Question #197671

1) The following data is obtained from a research


a) Find the best equation fitting this data by using regression analyses method

b) Find the correlation coefficient of the equation

c) Find the y value for the x=5.5

X Y

5 784

10 1043

15 1666

20 2478

25 4060

30 6069



Expert's answer

XYXYX2Y257843920256146561010431043010010878491516662499022527755562024784956040061404842540601015006251648360030606918207090036832761∑=   10516100372470227563934906\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} X & Y & XY & X^2 & Y^2 \\ \hline 5 & 784 & 3920 & 25 & 614656 \\ \hdashline 10 & 1043 & 10430 & 100 & 1087849 \\ \hdashline 15 & 1666 & 24990 & 225 & 2775556 \\ \hdashline 20 & 2478 & 49560 & 400 & 6140484 \\ \hdashline 25 & 4060 & 101500 & 625 & 16483600 \\ \hdashline 30 & 6069 & 182070 & 900 & 36832761 \\ \hline \sum=\ \ \ 105 & 16100 & 372470 & 2275 & 63934906 \\ \end{array}



Xˉ=1n∑i=1nXi=1056=17.5\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=\dfrac{105}{6}=17.5

Yˉ=1n∑i=1nYi=161006=2683.333333\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=\dfrac{16100}{6}=2683.333333

SSXX=∑i=1nXi2−1n(∑i=1nXi)2SS_{XX}=\displaystyle\sum_{i=1}^nX_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)^2

=2275−10526=437.5=2275-\dfrac{105^2}{6}=437.5

SSYY=∑i=1nYi2−1n(∑i=1nYi)2SS_{YY}=\displaystyle\sum_{i=1}^nY_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nY_i)^2

=63934906−1610026=20733239.333333=63934906-\dfrac{16100^2}{6}=20733239.333333

SSXY=∑i=1nXiYi−1n(∑i=1nXi)(∑i=1nYi)SS_{XY}=\displaystyle\sum_{i=1}^nX_iY_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nY_i)

=372470−105×161006=90720=372470-\dfrac{105\times16100}{6}=90720


m=SSXYSSXX=90720437.5=207.36m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{90720}{437.5}=207.36

n=Yˉ−m⋅Xˉ=2683.333333.2−207.36⋅17.5n=\bar{Y}-m\cdot \bar{X}=2683.333333.2-207.36\cdot17.5


=−945.466667=−945.466667

The regression equation is:


Y=−945.466667+207.36XY=−945.466667+207.36X

b)


r=SSXYSSXXSSYYr=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}

=90720437.520733239.333333≈0.952534=\dfrac{90720}{\sqrt{437.5}\sqrt{20733239.333333}}\approx0.952534

0.7<0.952534≤10.7<0.952534\leq1

Strong positive correlation,


c)

X=5.5X=5.5

Y=−945.466667+207.36(5.5)=195Y=−945.466667+207.36(5.5)=195






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