CE Systems HW5

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 1 of 9 Problem 1: (10) Given: The following data, for a certain type of civil engineering system, shows the variation of the project size (using cost as a size surrogate, in millions of dollars) to the duration of the contracts. Project size, $millions 1.9 4.2 5.8 7.8 10.1 12.0 13.8 15.6 17.7 19.5 Contract duration, months 0.6 1 1.4 2.9 4.9 7.5 8.9 9.1 8.9 9.0 Find: Three functional forms that provide a close fit to the data. Method: Using Excel Graphing functions and Line of Best Fit; Linear, Second Order Polynomial, and Power Expressions Solution:

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 2 of 9 Problem 2: (10) Given: The data below X 0.25 0.1 0.51 0.15 0.06 0.83 0.27 0.82 0.21 0.25 0.71 0.05 0.62 0.32 0.89 0.55 0.95 0.44 Y 16 28 10 24 32 12 17 10 22 12 11 36 10 10 14 12 17 12 Find: The best functional form that closely fits the data. Method: Excel Graphing Function and line of best fit- Third-order polynomial function

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 3 of 9 Bonus: Determine the RMSE of the estimated logarithmic model and the following newly observed data. (10) X 0.67 0.30 0.99 0.66 0.19 0.02 0.69 0.11 0.61 0.87 Y 12 18 9 11 22 39 12 26 12 9 Problem 3: (15) Given: D ata that is shown in the table below. Find: Plot the data to ascertain if the data are inherently linear. Develop an appropriate intrinsically linear model using a transformation of the X variable. Method: Using graphing functions on Excel, and the transforms 1/x, ln(x), and x 2 Solution: y = -0.0266x + 2.0126 R² = 0.7073 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 0 2 4 6 8 10 12 Correlation Graph X 1.7 10.0 9.1 7.2 6.4 4.2 8.1 1.5 3.1 2.6 5.1 1.8 Y 2.012 1.797 1.813 1.820 1.802 1.827 1.797 2.050 1.869 1.952 1.802 1.994

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 4 of 9 y = 0.4609x + 1.7404 R² = 0.9615 1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Correlation Graph - 1/x transform y = -0.1309x + 2.0645 R² = 0.8789 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 0 0.5 1 1.5 2 2.5 Correlation Graph - ln(x) transform

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 5 of 9 As seen in the graphs above, the best fit for a linear regression model for this data comes from a 1/x transform. It can be determined from the R 2 value that is associated with the linear best line of fit, as this represents the proportion of variance among the data from the regression model. Both the 1/x and ln(x) transforms were more accurate than the original data, but the x 2 model had a lower R 2 value than the original data, meaning it is less intrinsically linear than the original model. Problem 4: (45) Given: The durability of any civil engineering system is typically determined by factors such as the system environment, constituent material(s), rate of system use, level of maintenance, etc. A water supply system is a typical example. It is hypothesized that the duration of any water pipe system depends on pipe attributes such as the pipe material (MATRL), the acidity of the soil (ACIDTY) in which the pipe is buried, the average daily flow in 100s of gallons per hour, (AVFLOW), the average gradient of the pipes (GRADE), and the average annual amount of pipe maintenance done in $100s per mile (MNTCE). Data on the above characteristics were collected from a random sample of water supply pipes in a certain state (see data below). Find : Develop a Multiple variable linear regression model and answer the following questions. Solution: (a) Using any of the above factors (independent variables), provide the general form of a multiple linear regression model and two hypothesis statements that could be used to test the statistical significance of each factor in affecting the pipe durability. (5) Y i =  0 +  1 X 1 +  2 X 2 + …… .  n X n +  i ----------- General Form of a Multiple Linear Regression Model Two hypothesis statements that could be used to state the statistical significance of each factor in affecting the pipe durability are, y = -0.0021x + 1.9484 R² = 0.533 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05 2.1 0 20 40 60 80 100 120 Correlation Graph - x 2 transform

