Answer to Question #93495 in Statistics and Probability for ragib

Question

Many people assume that the total amount of money saved depends on their total income. The following data shows the average saving per month (RM’00) and average income per month (RM’00) for various groups of employees.



Income

19

22

27

30

36

43

47

51

61

64

Saving

1.0

1.4

1.8

2.4

3.0

3.8

4.3

4.5

5.8

6.3



Fit a simple linear regression model for this data.
How much money will a person save if his monthly income is RM4,500?

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c:c:c:c:c:} x & 19 & 22 & 27 & 30 & 36 & 43 & 47 & 51 & 61 & 64\\ \hline y & 1.0 & 1.4 & 1.8 & 2.4 & 3.0 & 3.8 & 4.3 & 4.5 & 5.8 & 6.3 \end{array}$

Linear regression: $y=A+Bx$

$$B={S_{xy} \over S_{xx}}, A=\bar{y}-B\bar{x}$$

$$mean: \bar{x}={\sum x_i \over n}=40, \bar{y}={\sum y_i \over n}=3.43$$ $$S_{xx}=\sum(x_i-\bar{x})^2=\sum x_i^2-n\bar{x}^2=2246$$ $$S_{yy}=\sum(y_i-\bar{y})^2=\sum y_i^2-n\bar{y}^2=29.821$$ $$S_{xy}=\sum(x_i-\bar{x})(y_i-\bar{y})=\sum x_iy_i-n\bar{x}\bar{y}=258.4$$

$$r={S_{xy} \over \sqrt{S_{xx}}\sqrt{S_{yy}}}={258.4 \over \sqrt{2246}\sqrt{29.821}}\approx$$ $$\approx0.99845, strong\ correlation$$

$$B={S_{xy} \over S_{xx}}={258.4 \over 2246}\approx0.115049$$

$$A=\bar{y}-B\bar{x}\approx3.43-0.115049\cdot 40\approx-1.171959$$

$$trend \ \ line: y=-1.171959+0.115049x$$

How much money will a person save if his monthly income is RM4,500?

$$y=-1.171959+0.115049\cdot45=4.01$$