Data 605 - Discussion 12

First we’ll take a quick look at our data - get the variable names and data set structure.

library(MASS)
library(tidyverse)
#fix(Boston)
names(Boston)

##  [1] "crim"    "zn"      "indus"   "chas"    "nox"     "rm"      "age"    
##  [8] "dis"     "rad"     "tax"     "ptratio" "black"   "lstat"   "medv"

str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

Visualize the Data with a scatter plot

plot(Boston$lstat, Boston$medv)

Next we develop a simple regression of medv (median home value) to lstat (percent of households with low socioeconomic status.). I would expect the coeffient for lsat to be negative. Let’s see if that the case.

lm.fit = lm(medv ~ lstat, data=Boston)
attach(Boston)
lm.fit = lm(medv~lstat)
lm.fit

## 
## Call:
## lm(formula = medv ~ lstat)
## 
## Coefficients:
## (Intercept)        lstat  
##       34.55        -0.95

The summary function can be utilized to see model results. The model yielded an intercept of 34.55 and lstat coefficient of -0.95. Both the intercept and the coefficient appear to be statistically significant.

summary(lm.fit)

## 
## Call:
## lm(formula = medv ~ lstat)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.168  -3.990  -1.318   2.034  24.500 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 34.55384    0.56263   61.41   <2e-16 ***
## lstat       -0.95005    0.03873  -24.53   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.216 on 504 degrees of freedom
## Multiple R-squared:  0.5441, Adjusted R-squared:  0.5432 
## F-statistic: 601.6 on 1 and 504 DF,  p-value: < 2.2e-16

The predict function can create confidence intervals and prediction intervals for the prediction of medv for given values of lstat. Let’s create some confidence intervals first.

predict(lm.fit, data.frame(lstat = c(5,10,15)),interval="confidence")

##        fit      lwr      upr
## 1 29.80359 29.00741 30.59978
## 2 25.05335 24.47413 25.63256
## 3 20.30310 19.73159 20.87461

Now for some prediction intervals:

predict(lm.fit, data.frame(lstat = c(5,10,15)),interval="prediction")

##        fit       lwr      upr
## 1 29.80359 17.565675 42.04151
## 2 25.05335 12.827626 37.27907
## 3 20.30310  8.077742 32.52846

The plot and abline plot call in question the linear relationship between medv and lstat

plot(lstat)
abline(lm.fit)

We can use the residuals function directly and in a plot to take a look at the residuals. The residuals seem to become form dense with higher prediction values.

plot(predict(lm.fit), residuals(lm.fit))

We can also take a look at the qq plots - this confirms our non-linearity concern.

qqnorm(resid(lm.fit))
qqline(resid(lm.fit))