what is the least - squares regression line with the point (20,8) included in the data set? (this is a…

Explanation:

Step1: Calculate sums

Let (n = 15) (including the new - point ((20,8))). Calculate (\sum_{i = 1}^{n}x_i=3 + 4+5+6+7+8+8+10+11+11+12+13+13+14 + 20=135), (\sum_{i = 1}^{n}y_i=6 + 5+7+4+6+9+8+8+9+7+10+12+10+11+8 = 112), (\sum_{i = 1}^{n}x_i^2=3^2+4^2+5^2+6^2+7^2+8^2+8^2+10^2+11^2+11^2+12^2+13^2+13^2+14^2+20^2=1447), (\sum_{i = 1}^{n}x_iy_i=3\times6 + 4\times5+5\times7+6\times4+7\times6+8\times9+8\times8+10\times8+11\times9+11\times7+12\times10+13\times12+13\times10+14\times11+20\times8=1089).

Step2: Calculate slope (b_1)

The formula for the slope of the least - squares regression line is (b_1=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}). Substitute (n = 15), (\sum_{i = 1}^{n}x_i = 135), (\sum_{i = 1}^{n}y_i = 112), (\sum_{i = 1}^{n}x_i^2=1447), (\sum_{i = 1}^{n}x_iy_i = 1089) into the formula: [ \begin{align*} b_1&=\frac{15\times1089-135\times112}{15\times1447-(135)^2}\ &=\frac{16335 - 15120}{21705-18225}\ &=\frac{1215}{3480}\ &\approx0.3491 \end{align*} ]

Step3: Calculate intercept (b_0)

The formula for the intercept is (b_0=\bar{y}-b_1\bar{x}), where (\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{135}{15}=9) and (\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{112}{15}\approx7.4667). [ \begin{align*} b_0&=7.4667-0.3491\times9\ &=7.4667 - 3.1419\ &=4.3248 \end{align*} ]

Answer:

(\hat{y}=0.3491x + 4.3248)