Find the equation of least square line which will enable to predict yield on the basis of temperature. Find also the degree of relationship between temperature & yield.

Subject: Applied Mathematics 4

Topic: Correlation & Regression

Difficulty: Medium

Change of scale: u = x - 140; v = y - 66

$ \bar{u} = \frac{1}{n} (\sum u) = \frac{1}{10}(50) = 5 \\ \bar{v} = \frac{1}{n} (\sum v) = \frac{1}{10}(13) = 1.3 \\ \sigma_u^2 = \frac{1}{n} (\sum u^2) - (\bar{u})^2 = \frac{1}{10}(8500) - (5^2) = 825 \\ \sigma_u = 28.7228 \\ \sigma_v^2 = \frac{1}{n} (\sum v^2) - (\bar{v})^2 = \frac{1}{10}(1949) - (1.3)^2 = 194.9 - 1.69 = 193.21 \\ \sigma_v = 13.9 \\ cov(u,v) = \frac{1}{n}(\sum uv) - \bar{u}\bar{v} \\ = \frac{1}{10}(4050) - (5)(1.3) \\ = 405 - 6.5 = 398.5 $

$ r_{uv} = \frac{cov(u,v)}{\sigma_u \sigma_v} = \frac{398.5}{(28.7228)}(13.9) = 0.998129 \\ \implies r = r_{xy} = r_{uv} = 0.998129 $

$ b_{uv} = r \frac{\sigma_u}{\sigma_v} = 0.998129 \times \frac{28.7228}{13.9} = 2.0625 \\ \implies b_{xy} = b_{uv} = 2.0625 $

$ b_{vu} = r \frac{\sigma_v}{\sigma_u} = 0.998129 \times \frac{13.9}{28.7228} = 0.48303 \\ \implies b_{yx} = b_{vu} = 0.48303 $

$ u = x - 140 \\ \bar{u} = \bar{x} - 140 \\ \implies \bar{x} = \bar{u} + 140 = 5 + 140 = 145 \\ v = y - 66 \\ \bar{v} = \bar{y} - 66 \\ \bar{y} = \bar{v} + 66 \\ \implies \bar{y} = 1.3 + 66 = 67.3 $

There are two equations of lines of regression:

(i) Equation of line of regression of 'y' on 'x' is given by:

$ (y - \bar{y}) = byx(x - \bar{x}) \\ (y - 67.3) = (0.48303)(x - 145) \\ y - 67.3 = 0.48303x - 70.0393 \\ \implies y = 0.48303x - 2.73935 $

(ii) Equation of line of regression of 'x' on 'y' is:

$ (x - \bar{x}) = bxy(y - \bar{y}) \\ (x - 145) = (2.0625)(y - 67.3) \\ x - 145 = 2.0625y - 138.80625 \\ \implies x = 2.0625y + 6.19375 $