Answer to Question #161289 in Statistics and Probability for Sunny

Question

Aerospace industry believes that the cost of a space project is a function of the weight of the major object being space sent into space. The following data represent weight in tons and the cost in $ millions.

Weights cost

1.59 54

3.20 150

0.56 20

0.89 25

1.45 40

1.90 120

2.57 124

calculate:

1) construct regression equation

2) What is the strength of the regression equation?

Answer 1

1)

The regression equation:

$y=ax+b$

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}$

$a=\frac{533\cdot26.18-12.16\cdot1203.99}{7\cdot26.18-12.16^2}=-19.4$

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

$b=\frac{7\cdot1203.99-12.16\cdot533}{7\cdot26.18-12.16^2}=55$

$y=-19.4x+55$

2)

Correlation coefficient:

$r=\frac{cov(x,y)}{\sqrt{s_x^2s_y^2}}$

$cov(x,y)=\frac{\sum (X-\overline{X})(Y-\overline{Y})}{n-1}$

$s_x^2=\frac{\sum (X-\overline{X})^2}{n-1}$

$s_y^2=\frac{\sum (Y-\overline{Y})^2}{n-1}$

$\overline{X}=\frac{\sum x}{n}=\frac{12.16}{7}=1.74$

$s_x^2=\frac{5.06}{6}=0.84$

$\overline{Y}=\frac{\sum y}{n}=\frac{533}{7}=76.14$

$s_y^2=\frac{17232.86}{6}=2872.14$

$cov(x,y)=278.1/6=46.35$

$r=\frac{46.35}{\sqrt{0.84\cdot2872.14}}=0.94$

The regression has a strong positive correlation.