Question

The Median Age of Vehicles on U. S. roads for seven different years.

Median Age in Years

Cars, x Trucks, y
8.1 7.8
7.7 7.6
6.5 6.5
6.9 7.6
6.0 6.3
5.4 5.8
4.9 5.9

1) Find the centroid (x-bar, y-bar)

2) Find the regression line (y-hat)

3) Find the coefficient of determination, what can you conclude?

4) Find the standard error of estimate

5) Construct a 95% prediction interval for the median age of trucks in use when the median age of cars in use is 7.3.

6) Find the a) explained variation, b) unexplained variation, and c) total variation

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
8.1	7.8	2.56	1.03	1.62
7.7	7.6	1.44	0.66	0.98
6.5	6.5	0.00	0.08	0.00
6.9	7.6	0.16	0.66	0.33
6	6.3	0.25	0.24	0.24
5.4	5.8	1.21	0.97	1.08
4.9	5.9	2.56	0.78	1.42

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	45.5	47.5	8.18	4.429	5.670
mean	6.500	6.786	SSxx	SSyy	SSxy

1)

sample size ,   n =   7      
here, x̅ = Σx / n=   6.500   ,     ȳ = Σy/n =   6.786

2)

SSxx =    Σ(x-x̅)² =    8.1800          
SSxy=   Σ(x-x̅)(y-ȳ) =   5.7          
                  
estimated slope , ß1 = SSxy/SSxx =   5.7   /   8.180   =   0.69315
                  
intercept,   ß0 = y̅-ß1* x̄ =   2.28021          
                  
so, regression line is   Ŷ =   2.280   +   0.693   *x

3)

      
R² =    (Sxy)²/(Sx.Sy) =    0.8875

88075% of variation in observation of y is explained by variable y

4)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    0.4984
      
std error of estimate ,Se =    √(SSE/(n-2)) =    0.3157

5)

X Value=   7.3                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   7                      
Degrees of Freedom,df=n-2 =   5                      
critical t Value=tα/2 =   2.571   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    6.50                      
Σ(x-x̅)² =Sxx   8.18                      
Standard Error of the Estimate,Se=   0.3157                      
                          
Predicted Y at X=   7.3   is                  
Ŷ =   2.2802   +   0.6932   *   7.3   =   7.340

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   0.35              
margin of error,E=t*std error=t*S(ŷ)=    2.5706   *   0.35   =   0.8968
                  
Prediction Interval Lower Limit=Ŷ -E =   7.340   -   0.90   =   6.443
Prediction Interval Upper Limit=Ŷ +E =   7.340   +   0.90   =   8.237

6)

Anova table
variation	SS	df	MS	F-stat	p-value
regression	3.9302	1	3.930	39.429	0.0015
error,	0.4984	5	0.100
total	4.4286	6

explained variation = 3.9302

uneplained=0.4984

total=4.4286