Pipes used in a water distribution network are susceptible to breakage due to a variety of factors. When pipes break, engineers must decide whether to repair or replace the broken pipe. A team of civil engineers used regression analysis to estimate y = the ratio of repair to replacement cost of commercial pipe in the IHS Journal of Hydraulic Engineering (Sept. 2012). The independent variable in the regression analysis was x = the diameter (in millimeters) of the pipe. Data for a sample of 13 different pipe sizes are provided in the table, followed by a MINITAB simple linear regression printout
Diameter ___________ Ratio
80 ............................. 6.58
100 .......................... 6.97
125 ......................... 7.39
150 ......................... 7.61
200 ......................... 7.78
250 ......................... 7.92
300 ......................... 8.20
350 ......................... 8.42
400 ......................... 8.60
450 ......................... 8.97
500 ......................... 9.31
600 ......................... 9.47
700 ......................... 9.72
a. Find the least squares line relating ratio of repair to replacement cost (y) to pipe diameter (x) on the printout.
b. Locate the value of SSE on the printout. Is there another line with an average error of 0 that has a smaller SSE than the line, part a? Explain.
c. Interpret practically the values β0 and β1.

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Regression Analysis: RATO versus DIAMETER The regression equation is RATO = 6.68 + 0.00479 DIAMETER SE Coef Predictor Coef Constant DIAMETER 6.6782 0.1208 55.29 0.000 0.0047856 0.0003218 14.87 0.000 S - 0.221034 R-Sq - 95.3% R-Sq (adj) - 94.88 Analysis of Variance Source DE MS 1 10.007 10.907 221.20 0.000 Regression Residual Error 11 Total 0.537 0.049 12 11.344