Question

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the meta

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Answer #1

A)

ΣX ΣY Σ(x-x̅)² Σ(y-ȳ)² Σ(x-x̅)(y-ȳ)
total sum 4380.00 39.73 504.00 9.81 46.62
mean 182.50 1.66 SSxx SSyy SSxy

Sample size,   n =   24      
here, x̅ = Σx / n=   182.500          
ȳ = Σy/n =   1.655          
SSxx =    Σ(x-x̅)² =    504.0000      
SSxy=   Σ(x-x̅)(y-ȳ) =   46.6      
              
estimated slope , ß1 = SSxy/SSxx =   46.615/504=   0.0925      
intercept,ß0 = y̅-ß1* x̄ =   1.6554- (0.0925 )*182.5=   -15.2240      
              
Regression line is, Ŷ=   -15.2240   + (   0.0925   )*x

............

B)

Predicted Y at X=   185   is              
Ŷ=   -15.22402   +   0.09249   *185=   1.8866   

....

C)

X Y PREDICTED Y RESIDUAL
185 0.84 1.886642 -1.05
185 1.85 1.886642 -0.04
185 1.92 1.886642 0.03
185 2.62 1.886642 0.73

These residuals do not all have the same sign because in the cases of the first two pairs of observations, the observed effic

OPTION D

...............

D)

R² =    (SSxy)²/(SSx.SSy) =    0.439
Approximately 0.439 Of variation in observations of variable Y, is explained by variable x

..............

Please let me know in case of any doubt.

Thanks in advance!


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