Simple Linear Regression
Modeling a response \(y\) using one predictor \(x\) with a straight line.
Simple Linear Regression
Modeling a response \(y\) using one predictor \(x\) with a straight line.
mtcars.head(mtcars)
## mpg cyl disp hp drat wt qsec vs am gear carb ## Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 ## Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 ## Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 ## Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 ## Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 ## Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
We fit: \( \text{mpg} = \beta_0 + \beta_1\,\text{wt} + \varepsilon \)
fit <- lm(mpg ~ wt, data = mtcars) summary(fit)
## ## Call: ## lm(formula = mpg ~ wt, data = mtcars) ## ## Residuals: ## Min 1Q Median 3Q Max ## -4.5432 -2.3647 -0.1252 1.4096 6.8727 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 37.2851 1.8776 19.858 < 2e-16 *** ## wt -5.3445 0.5591 -9.559 1.29e-10 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 3.046 on 30 degrees of freedom ## Multiple R-squared: 0.7528, Adjusted R-squared: 0.7446 ## F-statistic: 91.38 on 1 and 30 DF, p-value: 1.294e-10
\[ y_i = \\beta_0 + \\beta_1 x_i + \\varepsilon_i, \\quad \\varepsilon_i \\sim \\mathcal{N}(0, \\sigma^2). \]
The fitted line is \(\hat{y}_i = b_0 + b_1 x_i\).
Slope estimator:
\[ b_1 = \\frac{\\sum (x_i-\\bar{x})(y_i-\\bar{y})}{\\sum (x_i-\\bar{x})^2} \]
95% CI for slope:
\[ b_1 \\pm t_{\\alpha/2,\\,n-2}\\,\\mathrm{SE}(b_1). \]