Question

Completion Status: QUESTION 2 Weight in tons) and fuel economy on mpo were measured for sample of seven diesel trucs. The results are presented in the following table Weight 8.002 4.502 7.001 4.502 8.500 2.752 1.250 Mileage 4.69 5.97 5.46 6.49 4.34 16.24 8.45 v What is the correlation coefficient for this data? A. 2.325 What is the equation of the least squares line for this data? 8. 41.36 Chat-8.3720.455 if the weight of two trucks differ by 5 tons, by how much would you predict their miles to differ? D. 1397 Predict the mileage for a truck weighing 5 tons. E 0.871 F. 6.047 Predict the mileage for a truck weighing 15 tons. Is this reasonable? if so give the G. 0.93 mileage, if not explain why it is not reasonable H-3372-04 14. it is not resonable to use this model to predict the mileage for a truck weighing 15 ton because this value is far removed from the data collected to build the model Extrapolationis risky

Answer 1

1)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	36.51	41.64	44.56	11.05	-20.70
mean	5.22	5.95	SSxx	SSyy	SSxy

Sample size,   n =   7      
here, x̅ = Σx / n=   5.216          
ȳ = Σy/n =   5.949          
SSxx =    Σ(x-x̅)² =    44.5593      
SSxy=   Σ(x-x̅)(y-ȳ) =   -20.7      
              
estimated slope , ß1 = SSxy/SSxx =   -20.7033/44.5593=   -0.4646      
intercept,ß0 = y̅-ß1* x̄ =   5.9486- (-0.4646 )*5.2157=   8.3719      
              
Regression line is, Ŷ=   8.37   + (   -0.465   )*x
              
SSE=   (SSxx * SSyy - SS²xy)/SSxx =    1.4267      
std error ,Se =    √(SSE/(n-2)) =    0.5342      
              
correlation coefficient ,    r = SSxy/√(SSx.SSy) =   -0.933

2)

Regression line is, Ŷ=   8.372 + (   -0.465   )*x

3)

Mileage difference = 5 * 0.465

= 2.325

4)

Predicted Y at X=   5   is          
Ŷ=   8.37191   +   -0.46462   *5=   6.047

5)

Predicted Y at X=   15   is          
Ŷ=   8.37191   +   -0.46462   *15=   1.397

Please let me know in case of any doubt.

Thanks in advance!

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