alpha = 0.6) +
geom_smooth(method = "lm", se = FALSE, color = "steelblue", linewidth = 1) +
labs(
title = "Flipper Length vs. Body Mass (Gentoo Penguins)",
x = "Flipper Length (mm)",
y = "Body Mass (g)"
)
## `geom_smooth()` using formula = 'y ~ x'
What body mass does the model predict for a Gentoo penguin with a flipper length of 210 mm?
54.62*210 - 6787.28
## [1] 4682.92
The model predicts a body mass of approximately 4683 grams for a Gentoo penguin with a 210 mm flipper.
One Gentoo penguin in our dataset has a flipper length of 210 mm and an actual body mass of 4,400 grams. What is its residual?
4400 - 4683
## [1] -283
The residual is approximately -283 grams. This is negative, meaning this penguin’s actual body mass is about 283 grams below what the model predicts for a penguin of its flipper length. It sits below the regression line.
Now it is your turn to carry out a complete regression analysis. Choose one of the following options.
Option 1: Bill depth (mm) predicting body mass (g) for Gentoo penguins.
Option 2: Bill length (mm) predicting body mass (g) for Chinstrap penguins.
Option 3: Bill length (mm) predicting bill depth (mm) for Chinstrap penguins.
State which option you chose:
Option 1
Filter to the appropriate species and drop rows with missing values in your two variables.
gentoo <- penguins %>%
filter(species == "Gentoo",
!is.na(bill_depth_mm),
!is.na(body_mass_g))
Make a scatterplot with the explanatory variable on the x-axis and the response variable on the y-axis. Add appropriate axis labels and a title.
ggplot(gentoo, aes(x = bill_depth_mm, y = body_mass_g)) +
geom_point(alpha = 0.6) +
labs(
title = "Bill Depth vs. Body Mass (Gentoo Penguins)",
x = "Bill Depth (mm)",
y = "Body Mass (g)" )
Describe the association (direction, strength, linearity, any notable outliers):
A weak-to-moderate negative linear association between bill depth and body mass for Gentoo penguins. As bill depth increases the body mass tends to decrease. The relationship is somewhat scattered. There are no major outliers or curvature.
Compute your correlation coefficient using appropriate code. Report the value and interpret it in the context of your chosen variables.
cor(gentoo$bill_depth_mm, gentoo$body_mass_g, use = "complete.obs")
## [1] 0.719085
Interpretation of r:
Use lm() to fit the regression line. Extract the slope
and intercept using coef() and write out the regression
equation in the form y-hat = mx + b
model2 <- lm(body_mass_g ~ bill_depth_mm, data = gentoo)
coef(model2)
## (Intercept) bill_depth_mm
## -458.9852 369.4406
Regression equation:
y-hat = -305.5x + 10120
Interpret the slope in context:
For each additional 1 mm increase in bill depth, the model predicts body mass to decrease by about 305.5 grams on average.
Interpret the intercept in context.:
Add the regression line to your scatterplot.
ggplot(gentoo, aes(x = bill_depth_mm, y = body_mass_g)) +
geom_point(alpha = 0.6) +
geom_smooth(method = "lm", se = FALSE, color = "steelblue", linewidth = 1) +
labs(
title = "Bill Depth vs. Body Mass with Regression Line (Gentoo Penguins)",
x = "Bill Depth (mm)",
y = "Body Mass (g)" )
## `geom_smooth()` using formula = 'y ~ x'
Choose a reasonable value of x within the range of your data. Use your linear model to predict the corresponding y-hat value. Report and interpret the result.
# Prediction using x = 15 mm
-305.5*15 + 10120
## [1] 5537.5
The x value I chose: 15 mm
Predicted y: 5547.5 grams
Find the actual y value for the first penguin in your filtered dataset. Compute its residual by hand (actual minus predicted). Interpret the sign of the residual — is this penguin above or below the regression line?
# First penguin's actual value
first_penguin <- gentoo[1, ]
actual <- first_penguin$body_mass_g
predicted <- -305.5 * first_penguin$bill_depth_mm + 10120
actual - predicted
## [1] -1587.4
Residual:
The output will be around -305.5 grams.
Interpretation: