Regression

Definition: Regression is the method used to predict values of one numerical variable (response) from values of another (explanatory).

Note: Regression can be done on data from an observational or experimental study.

We will discuss 3 types:

• Linear regression
• Nonlinear regression
• Logistic regression

Definition: Linear regression draws a straight line through the data to predict the response variable from the explanatory variable.

Slope determines rate of change of response with explanatory - humans lose 0.076 units of genetic diversity with every 10,000 km from East Africa.

Definition: For the population, the regression line is

\[ Y = \alpha + \beta X, \]
where \( \alpha \) (the intercept) and \( \beta \) (the slope) are population parameters.

Definition: For a sample, the regression line is

\[ Y = a + b X, \]
where \( a \) and \( b \) are estimates of \( \alpha \) and \( \beta \), respectively.

• \( a \): intercept
• \( b \): slope

Note: At each value of \( X \), there is a population of \( Y \)-values whose mean lies on the true regression line (this is the linear assumption).

• At each value of \( X \), the \( Y \)-measurements represent a random sample from the population of possible \( Y \)-values.
• At each value of \( X \), the distribution of possible \( Y \)-values is normal.
• The variance of \( Y \)-values is the same at all values of \( X \).

Technically, the linear regression equation is

\[ \mu_{Y\, |\, X=X^{*}} = \alpha + \beta X^{*}, \]

were \( \mu_{Y\, |\, X=X^{*}} \) is the mean of \( Y \) in the sub-population with \( X=X^{*} \) (called predicted values).

You are predicting the mean of Y given X.

Method of least squares

Definition: The least-squares regression line is the line for which the sum of all the squared deviations in \( Y \) is smallest.

The method of least-squares leads to the following estimates for intercept and slope:

\[ \begin{align} b & = \frac{\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sum_{i}(X_{i}-\bar{X})^2} \\ a & = \bar{Y}-b\bar{X} \end{align} \]

Note:

\[ b = \frac{\mathrm{Covariance(X,Y)}}{s_{X}^2} = r\frac{s_{Y}}{s_{X}}, \]

where \( r \) is the correlation coefficient!

Practice Problem #12

Male lizards in the species Crotaphytus collaris use their jaws as weapons during territorial interactions. Lappin and Husak (2005) tested whether weapon performance (bite force) predicted territory size in this species.

Compute best-fit line: Slope

\[ b = \frac{\mathrm{Covariance(X,Y)}}{s_{X}^2} \]

# Slope
(b <- cov(biteData$bite, biteData$territory.area)/var(biteData$bite))

[1] 11.6773

Compute best-fit line: Intercept

\[ a = \bar{Y}-b\bar{X} \]

# Intercept
(a <- mean(biteData$territory.area) - b*mean(biteData$bite))

[1] -31.53929

Faster!!! Use lm…

(biteRegression <- lm(territory.area ~ bite, data = biteData))

Call:
lm(formula = territory.area ~ bite, data = biteData)

Coefficients:
(Intercept)         bite  
     -31.54        11.68  

Remember: This was used for ANOVA too!!

Bonus!!! With lm, can add best-fit line to plot.

# Need to adjust margins to see axis labels
par(mar=c(4.5,5.0,2,2))

# Scatter plot
plot(biteData, pch=16, col="firebrick", cex=1.5, cex.lab=1.5, xlab="Bite force (N)", ylab=expression("Territory area" ~ (m^2)))

# Add in the best-fit line
abline(biteRegression, lwd=3)

Bonus!!! With lm, can add best-fit line to plot.

Definition: The predicted value of \( Y \) (denoted \( \hat{Y} \), or \( \mu_{Y\, |\, X} \)) from a regression line estimates the mean value of \( Y \) for all individuals having a given value of \( X \).

Definition: Residuals measure the scatter of points above and below the least-squares regression line, and are denoted by

\[ r_{i} = Y_{i} - \hat{Y}_{i}, \]
where \( \hat{Y}_{i} = a + bX_{i} \).

We can predict what the mean value of \( Y \) is for values of the explanatory variable \( X \) not represented in our data, as long as we are within the range of values of the data.

The function predict accomplishes this, and even gives us a standard error for our estimate.

(pred_5.1 <- predict(biteRegression, data.frame(bite = 5.1), se.fit = TRUE))

$fit
       1 
28.01492 

$se.fit
[1] 2.163259

$df
[1] 9

$residual.scale
[1] 5.788413

Definition: Extrapolation is the prediction of the value of a response variable outside the range of \( X \)-values in the data.

Regression should not be used to predict the value of the response values for an \( X \)-value that lies well outside the range of the data.

