Question Bank - Mathematics

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For 4 data points of two correlated variables x and y, it is given that? x = 24, ? y = 11, ? x2 = 202, ? xy = 84, ? y2 = 39Fit a least squares line to this data using x as independent variable.

A.
\(y=\dfrac{1}{116}(36+103x)\)
B.
\(x=\dfrac{1}{116}(36+103y)\)
C.
\(x=\dfrac{1}{116}(103+36y)\)
D.
\(y=\dfrac{1}{116}(103+36x)\)

Solution:

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 + bGiven:? x = 24, ? y = 11, ? x2 = 202, ? xy = 84, ? y2 = 39 and N = 4Analysis:\(m = \frac{ N \sum (xy) - \sum x \sum y}{N \sum (x^2) - (\sum x)^2}\)\(\Rightarrow m = \frac{(4 \times 84) - (24 \times 11)}{(4\times 202) - (24)^2}\)m = 72/232Now, \( b = \frac{\sum y - m \sum x}{N}\)\(\Rightarrow b = \frac{11-24~m}{4}\)\(\Rightarrow b = \frac{11 - \frac{(72 \times 24)}{232}}{4}\)\( \Rightarrow b = \frac{824}{232 \times 4}\)The equation of the line is:\(y = \dfrac{72}{232} x~ +~ \dfrac{824}{4 \times 232}\)\(\Rightarrow y=\dfrac{1}{116}(103+36x)\)

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