Answer to Question #263780 in Statistics and Probability for Nishu

Question

The following data gives the information on the ages (in years) and the number of breakdowns during the past month for a sample of 7 machines at a large industrial company.

Age (X): 12, 7, 2, 8, 13, 9, 4

Number of breakdowns (Y): 9, 5, 1, 4, 11, 7, 2

(a) Draw a scatter diagram to represent the above data.

(b) Calculate the value of Pearson’s correlation coefficient r. Interpret the value of r.

(c) Calculate coefficient of determination. Interpret your answer.

(d) Determine the equation of the regression line using least squares method.

(e) What is the expected breakdown for an eleven year old machine? 

Answer 1

(a)

(b)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2\\ \hline & 12 & 9 & 108.2 & 144 & 81\\ & 7 & 5 & 35 & 49 & 25\\ & 2 & 1 & 2.8 & 4 & 1\\ & 8 & 4 & 32 & 64 & 16\\ & 13 & 11 & 143 & 169 & 121\\ & 9 & 7 & 63 & 81 & 49\\ & 4 & 2 & 8 & 16 & 4\\ Sum= & 55 & 39 & 391 & 527 & 297\\ \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum_iX_i=\dfrac{55}{7}=7.8571428571429$$

$$\bar{Y}=\dfrac{1}{n}\sum_iY_i=\dfrac{39}{7}=5.5714285714286$$

$$SS_{XX}=\sum_i(X_i-\bar{X})^2=94.857142857143$$

$$SS_{YY}=\sum_i(Y_i-\bar{Y})^2=79.714285714286$$

$$SS_{XY}=\sum_i(X_i-\bar{X})(Y_i-\bar{Y})=84.571428571429$$

Correlation coefficient:

$$r=\dfrac{84.571428571429}{\sqrt{94.857142857143}\sqrt{79.714285714286}}$$

$r=0.9725693247$

We have strong positive correlation.

(c)

$$r^2=(0.9725693247)^2=$$

$$=0.9458910913$$

A coefficient of determination of 0.945891 shows that 94.5891% of the data fit the regression model.

(d)

$$m=slope=\dfrac{SS_{XY}}{SS_{XX}}$$

$$=\dfrac{84.571428571429}{94.857142857143}=0.8915662651$$

$$n=\bar{Y}-m\bar{X}$$

$$=5.5714285714-0.89156626514(7.8571428571)$$

$$=-1.433735$$

Therefore, we find that the regression equation is:

$$Y=-1.433735+0.891566X$$

(e)

$$Y=-1.433735+0.891566(11)=8.373491$$

The expected breakdown for an eleven year old machine is $8.4$