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Run a regression analysis on the following data set, where y is the final grade in a math class and x is the average number o
Research was conducted on the amount of training for 5K and the time a contestant took to run the race. The researcher record
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Answer #1

a)

ΣX ΣY Σ(x-x̅)² Σ(y-ȳ)² Σ(x-x̅)(y-ȳ)
total sum 94.00 674.60 128.40 1397.60 352.56
mean 9.40 67.46 SSxx SSyy SSxy

Sample size,   n =   10      
here, x̅ = Σx / n=   9.400          
ȳ = Σy/n =   67.460          
SSxx =    Σ(x-x̅)² =    128.4000      
SSxy=   Σ(x-x̅)(y-ȳ) =   352.6      
              
estimated slope , ß1 = SSxy/SSxx =   352.56/128.4=   2.7458      
intercept,ß0 = y̅-ß1* x̄ =   67.46- (2.7458 )*9.4=   41.6495      
              
Regression line is, Ŷ=   41.6   + (   2.75 )*x

b)

Predicted Y at X=   12   is          
Ŷ=   41.64953   +   2.74579   *12=   74.6

-----------------------------

a)

ΣX ΣY Σ(x-x̅)² Σ(y-ȳ)² Σ(x-x̅)(y-ȳ)
total sum 359.00 150.00 1938.83 137.76 -22.20
mean 59.83 25.00 SSxx SSyy SSxy

Sample size,   n =   6      
here, x̅ = Σx / n=   59.833          
ȳ = Σy/n =   25.000          
SSxx =    Σ(x-x̅)² =    1938.8333      
SSxy=   Σ(x-x̅)(y-ȳ) =   -22.2      
              
estimated slope , ß1 = SSxy/SSxx =   -22.2/1938.8333=   -0.0115      
intercept,ß0 = y̅-ß1* x̄ =   25- (-0.0115 )*59.8333=   25.6851      
              
Regression line is, Ŷ=   25.7   + (   -0.011   )*x
              
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    137.5058      
std error ,Se =    √(SSE/(n-2)) =    5.8631      
              
correlation coefficient ,    r = SSxy/√(SSx.SSy) =   -0.043

b)

Regression line is, Ŷ=   25.6851 + (   -0.0115 )*x

c)

Predicted Y at X=   62   is          
Ŷ=   25.68510   +   -0.01145   *62=   24.975

d)

x y (x-x̅)² (y-ȳ)² (x-x̅)(y-ȳ) residual,ei=y-y^
62 20.5 4.6944 20.2500 -9.750 24.98 -4.4752

Please let me know in case of any doubt.

Thanks in advance!


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