linear_fit <- lm(ave_acres_burned ~ years_since_1985,
  data = wildfires)
coef(linear_fit)

##      (Intercept) years_since_1985 
##        19.616453         2.221771

• Predicted average number of acres burned in 1985 is 19.6

linear_fit_year <- lm(ave_acres_burned ~ year,
  data = wildfires)
coef(linear_fit_year)

##  (Intercept)         year 
## -4390.598311     2.221771

• Predicted average number of acres burned in year 0 is -4390

An Important Message

• Explanatory: Concentration of calcium in drinking water
• Response: Annual mortality rate per 100,000 population
• Equation: Predicted Mortality = 1676.36 - 3.23 \(*\) Calcium

• Residual = Observed - Predicted

• \(\definecolor{residual}{RGB}{230,159,0}\color{residual}e_i\) = \(\definecolor{observed}{RGB}{0,158,115}\color{observed}y_i\) - \(\definecolor{predicted}{RGB}{86,180,233}\color{predicted}\widehat{y}_i\)

• How much information does the model give us about Mortality rate?
• How big do the residuals tend to be?

• How big do the residuals tend to be?
• One way to measure this: The residual standard deviation

\[s_e = \sqrt{\frac{\sum_{i = 1}^n e_i^2}{n - 2}} = \sqrt{\frac{\sum_{i = 1}^n (y_i - \hat{y}_i)^2}{n - 2}}\]

• A reasonable guess for mortality in this town: the average mortality rate in our sample, \(\bar{y} = 1524\)

• If we know about Calcium: Guess \(\hat{y}_i = 1676.36 - 3.23 x_i\)
  • For about 95% of sample, \(y_i\) is within \(\pm 286\) of \(\hat{y}_i\)

• If we don't know about Calcium: Guess \(\bar{y} = 1524\)
  • For about 95% of sample, \(y_i\) is within \(\pm 376\) of \(\bar{y}\)

• Without knowing calcium:
  • Guess \(\bar{y} = 1524\) people per 100,000 population
  • About 95% of our sample fell within \(\pm 2 s_y\) (\(\pm 376\)) of this guess
• If we know Calcium:
  • Guess \(\hat{y} = 1676.36 - 3.23 * Calcium\)
  • About 95% of our sample fell within \(\pm 2 s_e\) (\(\pm 286\)) of this guess
• Knowing calcium narrowed down range of values of Mortality a fair amount!

• Let's compare the spreads of these distributions of deviations from the predictions with and without knowing calcium.

• Let's compute \(\frac{\text{spread of residuals if we DO know calcium}}{\text{spread of deviations if we DON'T know about calcium}}\)

• Sample variance of Mortality: \[s_y^2 = \left[\sqrt{\frac{\sum_{i = 1}^n (y_i - \bar{y})^2}{n - 1}}\right]^2 = \frac{\sum_{i = 1}^n (y_i - \bar{y})^2}{n - 1} = \frac{SST}{n - 1}\]
• \(SST = \sum_{i = 1}^n (y_i - \bar{y})^2\). \(SST\) stands for Sum of Squares Total (used in measuring the total variability in the response)
• Residual Variance: \[s^2_e = \left[\sqrt{\frac{\sum_{i = 1}^n e_i^2}{n - 2}}\right]^2 = \frac{\sum_{i = 1}^n e_i^2}{n - 2} = \frac{\sum_{i = 1}^n (y_i - \hat{y}_i)^2}{n - 2} = \frac{SSE}{n - 2}\]
• \(SSE = \sum_{i = 1}^n (y_i - \hat{y}_i)^2\). \(SSE\) stands for Sum of Squared Errors (used in measuring the amount of variability in the errors, or residuals, from the model fit)

• Let's summarize spread of residuals by \(SSE = \sum_{i = 1}^n (y_i - \hat{y}_i)^2\)
• … and summarize spread of Mortality by \(SST = \sum_{i = 1}^n (y_i - \bar{y})^2\)
• So \(\frac{\text{spread of residuals if we DO know calcium}}{\text{spread of deviations if we DON'T know about calcium}} = \frac{SSE}{SST}\)
• Two extreme cases:
  • If the error goes to zero, we'd have a "perfect fit". \(\frac{SSE}{SST} = 0\)
  • If \(SSE = SST\), \(x\) has told us nothing about \(y\). \(\frac{SSE}{SST} = 1\)
• Interpret \(SSE / SST\) as the fraction of the total variation in \(y\) that is still in the residuals
• \(R^2 = 1 - \frac{SSE}{SST}\) is the fraction of the total variation in \(y\) "accounted for" by the model in \(x\).

• Let's summarize spread of residuals by the residual variance \(s_e^2 = \frac{SSE}{n - 2}\)
• … and summarize spread of Mortality by its sample variance \(s_y^2 = \frac{SST}{n - 1}\)
• So \(\frac{\text{spread of residuals if we DO know calcium}}{\text{spread of deviations if we DON'T know about calcium}} = \frac{s_e^2}{s_y^2}\)
• Two extreme cases:
  • If the error goes to zero, we'd have a "perfect fit". \(\frac{s_e^2}{s_y^2} = 0\)
  • If \(s_e^2 = s_y^2\), \(x\) has told us nothing about \(y\). \(\frac{s_e^2}{s_y^2} = 1\)
• \(R^2_{adj} = 1 - \frac{s_e^2}{s_y^2}\) is the fraction of the total variation in \(y\) "accounted for" by the model in \(x\).

summary(linear_fit)

## 
## Call:
## lm(formula = Mortality ~ Calcium, data = mortality_water)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -348.61 -114.52   -7.09  111.52  336.45 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 1676.3556    29.2981  57.217  < 2e-16 ***
## Calcium       -3.2261     0.4847  -6.656 1.03e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 143 on 59 degrees of freedom
## Multiple R-squared:  0.4288, Adjusted R-squared:  0.4191 
## F-statistic:  44.3 on 1 and 59 DF,  p-value: 1.033e-08

• Sample variance of Mortality: \[s_y^2 = \frac{SST}{n - 1}\]
• Residual Variance: \[s^2_e = \frac{SSE}{n - 2}\]
• The \(n - 1\) and \(n - 2\) have to do with the number of parameters used to make our prediction:
  • Normal model: 1 parameter (\(\mu\)), use \(n - 1\)
  • Linear model: 2 parameters (\(b_0\), \(b_1\)), use \(n - 2\)
• We'll talk about why in a month or so