Question

For 4 data points of two correlated variables x and y, it is given that

∑ x = 24, ∑ y = 11, ∑ x² = 202, ∑ xy = 84, ∑ y² = 39

Fit a least squares line to this data using x as independent variable.

1. \(y=\dfrac{1}{116}(36+103x)\)
2. \(x=\dfrac{1}{116}(36+103y)\)
3. \(x=\dfrac{1}{116}(103+36y)\)
4. \(y=\dfrac{1}{116}(103+36x)\)

Answer 1

Correct Answer - Option 4 : \(y=\dfrac{1}{116}(103+36x)\)

Concept:

To solve this type of problems we need to follow several steps:

1) Calculate the slope m:

\(\Rightarrow m = \frac{ N \sum (xy) - \sum x \sum y}{N \sum (x^2) - (\sum x)^2}\)

N is the number of points. 

2) Calculate the intercept b:

\(\Rightarrow b = \frac{\sum y - m \sum x}{N}\)

3) The equation of the line is: y = mx + b

Given:

∑ x = 24, ∑ y = 11, ∑ x2 = 202, ∑ xy = 84, ∑ y2 = 39 

and N = 4

Analysis:

\(m = \frac{ N \sum (xy) - \sum x \sum y}{N \sum (x^2) - (\sum x)^2}\)

\(\Rightarrow m = \frac{(4 \times 84) - (24 \times 11)}{(4\times 202) - (24)^2}\)

m = 72/232

Now, 

\( b = \frac{\sum y - m \sum x}{N}\)

\(\Rightarrow b = \frac{11-24~m}{4}\)

\(\Rightarrow b = \frac{11 - \frac{(72 \times 24)}{232}}{4}\)

\( \Rightarrow b = \frac{824}{232 \times 4}\)

The equation of the line is:

\(y = \dfrac{72}{232} x~ +~ \dfrac{824}{4 \times 232}\)

\(\Rightarrow y=\dfrac{1}{116}(103+36x)\)