Definition: a residual plot is a scatter plot of the residuals \( (Y_{i}-\hat{Y}_{i}) \) against the \( X_{i} \), the values of the explanatory variable.

# Get residuals from regression output
biteData$res = resid(biteRegression)

# Plot residuals
plot(res ~ bite, data=biteData, pch=16, cex=1.5, cex.lab=1.5, col="firebrick", xlab="Bite force (N)", ylab="Residuals")

# Add a horizontal line at zero
abline(h=0, lty=2)

Definition: The variance of the residuals is known as \( \mathrm{MS}_{\mathrm{residuals}} \) and is given by

\[ \mathrm{MS}_{\mathrm{residuals}} = \frac{\sum_{i}\left(Y_{i}-\hat{Y}_{i}\right)^2}{n-2}. \]
This is also known as residual mean square.

\( \mathrm{MS}_{\mathrm{residuals}} \) is the average variance in the data around the regression line. This is completely analogous to the error mean square in ANOVA.

\( \mathrm{MS}_{\mathrm{residuals}} \) has \( n-2 \) degrees of freedom since we needed to estimate slope and intercept for the predicted values.

Use anova command on biteRegression

anova(biteRegression)

Analysis of Variance Table

Response: territory.area
          Df Sum Sq Mean Sq F value  Pr(>F)  
bite       1 194.37 194.374  5.8012 0.03934 *
Residuals  9 301.55  33.506                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual mean square is 33.506.

Can calculate residual mean square manually: \[ \mathrm{MS}_{\mathrm{residuals}} = \frac{\sum_{i}\left(Y_{i}-\hat{Y}_{i}\right)^2}{n-2}. \]

n <- nrow(biteData)
(MSresid <- ((n-1)/(n-2))*var(biteData$res))

[1] 33.50573

Same answer as ANOVA gave!!!

Two sources of variation:

• Residual part: Deviation between \( Y_{i} \) and predicted values \( \hat{Y}_{i} \) (analogous to error component in ANOVA).
• Regression part: Deviation between predicted values \( \hat{Y}_{i} \) and grand mean \( \bar{Y} \) (analogous to groups component in ANOVA).

In the ANOVA table, we can compute the \( F \)-ratio

\[ F = \frac{\mathrm{MS}_{\mathrm{regression}}}{\mathrm{MS}_{\mathrm{residual}}} \]

which is a test on the slope with \( H_{0}: \beta = 0 \).

Analysis of Variance Table

Response: territory.area
          Df Sum Sq Mean Sq F value  Pr(>F)  
bite       1 194.37 194.374  5.8012 0.03934 *
Residuals  9 301.55  33.506                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

\( P \)-value is less than 0.05, so we can reject the null hypothesis that the slope \( \beta = 0 \).

We can measure how well the line “fits” the data by estimating the \( R^{2} \) value using ANOVA, i.e.

\[ R^{2} = \frac{\mathrm{SS}_{\mathrm{regression}}}{\mathrm{SS}_{\mathrm{total}}}. \]

This also can be said to measure the fraction of variation in \( Y \) that is “explained” by \( X \).

Basic idea is:

• If \( R^{2} \) is close to 1, then \( X \) is explaining most of the variation in \( Y \), and any other variation which could be caused by other sources is negligible in comparison.
• If \( R^{2} \) is close to 0, then \( X \) is explaining very little of the variation in \( Y \), and the remaining variation is caused by other sources not accounted for or measured in the system of study.

For the lizard example,

biteRegSummary <- summary(biteRegression)
biteRegSummary$r.squared

[1] 0.3919418

Thus, 39% of the variation in territory area is explained by bite force.

# Need to adjust margins to see axis labels
par(mar=c(4.5,5.0,2,2))

# Scatter plot
plot(territory.area ~ bite, pch=16, col="firebrick", cex=1.5, cex.lab=1.5, xlab="Bite force (N)", ylab=expression("Territory area" ~ (m^2)), data=biteData)

# Add in the best-fit line
abline(biteRegression, lwd=3)

# Text
text(5, 30.5, expression(R^{2} ~ "=" ~ 0.39), cex=1.5)
text(5.14, 31.7, expression(y ~ "=" -31.54 + 11.68), cex=1.5)

summary(biteRegression)

Call:
lm(formula = territory.area ~ bite, data = biteData)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.0472 -4.5101 -0.5504  3.6689 10.2237 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  -31.539     23.513  -1.341   0.2127  
bite          11.677      4.848   2.409   0.0393 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 5.788 on 9 degrees of freedom
Multiple R-squared:  0.3919,    Adjusted R-squared:  0.3244 
F-statistic: 5.801 on 1 and 9 DF,  p-value: 0.03934