Linear Regression

• Biology Seminar today, 4:00 pm in Millington 150
• To celebrate, NO READING QUIZ FOR MONDAY
• I will start grading stuff soon. Sorry for the delay!
• Homework #2
  • OpenStats, Chapter 4: 4.6.3 Hypothesis testing (p. 209) - #4.18, 4.20, 4.22, 4.24, 4.28, 4.30
  • OpenStats, Chapter 7: 7.5.1 Line fitting, residuals, and correlation (p. 356) - #7.1-7.10 (even)
  • OpenStats, Chapter 7: 7.5.2 Fitting a line by least squares regression (p. 362) - #7.24, 7.26, 7.30
  • OpenStats, Chapter 7: 7.5.4 Inference for linear regression (p. 367) - #7.36

Linear regression is a model formulation

Usually (but not always) it is reserved for situations where you assert evidence of causation (e.g. A causes B)

Correlation, in contrast, describes relationships (e.g. A and B are positively correlated)

Variables: For a correlation, our data consist of two numerical variables (continuous or discrete).

Definition: The (linear) correlation coefficient \( \rho \) measures the strength and direction of the association between two numerical variables in a population.

The linear (Pearson) correlation coefficient measures the tendency of two numerical variables to co-vary in a linear way.

The symbol \( r \) denotes a sample estimate of \( \rho \).

\[ r = \frac{\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sqrt{\sum_{i}(X_{i}-\bar{X})^2}\sqrt{\sum_{i}(Y_{i}-\bar{Y})^2}} \]

\[ -1 \leq r \leq 1 \]

\[ r = \frac{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})(Y_{i}-\bar{Y})}{\sqrt{\frac{1}{n-1}\sum_{i}(X_{i}-\bar{X})^2}\sqrt{\frac{1}{n-1}\sum_{i}(Y_{i}-\bar{Y})^2}} \]

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

http://www.tylervigen.com/spurious-correlations

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!

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  

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

\[ r_{i} = \hat{Y}_{i} - Y_{i}, \]
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 \( (\hat{Y}_{i}-Y_{i}) \) 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)

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

\( P \)-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, i.e.

\[ R^{2} = \frac{\sigma^{2}_{\mathrm{regression}}}{\sigma^{2}_{\mathrm{response}}} = \frac{\sigma^{2}_{\mathrm{response}} - \sigma^{2}_{\mathrm{residual}}}{\sigma^{2}_{\mathrm{response}}}. \]

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