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 (USA Today, January 7, 2003).
City Miles of Track Ridership (1000s)
Cleveland 15 15
Denver 17 35
Portland 38 81
Sacramento 21 31
San Diego 47 75
San Jose 31 30
St. Louis 34 42
Use these data to develop an estimated regression equation that could be used to predict the ridership given the miles of track.
Did the estimated regression equation provide a good fit? Explain.
Develop a 95% confidence interval for the mean weekday ridership for all light-rail systems with 30 miles of track.
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. Do you think that the prediction interval you developed would be of value to Charlotte planners in anticipating the number of weekday riders for their new light-rail system? Explain.

Answer 1

Answer:

Step-by-step explanation:

Hello!

Given the variables:

Y: Weekday ridership for six light-rail systems (measured in thousands of passengers)

X: Miles of track of six light-rail systems

You have to estimate the regression equation to predict the ridership given the miles of track

^Yi= a + bXi

Where a is the estimation of the intercept and b is the estimation of the slope.

To calculate it you have to use the following formulas:

a= Y[bar] - bX[bar]

b=$\frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

∑X= 203; ∑X²= 6725; ∑Y= 309; ∑Y²= 17261; ∑XY= 17261

X[bar]= ∑X/n= 203/7= 29

Y[bar]= ∑Y/n= 309/7= 44.14

$b= \frac{17261-\frac{203*309}{7} }{6725-\frac{(203)^2}{7} }= 1.76$

$a= 44.14 - 1.76*29= -6.76$

The estimated regression equation is ^Yi= -6.76 + 44.14Xi

The formula for the 95% CI for the mean weekday ridership for all light-rail systems with 30 miles of track is:

95% CI for E(Y/X=30)

(a+bX₀) ± $t_{n-2;1-\alpha /2}$ * $\sqrt{Se^2(\frac{1}{n}+\frac{(X_0-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } )}$

Se²= 207.74

$t_{n-2;1-\alpha /2}= t_{5;0.975}= 2.571$

(-6.76 + 44.14*30) ± 2.571 * $\sqrt{207.74(\frac{1}{7}+\frac{(30-29)^2}{6725-\frac{(203)^2}{7} } )}$

1317.44 ± 2.571 * 5.47

[1303.38; 1331.50]

Using a 95% confidence level you'd expect that the interval [1303.38; 1331.50] includes the population average of weekday ridership for a 30 miles light-rail system.

95% Prediction interval: Y/$X_{n+1}$=30miles

(a+b$X_{n+1}$) ± $t_{n-2;1-\alpha /2}$ * $\sqrt{Se^2(1+\frac{1}{n}+\frac{(X_{n+1}-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } )}$

(-6.76 + 44.14*30) ± 2.571 * $\sqrt{207.74(1+\frac{1}{7}+\frac{(30-29)^2}{6725-\frac{(203)^2}{7} } )}$

1317.44±2.571*15.41

[1277.82; 1357.06]

With a 95% level, you can expect that the interval [1277.82; 1357.06] contains the possible value of weekday riders for a 30 miles light-rail system.

I hope this helps!