• Debabrata Kabiraj (6.11) http://rpubs.com/destination4debabrata/Chapter6-InferenceForCategoricalData-Presentation
• Samriti Malhotra (6.27) http://rpubs.com/samriti0202/483129
• Ryan Gordon (6.43)

We will use the SAT data for 162 students which includes their verbal and math scores. We will model math from verbal. Recall that the linear model can be expressed as:

\[y = mx + b\]

Or alternatively as:

\[y = {b}_{1}x + {b}_{0}\]

Where m (or \(b_1\)) is the slope and b (or \(b_0\)) is the intercept. Therefore, we wish to model:

\[SAT_{math} = {b}_{1}SAT_{verbal} + {b}_{0}\]

To begin, we read in the CSV file and convert the Verbal and Math columns to integers. The data file uses . (i.e. a period) to denote missing values. The as.integer function will automatically convert those to NA (the indicator for a missing value in R). Finally, we use the complete.cases eliminate any rows with any missing values.

sat <- read.csv('../Data/SAT_scores.csv', stringsAsFactors=FALSE)
names(sat) <- c('Verbal','Math','Sex')
sat$Verbal <- as.integer(sat$Verbal)
sat$Math <- as.integer(sat$Math)
sat <- sat[complete.cases(sat),]

The first step is to draw a scatter plot. We see that the relationship appears to be fairly linear.

Calcualte z-scores (standard scores) for the verbal and math scores.

\[ z=\frac { y-\overline { y } }{ s } \]

sat$Verbal.z <- (sat$Verbal - verbalMean) / verbalSD
sat$Math.z <- (sat$Math - mathMean) / mathSD
head(sat)

##   Verbal Math Sex    Verbal.z      Math.z
## 1    450  450   F -1.47002058 -1.65180456
## 2    640  540   F  0.43914539 -0.73469449
## 3    590  570   M -0.06326671 -0.42899113
## 4    400  400   M -1.97243268 -2.16131016
## 5    600  590   M  0.03721571 -0.22518889
## 6    610  610   M  0.13769813 -0.02138665

Scatter plot of z-scores. Note that the pattern is the same but the scales on the x- and y-axes are different.

\[m = r\frac{S_y}{S_x} = r\frac{S_{math}}{S_{verbal}}\]

m <- r * (mathSD / verbalSD)
m

## [1] 0.6750748

Recall that the point where the mean of x and mean of y intersect will be on the line of best fit). Therefore,

\[b = \overline{y} - m \overline{x} = \overline{SAT_{math}} - m \overline{SAT_{verbal}}\]

b <- mathMean - m * verbalMean
b

## [1] 209.5542

We can now add the regression line to the scatter plot. The vertical and horizontal lines represent the mean Verbal and Math SAT scores, respectively.

To examine the residuals, we first need to calculate the predicted values of y (Math scores in this example).

sat$Math.predicted <- m * sat$Verbal + b
head(sat, n=4)

##   Verbal Math Sex    Verbal.z     Math.z Math.predicted
## 1    450  450   F -1.47002058 -1.6518046       513.3378
## 2    640  540   F  0.43914539 -0.7346945       641.6020
## 3    590  570   M -0.06326671 -0.4289911       607.8483
## 4    400  400   M -1.97243268 -2.1613102       479.5841

The residuals are simply the difference between the observed and predicted values.

sat$residual <- sat$Math - sat$Math.predicted
head(sat, n=4)

##   Verbal Math Sex    Verbal.z     Math.z Math.predicted   residual
## 1    450  450   F -1.47002058 -1.6518046       513.3378  -63.33782
## 2    640  540   F  0.43914539 -0.7346945       641.6020 -101.60203
## 3    590  570   M -0.06326671 -0.4289911       607.8483  -37.84829
## 4    400  400   M -1.97243268 -2.1613102       479.5841  -79.58408

Plot our regression line with lines representing the residuals. The line of best fit minimizes the residuals.

To show that \(m = r \frac{S_y}{S_x}\) minimizes the sum of squared residuals, this loop will calculate the sum of squared residuals for varying values of r above and below the calculated value.

results <- data.frame(r=seq(r - .2, r + .2, by=.01), 
                      m=as.numeric(NA),
                      b=as.numeric(NA),
                      sumsquares=as.numeric(NA))
for(i in 1:nrow(results)) {
    results[i,]$m <- results[i,]$r * (mathSD / verbalSD)
    results[i,]$b <-  mathMean - results[i,]$m * verbalMean
    predicted <- results[i,]$m * sat$Verbal + results[i,]$b
    residual <- sat$Math - predicted
    sumsquares <- sum(residual^2)
    results[i,]$sumsquares <- sum(residual^2)
}

Plot the sum of squared residuals for different slopes (i.e. r’s). The vertical line corresponds to the r (slope) calcluated above and the horizontal line corresponds the sum of squared residuals for that r. This should have the smallest sum of squared residuals.

To exemplify how the residuals change, the following scatter plot picks one of the “bad” models and plot that regression line with the original, best fitting line. Take particular note how the residuals would be less if they ended on the red line (i.e. the better fitting model). This is particularly evident on the far left and far right, but is true across the entire range of values.

b.bad <- results[1,]$b
m.bad <- results[1,]$m
sat$predicted.bad <- m.bad * sat$Verbal + b.bad

r ^ 2

## [1] 0.4686855

This model accounts for 46.9% of the variance math score predicted from verbal score.

