Week 12 Discussion

Tree Girth, Volume and Height Analysis

Let’s view the data we’re working with:

head(trees)

##   Girth Height Volume
## 1   8.3     70   10.3
## 2   8.6     65   10.3
## 3   8.8     63   10.2
## 4  10.5     72   16.4
## 5  10.7     81   18.8
## 6  10.8     83   19.7

Let’s see if there appears to be a linear relationship between girth and height where the girth of the tree can predict the height of a tree:
* There appear to be a linear relationships of varying strength between girth and height and volume and height

pairs(trees)

Using the lm() function we see that the linear function is:
Height = 83.2958 - Girth * 1.8615 + Volume * 0.5756

attach(trees)
trees_lm <- lm(Height ~ Girth + Volume, data=trees)
trees_lm

## 
## Call:
## lm(formula = Height ~ Girth + Volume, data = trees)
## 
## Coefficients:
## (Intercept)        Girth       Volume  
##     83.2958      -1.8615       0.5756

Summary

• The residuals are roughly balanced around the median, which is close to 0
• The Adjusted R^2 is low, 37.03%, so we see that volume and girth together don’t predict Height well
• Looking at the p-value, we see Volume is statistically significant while Girth is not

summary(trees_lm)

## 
## Call:
## lm(formula = Height ~ Girth + Volume, data = trees)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.7855 -3.3649  0.5683  2.3747 11.6910 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  83.2958     9.0866   9.167 6.33e-10 ***
## Girth        -1.8615     1.1567  -1.609   0.1188    
## Volume        0.5756     0.2208   2.607   0.0145 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.056 on 28 degrees of freedom
## Multiple R-squared:  0.4123, Adjusted R-squared:  0.3703 
## F-statistic:  9.82 on 2 and 28 DF,  p-value: 0.0005868

Backward Elimination

• Using backward elimination, let’s remove Girth to see how Volume alone predicts Girth
• The residuals are still roughly normally distributed around the median, which is close to 0
• Doing this we see that volume is statistically significant and the adjusted R^2 value is still low
• This is a weak linear relationship

trees_lm <- update(trees_lm, .~. - Girth, data = trees)
summary(trees_lm)

## 
## Call:
## lm(formula = Height ~ Volume, data = trees)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.7777  -2.9722  -0.1515   2.0804  10.6426 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 69.00336    1.97443  34.949  < 2e-16 ***
## Volume       0.23190    0.05768   4.021 0.000378 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.193 on 29 degrees of freedom
## Multiple R-squared:  0.3579, Adjusted R-squared:  0.3358 
## F-statistic: 16.16 on 1 and 29 DF,  p-value: 0.0003784

Now we’ll look more into residual analysis.

• From the summary(), the residuals appear to be roughly centered around the median, which is close to 0 and suggest they are normally distributed, which is good
• Plotting the residuals in a histogram we can also see that the residuals are normally distributed

hist(resid(trees_lm))

The residuals in our plot are not consistent. The majority of the residuals are on the left-center side of the graph. The volume isn’t a great predictor of height. Having additional predictors would make a better model.

plot(fitted(trees_lm),resid(trees_lm))
abline(0, 0, col = "red")

QQ Plots

If the residuals are normally distribuated, we would know the model is good. The points diverge from the line on the left and especially on the right side of this graph. The volume isn’t a good predictor.

qqnorm(resid(trees_lm))
qqline(resid(trees_lm),col="red")