Answer to Question #112905 in Statistics and Probability for Nimra

Question

Annual income in thousands:(12),(13),(14),(15),(16),(17),(18),(19),(20).
Annual savings in thousands:(0),(0.1),(0.2),(0.2),(0.5),(0.5),(0.6),(0.7),(0.8).
Find the following:
1) equation line y on x
2) equation line x on y
3) find the annual saving when annual income is 25000.
4)find the annual income when annual saving is 550.

Answer 1

Let $x=$ annual income in thousands, $y=$ annual savings in thousands.

$$\begin{matrix} & x & y & xy & x^2 & y^2 \\ & 12 & 0 & 0 & 144 & 0 \\ & 13 & 0.1 & 1.3 & 169 & 0.01 \\ & 14 & 0.2 & 2.8 & 196 & 0.04 \\ & 15 & 0.2 & 3 & 225 & 0.04 \\ & 16 & 0.5 & 8 & 256 & 0.25 \\ & 17 & 0.5 & 8.5 & 289 & 0.25 \\ & 18 & 0.6 & 10.8 & 324 & 0.36 \\ & 19 & 0.7 & 13.3 & 361 & 0.49 \\ & 20 &0.8 & 16 & 400 & 0.64 \\ Sum= & 144 & 3.6 & 63.7 & 2364 & 2.08 \end{matrix}$$

$$mean: \bar{x}={\sum x_i\over n}={144\over 9}=16, \bar{y}={\sum y_i\over n}={3.6\over 9}=0.4$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-{1\over n}\big(\displaystyle\sum_{i=1}^nx_i\big)^2=$$ $$=2364-{(144)^2\over 9}=60$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-{1\over n}\big(\displaystyle\sum_{i=1}^ny_i\big)^2=$$ $$=2.08-{(3.6)^2\over 9}=0.64$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-{1\over n}\big(\displaystyle\sum_{i=1}^nx_i\big)\big(\displaystyle\sum_{i=1}^ny_i\big)=$$ $$=63.7-{144(3.6)\over 9}=6.1$$

1) Find the equaition line y on x 

$$B={SS_{xy} \over SS_{xx}}={6.1\over 60}\approx0.101667$$

$$A=\bar{y}-B\cdot\bar{x}=0.4-({6.1 \over 60})(16)\approx-1.226667$$

$$y=-1.226667+0.101667x$$

2) Find the equaition line x on y 

$$M={SS_{xy} \over SS_{yy}}={6.1 \over 0.64}=9.53125$$

$$N=\bar{x}-M\cdot\bar{y}=16-({6.1\over 0.64})(0.4)=12.1875$$

$$x=12.1875+9.53125y$$

3) Find the Annual saving when Annual income is 25000. 

$$y=-1.226667+0.101667(25)=1.315008$$

The Annual saving is $1315.$

4) Find the Annual income when Annual saving is 550.

$$x=12.1875+9.53125(0.550)=17.4296875$$

The Annual income is $17430.$