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d. Use the regression line to predict the ratio of repair to replacement cost of pipe with a diameter of 800 millimeters. e. Comment on the reliability of the prediction, part d.

d. Use the regression line to predict the ratio of repair to replacement cost of pipe with a diameter of 800 millimeters. e. Comment on the reliability of the prediction, part d.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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d and e

## 11.21 Repair and Replacement Costs of Water Pipes **Context:** Refer to the IHS Journal of Hydraulic Engineering (September 2012) study of water pipes. Recall that a team of civil engineers used regression analysis to estimate \( y = \) the ratio of repair to replacement cost of commercial pipe. 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 below, followed by a Minitab simple linear regression printout. **Data Table:** | Diameter (mm) | Ratio | |---------------|-------| | 80 | 6.58 | | 100 | 6.97 | | 125 | 7.39 | | 150 | 7.61 | | 200 | 7.92 | | 250 | 7.92 | | 300 | 8.20 | | 350 | 8.42 | | 400 | 8.60 | | 450 | 8.97 | | 500 | 9.31 | | 600 | 9.47 | | 700 | 9.72 | **Source:** Based on ISH Journal of Hydraulic Engineering, Volume 18, Issue 3, pp. 241–251. Copyright: September 2012. **Tasks:** a. **Linear Regression Line**: Use the printout to find the least squares line relating the ratio of repair to replacement cost (\( y \)) to pipe diameter (\( x \)). b. **Sum of Squares due to Error (SSE)**: Locate the SSE on the printout. Consider if there is another line with an average error of 0 that has a smaller SSE than the line in part a. Explain your findings. c. **Interpret Coefficients**: Provide a practical interpretation of \(\beta_0\) and \(\beta_1\) values. d. **Prediction**: Use the regression line to predict the ratio for a pipe with a diameter of 800 millimeters. e. **Reliability of Prediction**: Comment on the reliability of the prediction made in part d. **Regression Analysis Summary:** - **Analysis of Variance:
Transcribed Image Text:## 11.21 Repair and Replacement Costs of Water Pipes **Context:** Refer to the IHS Journal of Hydraulic Engineering (September 2012) study of water pipes. Recall that a team of civil engineers used regression analysis to estimate \( y = \) the ratio of repair to replacement cost of commercial pipe. 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 below, followed by a Minitab simple linear regression printout. **Data Table:** | Diameter (mm) | Ratio | |---------------|-------| | 80 | 6.58 | | 100 | 6.97 | | 125 | 7.39 | | 150 | 7.61 | | 200 | 7.92 | | 250 | 7.92 | | 300 | 8.20 | | 350 | 8.42 | | 400 | 8.60 | | 450 | 8.97 | | 500 | 9.31 | | 600 | 9.47 | | 700 | 9.72 | **Source:** Based on ISH Journal of Hydraulic Engineering, Volume 18, Issue 3, pp. 241–251. Copyright: September 2012. **Tasks:** a. **Linear Regression Line**: Use the printout to find the least squares line relating the ratio of repair to replacement cost (\( y \)) to pipe diameter (\( x \)). b. **Sum of Squares due to Error (SSE)**: Locate the SSE on the printout. Consider if there is another line with an average error of 0 that has a smaller SSE than the line in part a. Explain your findings. c. **Interpret Coefficients**: Provide a practical interpretation of \(\beta_0\) and \(\beta_1\) values. d. **Prediction**: Use the regression line to predict the ratio for a pipe with a diameter of 800 millimeters. e. **Reliability of Prediction**: Comment on the reliability of the prediction made in part d. **Regression Analysis Summary:** - **Analysis of Variance:
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