Question #199207

Calculate the residual for x=3 when given

Hours,x and Scores, y (3,60),(7,92),(6,88),(5,81),(6,85),(4,74),(5,82),(7,94),(4,76),(5,84)


1
Expert's answer
2021-05-31T00:12:27-0400



Mean xˉ=xin=3+7+6+5+6+4+5+7+4+510=5210=5.2\bar x=\dfrac{\sum x_i}{n}=\dfrac{3+7+6+5+6+4+5+7+4+5}{10}=\dfrac{52}{10}=5.2


Mean yˉ=yin=81610=81.6\bar y =\dfrac{\sum y_i}{n}=\dfrac{816}{10}=81.6



xˉ=5.2\bar x=5.2 is not an integer , use assumed mean A=5

yˉ=81.6\bar y = 81.6 is not an integer , use assumed mean B = 82

Now,




From table:

byx=ndxdy(dx)(dy)ndx2(dx)2 =(10×111)2×(4)(10×16)(2)2 =1110+81604 =1118156=7.1667b_{yx}=\dfrac{n\sum dxdy-(\sum dx)(\sum dy)}{n\sum dx^2-(\sum dx)^2}\\\ \\=\dfrac{(10\times 111)-2\times (-4)}{(10\times 16)-(2)^2}\\\ \\=\dfrac{1110+8}{160-4}\\\ \\=\dfrac{1118}{156}=7.1667



Regression line y on x :

yyˉ=byx(xxˉ)y81.6=7.1667(x5.2)y=7.1667x+44.3333y-\bar y =b_{yx}(x-\bar x)\\\Rightarrow y-81.6=7.1667(x-5.2)\\\Rightarrow \boxed{ y=7.1667x+44.3333}



Now estimate y for x=3

y=7.1667×3+44.3333y=21.5+44.3333y=65.8333y=7.1667\times 3+44.3333\\\Rightarrow y=21.5+44.3333\\\Rightarrow \boxed{y=65.8333}


So, the residual at x=3

ϵ0=yy^\epsilon_0=|y-\hat y|

From data when x=3 y= 60

And residual is the absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y

So residual at x=3

65.83360=5.833\Rightarrow |65.833-60|=5.833


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