Answer to Question #112804 in Statistics and Probability for Nimra

Question

Find y on x and x on y
Price in RS :(12),(15),(18),(25),(22),(18),(30). Demand in kg : (65),(60),(50),(41),(40),(56),(45).
1)Find the equaition line y on x
2) find the equation line on x on y
3)find the demand when price is 35
4)find the price when demand is 25.

Answer 1

Let $x=$ price in RS, $y=$ demand in kg

$$\begin{matrix} & x & y & xy & x^2 & y^2 \\ & 12 & 65 & 780 & 144 & 4225 \\ & 15 & 60 & 900 & 225 & 3600 \\ & 18 & 50 & 900 & 324 & 2500 \\ & 25 & 41 & 1025 & 625 & 1681 \\ & 22 & 40 & 880 & 484 & 1600 \\ & 18 & 56 & 1008 & 324 & 3136 \\ & 30 & 45 & 1350 & 900 & 2025 \\ Sum= & 140 & 357 & 6843 & 3026 & 18767 \end{matrix}$$

$$mean: \bar{x}={\sum x_i\over n}={140\over 7}=20, \bar{y}={\sum y_i\over n}={357\over 7}=51$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-{1\over n}\big(\displaystyle\sum_{i=1}^nx_i\big)^2=$$ $$=3026-{(140)^2\over 7}=226$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-{1\over n}\big(\displaystyle\sum_{i=1}^ny_i\big)^2=$$ $$=18767-{(357)^2\over 7}=560$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-{1\over n}\big(\displaystyle\sum_{i=1}^nx_i\big)\big(\displaystyle\sum_{i=1}^ny_i\big)=$$ $$=6843-{140(357)\over 7}=-297$$

1) Find the equaition line y on x 

$$B={SS_{xy} \over SS_{xx}}={-297 \over 226}\approx-1.314159$$

$$A=\bar{y}-B\cdot\bar{x}=51-({-297 \over 226})(20)\approx77.283186$$

$$y=77.283186-1.314159x$$

2) Find the equaition line x on y 

$$M={SS_{xy} \over SS_{yy}}={-297 \over 560}\approx-0.530357$$

$$N=\bar{x}-M\cdot\bar{y}=20-({-297 \over 560})(51)\approx47.048214$$

$$x=47.048214-0.530357y$$

3) Find the demand when price is 35

$$y=77.283186-1.314159(35)=31.3 \ (kg)$$

4) Find the price when demand is 25.

$$x=47.048214-0.530357(25)=33.79 \ (RS)$$