$$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i,\qquad i = 1,\dots,n$$
$$\beta_1 = \frac{\sum_{i=1}^{n}{(x_i-\bar{x}) \cdot (y_i-\bar{y})}}{\sum_{i=1}^{n}{(x_i-\bar{x})^2}}$$
$$\beta_0 = \bar{y} - \beta_1 \cdot \bar{x}$$
Predict: $$\hat{y} = \hat{\beta}_0 + \hat{\beta}_1 x$$
Standard Error: $$\hat{\sigma}^2 = \frac{1}{n-2} \sum_{i=1}^n (y_i - \hat{y}_i)^2$$
$$ \hat{\sigma}^2 = \frac{1}{n-2}\sum_{i=1}^n (y_i - \hat{y}_i)^2 $$
$$ SE(\hat{\beta}_1) = \sqrt{ \frac{\hat{\sigma}^2} {\sum_{i=1}^n (x_i - \bar{x})^2}} $$ $$ SE(\hat{\beta}_0) = \sqrt{ \hat{\sigma}^2 \left( \frac{1}{n} + \frac{\bar{x}^2}{\sum_{i=1}^n (x_i - \bar{x})^2} \right)} $$ $$ t = \frac{\hat{\beta}_1}{SE(\hat{\beta}_1)} $$ $$ \hat{\beta}_1 \pm t_{0.975, n-2} \cdot SE(\hat{\beta}_1) $$ $$ \hat{\beta}_0 \pm t_{0.975, n-2} \cdot SE(\hat{\beta}_0) $$| Dataset I | Dataset II | Dataset III | Dataset IV | ||||
|---|---|---|---|---|---|---|---|
| x | y | x | y | x | y | x | y |
| 10.0 | 8.04 | 10.0 | 9.14 | 10.0 | 7.46 | 8.0 | 6.58 |
| 8.0 | 6.95 | 8.0 | 8.14 | 8.0 | 6.77 | 8.0 | 5.76 |
| 13.0 | 7.58 | 13.0 | 8.74 | 13.0 | 12.74 | 8.0 | 7.71 |
| 9.0 | 8.81 | 9.0 | 8.77 | 9.0 | 7.11 | 8.0 | 8.84 |
| 11.0 | 8.33 | 11.0 | 9.26 | 11.0 | 7.81 | 8.0 | 8.47 |
| 14.0 | 9.96 | 14.0 | 8.10 | 14.0 | 8.84 | 8.0 | 7.04 |
| 6.0 | 7.24 | 6.0 | 6.13 | 6.0 | 6.08 | 8.0 | 5.25 |
| 4.0 | 4.26 | 4.0 | 3.10 | 4.0 | 5.39 | 19.0 | 12.50 |
| 12.0 | 10.84 | 12.0 | 9.13 | 12.0 | 8.15 | 8.0 | 5.56 |
| 7.0 | 4.82 | 7.0 | 7.26 | 7.0 | 6.42 | 8.0 | 7.91 |
| 5.0 | 5.68 | 5.0 | 4.74 | 5.0 | 5.73 | 8.0 | 6.89 |