Question

2. You are a consultant hired by the city government of a community considering the development of a light rail mass transit system. The data you have to help you make recommendations to the city are given in the table of data on similar mass transit systems in 11 different U.S. cities shown below (USA Today - 12/28/04): Stations 38 14 53 City Atlanta Baltimore Boston Chicago Cleveland Los Angeles Miami New York Philadelphia San Francisco Washington 144 18 16 22 468 53 43 86 Vehicles 252 100 408 1190 60 102 136 6333 371 669 950 Track (mi) 193 34 108 288 42 34 57 835 102 246 226 a) Are there any cities above that might be regarded as outliers/fliers? If so what city or cities? b) Can you justify removal of this city (or cities) from the data to be used for further evaluation for the proposed mass transit system? Create box plots for each of the three variables (ie, Stations, Vehicles, and Track). Consider an Upper Flier Limit (UFL) of the upper quartile (03) plus two times the inter-quartile range (IQR = 23-01). Remove the one city that has results above these limits for all three variables. c) Create a scatter plot of Track (x) and Stations (y) using the data after removing any fliers. d) Is there evidence of a linear relationship? What is the correlation coefficient (ie,r)? e) Estimate the line of best fit (ie, least squares) for the scatter plot in c), and plot it on the chart. f) Provide a 95% confidence interval for the number of stations to service a given number of miles of track? g) Create a scatter plot of Track (x) and Vehicles (y) again using the same data as used for c)? h) Is there evidence of a linear relationship? What is the correlation coefficient (ie, r)? i) Estimate the line of best fit (ie, least squares) for the scatter plot ing), and plot it on the chart. i) Provide a 95% confidence interval for the number of vehicles to service a given number of miles of track?

Answer 1

Note : Allowed to solve only 4 questions in one post. Please post the remaining in a new post.

a. We see that the number of stations in New York as well as number of vehicles are far greater than the remaining city. Hence New York could be an outlier.

b.
Box plot for Stations

Q1 Q2 Q3 14 16 1 22 38 43 53 53 56 144 408

Arrange the data in ascending order To find the find the median: We see that there odd number of observation. Hence we take middle value as shown above. Quartile 2 or Median = 43 To find the find the first quartile : We take the upper half of the data from the median value and take the middle value as shown. Quartile 1 or 25th percentile = 18 To find the find the third quartile : We take the lower half of the data from the median value and take the middle value as shown. Quartile 3 or 75th percentile = 86 The five point summary min = 14 1st quartile =18 median = 43 3rd quartile = 86 max = 468 3(86.8182-43) skewness 3(mean-median) 0.9953667 o 132.0665 Interquartile Range (IQR) IQR = Q3 - Q1 = 86-18=68 To find the outlier : lowerlimit = Q1 - 2 IQR = 18 - (1.5 X 68)= -118 upperlimit = Q3 + 2 IQR = 86 + (1.5 X 68)= 222 Hence any value lower than-118 or greater than 222 is an outlier.

Boxplot o 468 400 300 200 0 100 86 43 18 0

Box plot for Vehicles

Q1 Q2 Q3 60 100 102 136 252 371 408 669 950 1190 6333

Arrange the data in ascending order To find the find the median : We see that there odd number of observation. Hence we take middle value as shown above. Quartile 2 or Median = 371 To find the find the first quartile : We take the upper half of the data from the median value and take the middle value as shown. Quartile 1 or 25th percentile = 102 To find the find the third quartile : We take the lower half of the data from the median value and take the middle value as shown. Quartile 3 or 75th percentile = 950 The five point summary min = 60 1st quartile = 102 median = 371 3rd quartile = 950 max = 6333 skewness 3(mean-median) 3(961-371) 1820.0118 0.9725212 Interquartile Range (IQR) IQR = Q3 - Q1 = 950- 102= 848 To find the outlier lowerlimit = Q1 - 2 IQR = 102 - (1.5 X 848)= - 1594 upperlimit = Q3 +2 IQR = 950 + (1.5 X 848)= 2646 Hence any value lower than -1594 or greater than 2646 is an outlier.

Boxplot 6333 5000 T 3000 1000 950 371 o

Box plot for Track

Q1 Q2 Q3 34 34 42 57 102 108 193 226 246 288 835

Arrange the data in ascending order To find the find the median : We see that there odd number of observation. Hence we take middle value as shown above. Quartile 2 or Median = 108 To find the find the first quartile : We take the upper half of the data from the median value and take the middle value as shown. Quartile 1 or 25th percentile = 42 To find the find the third quartile : We take the lower half of the data from the median value and take the middle value as shown. Quartile 3 or 75th percentile = 246 The five point summary min = 34 1st quartile =42 median = 108 3rd quartile = 246 max = 835 3(mean-median) skewness 3(196.8182-108) 230.7145 1.1549105 Interquartile Range (IQR) IQR = Q3 - Q1 = 246-42= 204 To find the outlier: lowerlimit = Q1 - 2 IQR = 42-(1.5 X 204)= -366 upperlimit = Q3 +2 IQR = 246 + (1.5 x 204)= 654 Hence any value lower than -366 or greater than 654 is an outlier.

Boxplot 835 800 600 400 246 200 108 84

We see that New York is an outlier in all three plot hence it must be removed.

c. Scatterplot

Track(x) Station(y) 193 38 34 14 Scatter plot 160 City altanta Baltimore Boston Chicago Cleveland 108 53 140 . 288 144 120 42 18 100 LA 34 16 . Station(y) 80 57 22 60 102 53 2 Miami Philadelphia SF Washington 40 246 43 226 86 20 0 0 50 100 150 200 250 300 350 Track(x)

d. Correlation coefficient
Yes there seem to be positive linear relationship between the two variables.

The correlation coefficient is given below.

Let x represent Track Let y represent Station Obs.No. X (x - 2) y Y-Y (y-y2 (x-7)(y-7) 1 193 50 3600 38 -10.7 114.49 -642 34 -99 9801 14 -34.7 1204.09 3435.3 N کا 108 -25 625 53 4.3 18.49 -107.5 4 288 155 24025 144 95.3 9082.09 14771.5 5 42 -91 8281 co -30.7 942.49 2793.7 6 34 -99 9801 16 -32.7 1069.29 3237.3 7 57 -76 5776 22 -26.7 712.89 2029.2 8 102 -31 961 53 4.3 18.49 -133.3 9 246 113 12769 43 -5.7 32.49 -644.1 10 226 93 8649 86 37.3 1391.29 3468.9 Total 1330 0 84288 487 O 14586.1 28209

n=10 x = 1330 Mean of Track 1330 10 133 72 x-=0 (x - 2)2 = 84288 y="487 Mean of Station 487 y 48.7 10 Sy-y=0 (y-7)? = 14586.1 (x - 2)(y- y) = 28209 Standard deviation of Track = 72 (2-1) 84288 96.7746523 72-1 3 Variance of Track o = 96.7746523 = 9365.3333333 Standard deviation of Station 14586.1 40.2576425 3 = 1620.6777778 12-1 Variance of Station o} = 40.25764252 Correlation (1-1)(y-7) P (1-7)(y-7) 28209 84288X14586.1 0.8045 V