Find the equation `y = β_0 + β_1x` of the least-squares line that best fits the data points (0, 1), (1, 1), (2, 2), (3, 2).

 

Find the equation `y = β_0 +  β_1x` of the least-squares line that best fits the data points (0, 1), (1, 1), (2, 2), (3, 2).

Solution:

To find the equation of the least-squares line that best fits the data points, we need to find the values of `β_0` and `β_1` that minimize the sum of the squares of the residuals (the difference between the observed y values and the predicted y values).

To do this, we can use the formulas for `β_0` and `β_1`:

`β_1 = (nΣ(xy) - ΣxΣy) / (nΣ(x^2) - (Σx)^2)`

`β_0 = ȳ - β_1x̄`

where:

n is the number of data points (in this case, 4)

xy is the product of the x and y values for each data point

x and y are the x and y values for each data point

`Σx, Σy, Σ(x^2), and Σ(xy)` are the sums of the x values, y values,`x^2` values, and xy values, respectively

x̄ and ȳ are the mean of the x values and y values, respectively

Using the given data points, we can calculate the values for `β_1` and `β_0`:

Σx = 0 + 1 + 2 + 3 = 6

Σy = 1 + 1 + 2 + 2 = 6

`Σx^2 = 0^2 + 1^2 + 2^2 + 3^2 = 14`

`Σxy = 0*1 + 1*1 + 2*2 + 3*2 = 0+1+4+6 = 11`

`x̄ = \sum x / n = 6 / 4 = 1.5`

`ȳ = \sum y / n = 6 / 4 = 1.5`

`β_1 = (nΣxy - ΣxΣy) / (nΣx^2 - Σx^2)`

`β_1 = (4*11 - 6*6) / (4*14 - 6^2) =(44-36)/(56-36)=8/20 =0.4`

`β_0 = ȳ - β_1x̄`

`β_0 = 1.5 - 0.4*1.5 = 1.5-0.6=2.1`

So, the equation of the least-squares line that best fits the data points is:

`y = 2.1- 0.4x`