Question

In the United States, almost the entire tram system uses electric cars that run on roads at street level. The Federal Transit Administration states that the tram is one of the means more safe transport, since the accident rate is 0.99 accidents per million miles- passenger compared to 2.29 on buses. The following data gives the miles of track and the number of passengers transported on weekdays, in thousands, of six systems of streetcars (USA Today, January 7, 2003).

Cities                     miles o road        pasanger( thousand)

Cleveland                            15                           15

Denver                                  17                           35

Portland                                38                           81

Sacramento                         21                           31

San Diego                            47                           75

San Jose                              31                          30

St. Louis                               34                           42

a.Use this data to obtain the estimated regression equation that could be used to predict the number of passengers given the miles of roads.

b. Does the estimated regression equation provide a good fit? Explain

c. Get a 95% confidence interval for the average number of passengers transported on weekdays in tram systems that have 30 miles of tracks.

d. Suppose Charlotte is considering the construction of a 30-mile tram system of roads. Give a 95% prediction interval for the number of passengers transported

on a weekday by the Charlotte system. Do you think that the prediction interval that he developed may be useful to those who are planning Charlotte to anticipate the amount

of passengers on a weekday in your tram system? Explain.

Answer 1

Solution

Let x = miles of road and y = number of passengers (thousands) transported on weekdays.

Back-up Theory

The linear regression model: Y = β₀ + β₁X + ε, ……………………………………..............................…..(1)

where ε is the error term, which is assumed to be Normally distributed with mean 0 and variance σ².

Estimated Regression of Y on X is given by: Y_cap = β_0cap + β_1capX, ………………........................…….(2)

where β_1cap = Sxy/Sxx and β_0cap = Ybar – β_1cap.Xbar..………...........................…………………….…..(3)

Mean X = Xbar = (1/n) Σ(i = 1 to n)x_i …………………………............................……………….……….….(4)

Mean Y = Ybar = (1/n) Σ(i = 1 to n)y_i …………………………............................……………….……….….(5)

Sxx = Σ(i = 1 to n)(x_i – Xbar)² …………………………………............................……………..…………....(6)

Syy = Σ(i = 1 to n)(y_i – Ybar)² ……………………………………………..…............................……………(7)

Sxy = Σ(i = 1 to n){(x_i – Xbar)(y_i – Ybar)} …………………………………............................…………….(8)

Estimate of σ² is given by s² = (Syy – β_1cap²Sxx)/(n - 2)…………................………………….………..(9)

Standard Error of yicap = s√[(1/n) + {(xi – Xbar)²/Sxx}] …………................………………….………..(10)

100(1 - α)% Confidence Interval (CI) for y_cap at x = x₀ is

(β_0cap + β_1capx₀) ± t_{n – 2, α/2}xs√[(1/n) + {(x₀ – Xbar)²/Sxx}] …………................………………….…..(11)

100(1 - α)% Prediction Interval (PI) for y_cap at x = x₀ is

(β_0cap + β_1capx₀) ± {t_{n – 2, α/2} x s√[1 + (1/n) + {(x₀ – Xbar)²/Sxx}]} ……........………………………..(12)

Now, to work out the solution,

Final answers are given below. Details of calculations follow at the end

Part (a)

Estimated regression equation that could be used to predict the number of passengers given the miles of roads: y_hat (number of passengers) = -6.7628 + 1.7553x (miles of roads) Answer

Part (b)

To answer this question, we will perform ANOVA, details given below:

ANOVA	α	0.05
Source	DF	SS	MS	F	Fcrit	p-value
Regression	1	2582.149165	2582.149	12.42962	6.607891	0.01681901
Error	5	1038.707978	207.7416
Total	6	3620.857143

Since, F > Fcrit or equivalently, since p- value < α, we conclude that

the estimated regression equation provides a good fit. Answer 2

Part (c)

95% confidence interval for the average number of passengers transported on weekdays in tram systems that have 30 miles of tracks: [31.836, 59.960] Answer 3

Part (d)

95% prediction interval for the number of passengers transported for a 30-mile tram system of roads:

[6.268, 85.527] Answer 4

DONE

Details of calculations

n	7
Xbar	29.0000
ybar	44.1429
Sxx	838
Syy	3620.857143
Sxy	1471
β1cap	1.75537
β0cap	-6.762870781
s^2	207.7415956
s	14.41324376
α	0.05
n-2	5
tn-2,α/2	2.570581835
x0	30
CIYcapLB	31.83611707
CIYcapUB	59.96033707
PI LB	6.268985956
PIUB	85.52746819
ycap at x0	45.89822707

Complete