The lm function in R will calculate everything above for us in one command.

sat.lm <- lm(Math ~ Verbal, data=sat)
summary(sat.lm)

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -173.590  -47.596    1.158   45.086  259.659 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 209.55417   34.34935   6.101 7.66e-09 ***
## Verbal        0.67507    0.05682  11.880  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 71.75 on 160 degrees of freedom
## Multiple R-squared:  0.4687, Adjusted R-squared:  0.4654 
## F-statistic: 141.1 on 1 and 160 DF,  p-value: < 2.2e-16

We can get the predicted values and residuals from the lm function

sat.lm.predicted <- predict(sat.lm)
sat.lm.residuals <- resid(sat.lm)

Confirm that they are the same as what we calculated above.

head(cbind(sat.lm.predicted, sat$Math.predicted), n=4)

##   sat.lm.predicted         
## 1         513.3378 513.3378
## 2         641.6020 641.6020
## 3         607.8483 607.8483
## 4         479.5841 479.5841

head(cbind(sat.lm.residuals, sat$residual), n=4)

##   sat.lm.residuals           
## 1        -63.33782  -63.33782
## 2       -101.60203 -101.60203
## 3        -37.84829  -37.84829
## 4        -79.58408  -79.58408

The histograms also show that the distribution are different across gender.

Upon careful examination of these two figures, there is some indication there may be a difference between genders. In the scatter plot, it appears that there is a cluster of males towoards the top left and a cluster of females towards the right. The residual plot also shows a cluster of males on the upper left of the cluster as well as a cluster of females to the lower right. Perhaps estimating two separate models would be more appropriate.

To start, we create two data frames for each gender.

sat.male <- sat[sat$Sex == 'M',]
sat.female <- sat[sat$Sex == 'F',]

Calculate the mean for Math and Verbal for both males and females.

(male.verbal.mean <- mean(sat.male$Verbal))

## [1] 590.375

(male.math.mean <- mean(sat.male$Math))

## [1] 626.875

(female.verbal.mean <- mean(sat.female$Verbal))

## [1] 602.0732

(female.math.mean <- mean(sat.female$Math))

## [1] 597.6829

Estimate two linear models for each gender.

sat.male.lm <- lm(Math ~ Verbal, data=sat.male)
sat.female.lm <- lm(Math ~ Verbal, data=sat.female)
sat.male.lm

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat.male)
## 
## Coefficients:
## (Intercept)       Verbal  
##    250.1452       0.6381

sat.female.lm

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat.female)
## 
## Coefficients:
## (Intercept)       Verbal  
##    158.9965       0.7286

Let’s compare the \(R^2\) for the three models.

cor(sat$Verbal, sat$Math) ^ 2

## [1] 0.4686855

cor(sat.male$Verbal, sat.male$Math) ^ 2

## [1] 0.4710744

cor(sat.female$Verbal, sat.female$Math) ^ 2

## [1] 0.5137626

The \(R^2\) for the full model accounts for approximately 46.9% of the variance. By estimating separate models for each gender we can account for 47.1% and 51.4% of the variance for males and females, respectively.

Re-examining the histogram of residuals, there is one data point with a residual higher than the rest. This is a possible outlier. In this section we’ll examine how that outlier may impact our linear model.

We see that excluding this point changes model slightly. With the outlier included we can account for 45.5% of the variance and by excluding it we can account for 47.9% of the variance. Although excluding this point improves our model, this is an insufficient enough reason to do so. Further explenation is necessary.

(sat.lm <- lm(Math ~ Verbal, data=sat))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat)
## 
## Coefficients:
## (Intercept)       Verbal  
##    209.5542       0.6751

(sat.lm2 <- lm(Math ~ Verbal, data=sat[sat$residual < 200,]))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat[sat$residual < 200, ])
## 
## Coefficients:
## (Intercept)       Verbal  
##    197.4697       0.6926

sat.lm$coefficients[2] ^ 2

##   Verbal 
## 0.455726

sat.lm2$coefficients[2] ^ 2

##    Verbal 
## 0.4797177

For the following two examples, we will add outliers to examine how they would effect our models. In the first example, we will add an outlier that is close to our fitted model (i.e. a small residual) but lies far away from the cluster of points. As we can see below, this single point increases our \(R^2\) by more than 5%.

outX <- 1200
outY <- 1150
sat.outlier <- rbind(sat[,c('Verbal','Math')], c(Verbal=outX, Math=outY))

(sat.lm <- lm(Math ~ Verbal, data=sat))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat)
## 
## Coefficients:
## (Intercept)       Verbal  
##    209.5542       0.6751

(sat.lm2 <- lm(Math ~ Verbal, data=sat.outlier))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat.outlier)
## 
## Coefficients:
## (Intercept)       Verbal  
##     186.372        0.715

unname(sat.lm$coefficients[2] ^ 2)

## [1] 0.455726

unname(sat.lm2$coefficients[2] ^ 2)

## [1] 0.5112893

Outliers can have the opposite effect too. In this example, our \(R^2\) is decreased by almost 16%.

outX <- 300
outY <- 1150
sat.outlier <- rbind(sat[,c('Verbal','Math')], c(Verbal=outX, Math=outY))
(sat.lm <- lm(Math ~ Verbal, data=sat))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat)
## 
## Coefficients:
## (Intercept)       Verbal  
##    209.5542       0.6751

(sat.lm2 <- lm(Math ~ Verbal, data=sat.outlier))

## 
## Call:
## lm(formula = Math ~ Verbal, data = sat.outlier)
## 
## Coefficients:
## (Intercept)       Verbal  
##    290.8915       0.5459

unname(sat.lm$coefficients[2] ^ 2)

## [1] 0.455726

unname(sat.lm2$coefficients[2] ^ 2)

## [1] 0.2979727