Question

Almost all U.S. light-rail systems use electric cars that run on tracks built at street level. The Federal Transit Administration claims light-rail is one of the safest modes of travel, with an accident rate of .99 accidents per million passenger miles as compared to 2.29 for buses. The following data show the miles of track and the weekday ridership in thousands of passengers for six light-rail systems.

City	Miles of Track	Ridership (1000s)
Cleveland	15	17
Denver	17	37
Portland	38	83
Sacramento	21	33
San Diego	47	77
San Jose	31	32
St. Louis	34	44

1. Use these data to develop an estimated regression equation that could be used to predict the ridership given the miles of track.
   
   Compute b₀ and b₁ (to 2 decimals).
   b₁   
   b₀   
   
   Complete the estimated regression equation (to 2 decimals).
   =   +  x
   
2. Compute the following (to 1 decimal):

   SSE
   SST
   SSR
   MSE

   
   
3. What is the coefficient of determination (to 3 decimals)? Note: report r² between 0 and 1.
     
   
   Does the estimated regression equation provide a good fit?
   SelectYes, it even provides an excellent fitYes, it provides a good fitNo, it does not provide a good fitItem 10  
   
4. Develop a 95% confidence interval for the mean weekday ridership for all light-rail systems with 30 miles of track (to 1 decimal).
   (  ,   )
   
5. Suppose that Charlotte is considering construction of a light-rail system with 30 miles of track. Develop a 95% prediction interval for the weekday ridership for the Charlotte system (to 1 decimal).
   (  ,   )
   
   

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
15	17	196.00	849.306	408.000
17	37	144.00	83.592	109.714
38	83	81.00	1358.449	331.714
21	33	64.00	172.735	105.143
47	77	324.00	952.163	555.429
31	32	4.00	200.020	-28.286
34	44	25.00	4.592	-10.714

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	203	323	838.000	3620.9	1471
mean	29.00	46.14	SSxx	SSyy	SSxy

sample size ,   n =   7          
here, x̅ = Σx / n=   29.00   ,     ȳ = Σy/n =   46.14  
                  
SSxx =    Σ(x-x̅)² =    838.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   1471.0          

a)

estimated slope , ß1 = SSxy/SSxx =   1471.0   /   838.000   =   1.76
                  
intercept,   ß0 = y̅-ß1* x̄ =   -4.76   
                  
so, regression line is   Ŷ =   -4.76 +   1.76 *x

b)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    1038.708  
SST=SSyy=3620.9

SSR=SSt-SSE=2582.149

MSE=SSE/(n-2) = 207.742

SSE	1038.7
SST	3620.9
SSR	2582.1
MSE	207.7

c)

R² =    (Sxy)²/(Sx.Sy) =    0.7131

71.31% of variation in observation of y is explained by x

So,

es, it provides a good fit

d)

X Value=   30                      
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 =    29.00                      
Σ(x-x̅)² =Sxx   838.0                      
Standard Error of the Estimate,Se=   14.41                      
                          
Predicted Y at X=   30   is                  
Ŷ =   -4.763   +   1.755   *   30   =   47.898
                          
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    5.470                      
margin of error,E=t*Std error=t* S(ŷ) =   2.5706   *   5.4704   =   14.0621      
                          
Confidence Lower Limit=Ŷ +E =    47.898   -   14.0621   =   33.8   
Confidence Upper Limit=Ŷ +E =   47.898   +   14.0621   =   62.0   
                          

e)

standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   15.4164              
margin of error,E=t*std error=t*S(ŷ)=    2.5706   *   15.42   =   39.6292
                  
Prediction Interval Lower Limit=Ŷ -E =   47.898   -   39.6292   =   8.3
Prediction Interval Upper Limit=Ŷ +E =   47.898   +   39.6292   =   87.5