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	17	16
Denver	19	36
Portland	40	82
Sacramento	23	32
San Diego	49	76
San Jose	33	31
St. Louis	36	43

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

The regression analysis is done in excel by following these steps,

Step 1: Write the data values in excel,

Step 2: DATA > Data Analysis > Regression > OK. The screenshot is shown below,

Step 3: Select Input Y Range: 'Y' column, Input X Range: 'X' column then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

_a)

b₁ = 1.7554
b₀ = -9.2736

y = =-9.2736 + 1.7554 x

b)

SSE	1038.708
SST	3620.857
SSR	2582.149
MSE	207.7416

c)

coefficient of determination,  r² = 0.7131

Yes, it provides a good fit.

The R-square value tells, how well the regression model fits the data values. The R-square value of the model is 0.7131 which means, the model explains approximately 71.31% of the variance of the data value. Based on this evidence we can conclude the model is a good fit.

d)

The standard error of the regression is,

For X = 30,

Now, the confidence interval for X_predictor = 30 is obtained using the formula,

From the data values,

	Miles of Track, X	X^2
	17	289
	19	361
	40	1600
	23	529
	49	2401
	33	1089
	36	1296
Sum	217	7565

The t-critical value is obtained from t-distribution table for significance level = 0.05 and degree of freedom = n - 2 = 7 - 2 = 5

e)

Now, the prediction interval for X_predictor = 30 is obtained using the formula,

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