Using the “cars” dataset in R, build a linear model for stopping distance as a function of speed and replicate the analysis of your textbook chapter 3 (visualization, quality evaluation of the model, and residual analysis.)

```
## speed dist
## 1 4 2
## 2 4 10
## 3 7 4
## 4 7 22
## 5 8 16
## 6 9 10
```

A scatter plot of stopping distance as a function of speed.

```
##
## Call:
## lm(formula = dist ~ speed)
##
## Coefficients:
## (Intercept) speed
## -17.579 3.932
```

- This model has an
`intercent`

of 17.5791 and a`slope`

of 3.9324, so for every additional unit of speed, the stopping distance increases by 3.9324. - The model has 48 degree of freedom, which is the number of samples minus the number of coefficents in the model (50 - 2).
- To determine the model quality, we look at the
`Multiple R-squared`

value. R-squared values closer to one indicate better model quality, so R^2 = 0.6511 indicates that this model is sufficient.

```
##
## Call:
## lm(formula = dist ~ speed)
##
## Residuals:
## Min 1Q Median 3Q Max
## -29.069 -9.525 -2.272 9.215 43.201
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -17.5791 6.7584 -2.601 0.0123 *
## speed 3.9324 0.4155 9.464 1.49e-12 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 15.38 on 48 degrees of freedom
## Multiple R-squared: 0.6511, Adjusted R-squared: 0.6438
## F-statistic: 89.57 on 1 and 48 DF, p-value: 1.49e-12
```

The residuals do not have a trend, which indicates that a linear model fits the data well. However, there are more residual values below 0 than above 0, which suggests that the data may have outliers.

```
dist_outlier <- boxplot.stats(cars$dist)$out
print(paste("There is a stopping distance outlier at",dist_outlier))
```

`## [1] "There is a stopping distance outlier at 120"`