Answer to Question #263780 in Statistics and Probability for Nishu

Question #263780

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? 



1
Expert's answer
2021-11-15T14:27:03-0500

(a)




(b)


XYXYX2Y2129108.21448175354925212.84184326416131114316912197638149428164Sum=5539391527297\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}




Xˉ=1niXi=557=7.8571428571429\bar{X}=\dfrac{1}{n}\sum_iX_i=\dfrac{55}{7}=7.8571428571429

Yˉ=1niYi=397=5.5714285714286\bar{Y}=\dfrac{1}{n}\sum_iY_i=\dfrac{39}{7}=5.5714285714286

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

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

SSXY=i(XiXˉ)(YiYˉ)=84.571428571429SS_{XY}=\sum_i(X_i-\bar{X})(Y_i-\bar{Y})=84.571428571429

Correlation coefficient:


r=84.57142857142994.85714285714379.714285714286r=\dfrac{84.571428571429}{\sqrt{94.857142857143}\sqrt{79.714285714286}}

r=0.9725693247r=0.9725693247

We have strong positive correlation.


(c)


r2=(0.9725693247)2=r^2=(0.9725693247)^2=

=0.9458910913=0.9458910913


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


(d)


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

=84.57142857142994.857142857143=0.8915662651=\dfrac{84.571428571429}{94.857142857143}=0.8915662651

n=YˉmXˉn=\bar{Y}-m\bar{X}

=5.57142857140.89156626514(7.8571428571)=5.5714285714-0.89156626514(7.8571428571)

=1.433735=-1.433735

Therefore, we find that the regression equation is:


Y=1.433735+0.891566XY=-1.433735+0.891566X



(e)


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

The expected breakdown for an eleven year old machine is 8.48.4


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