Find the equation `y = β_0 + β_1x` of the least-squares line that best fits the data points (-2, 0), (-1, 0), (0, 2,),(1, 4),(2, 4), assuming that the first and last data points are less reliable. Weight them half as much as the three interior points.

Find the equation `y = β_0 +  β_1x` of the least-squares line that best fits the data points (-2, 0), (-1, 0), (0, 2,),(1, 4),(2, 4), assuming that the first and last data points are less reliable. Weight them half as much as the three interior points.

Solution:

The least-squares line is a line that best fits a set of data points such that the sum of the squares of the distances between the line and the data points is minimized. To find the equation of this line, we can use weighted least squares.

We will first assign different weights to each data point, with the first and last data points having a weight of `1/2`, and the three interior points having a weight of 1. The equation of the least-squares line is given by:

`y = β_0 + β_1x`

Where

`β_1 = (Σ(w_i * x_i * y_i) - (Σw_i * x_i) * (Σw_i * y_i)) / (Σ(w_i * x_i^2) - (Σw_i * x_i)^2)`

`β_0 = (Σw_i * y_i) - β_1 * (Σw_i * x_i)`

Let's find the values of `β_1` and `β_0` using these equations

`x_i = [-2,-1,0,1,2]`

`y_i = [0,0,2,4,4]`

`w_i = [1/2,1,1,1,1/2]`

`Σ(w_i * x_i * y_i) = -1 + 0 + 0 + 4 + 4 = 7`

`Σ(w_i * x_i) = -1 + 0 + 0 + 1 + 1 = 1`

`Σ(w_i * y_i) = 0 + 0 + 2 + 4 + 2 = 8`

`Σ(w_i * x_i^2) = 1 + 1 + 0 + 1 + 1 = 4`

`β_1 = (7 - (1*8)) / (4 - (1)^2) = -1/3`

`β_0 = (8 - (-1/3 * 1)) = 8 + 1/3 = 8.333`

So the equation of the least-squares line is 

`y = 8.333 - (1/3)x`

Therefore, the equation of the least-squares line that best fits the data points (-2, 0), (-1, 0), (0, 2,),(1, 4),(2, 4) while weighting the first and last data points half as much as the three interior points is:

`y = 8.333 - (1/3)x`