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 6 of 9 H 1 : B 1 = 0; the variable X 1 is not significant within the specified level of confidence; or H 2 : B 1  0; the variable X 1 is significant within the specified level of confidence. Each hypothesis can be written depending on the factor that is affecting pipe durability, X 1 – X n . (b) Using the hypothesis, on what basis can we say that any given variable is or is not significant (i.e., it has or does not have any significant influence on the durability of the water pipe)? (5) Based on the results from the multiple linear regression, rejection of the null hypothesis would imply that the intercept of the upper 90% confidence limit (  62) is not statistically equal to zero. Failure to reject the null hypothesis would imply that the independent variable is not significant and is therefore not optimal for describing the dependent variable. The rejection region for this sample is defined by the confidence level being 90%, resulting in a 0.05 level of significance. From this level of significance, the rejection region of the sample can be determined using t-statistic tables. The critical value for this data set was determined to be ±1.69. (c) Using the supplied dataset, define multiple linear regression model coefficients and t-statistics using Excel or MINTAB. Fill in Columns 2 and 3 in the table below. Attach a copy of your results below. (5) Regression Statistics Multiple R 0.521010827 R Square 0.271452282 Adjusted R Square 0.161066264 Standard Error 18.76294063 Observations 39

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 7 of 9 (d) From your output, determine which variables are significant at 90% confidence. (Hint: This implicitly involves hypothesis testing where we are simply comparing the absolute value of the calculated t statistic with the critical t statistic). Conduct a correlation analysis of the independent variables, comment on the relative strength of the relationship of each variable to pipe durability, and attach results. To answer, fill in Column 4 of the table below. (5) From the correlation matrix that was created, it can be determined that there is some correlation between the independent variables, suggesting the effects of interaction. Grade and Maintenance Cost have the highest positive correlation, while grade and average flow have the highest negative correlation. A high correlation signifies a direct relationship between the two variables, while a negative correlation signifies an inverse relationship. The correlations were determined in a stepwise fashion. (i.) What would you suggest to improve the overall significance of the model? (5) Two ways to improve the overall significance of the model would be to (1) increase the size of the sample population or (2) use more accurate methods of measurement. By increasing the sample population size, there will be less deviation in the data overall, as a wider pool of numbers will be added to remove outliers. Increasing the accuracy of measurements also reduces discrepancies in the data due to errors, and as a result, an improvement in the overall significance of the model. (e) Assess the intuitiveness of the result for each variable. Does each sign make sense to you? Explain. To answer, fill in Columns 5 and 6 of the table below. (5) (f) From your output, indicate the goodness of fit of the model: Explain. (5) The overall fit of the model is good because the significance of F is 0.053, which is less than the required level of significance of 0.10. The hypothesis of no relationship between any of the independent variables to the durability of the pipe is rejected. However, there is a poor linear model fit with the given data set, because the R square value is less than 0.50 at just 0.27. (g) A water distribution system is planned for a new suburb on the outskirts of Lowell. It is intended to use concrete pipes sloping along a gentle terrain with an average gradient of 1.25%, and soil tests indicate that the soil in that area is acidic. From population projections, the expected average hourly flow rate is 200 gallons, and the pipes are expected to receive an annual maintenance of $150 per mile. (i) Using the model you have developed, predict how long the proposed pipe system is expected to last. (5)

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 8 of 9 Durability (years) = (34.61) – [(2) * (2.017)] + [(1.50) * (9.757)] – [(321.7) *(0.0125)] – [(5.667) * (0)] – [(13.81) * (1)] ; Durability = 27.38 years (ii) Assuming everything else remains the same, what level of maintenance would be needed if the system were to last for 50 years? (5) Durability (years) , 50 = (34.61) – [(2) * (2.017)] + [(maintenance) * (9.757)] – [(321.7) *(0.0125)] – [(5.667) * (0)] – [(13.81) * (1)] ; maintenance= 3.86 ; Expected Cost of Maintenance = 3.86 x 100 = $386 per mile of pipe Answers to Questions (c), (d), and (e) must be provided in the table below. Symbol (Independent Variables) Coefficient Estimate t-statistic Is the Variable significant? Intuitiveness of the result (Yes/No) Reasons for Intuitiveness or Non-Intuitiveness Constant Term MATRL = 0 if concrete = 1 if steel 34.61 2.146 Yes No It would be expected that when moving from concrete to steel, the durability of the material would decrease as it is more susceptible to corrosion. A negative coefficient would then be expected, and a positive correlation was produced. ACIDITY = 0 if not acidic = 1 if acidic -13.81 -2.103 No Yes Between non-acidic to acidic soils, one can expect a durability decrease due to chemical corrosion. So, the negative can be considered intuitive. AVFLOW (100’s gallons per hr.) -2.017 -2.175 Yes Yes As average flow increases, it can be expected that the duration of the pipe-line system will decrease. As a result, a negative coefficient is expected, and the result is intuitive. GRADE (%) -321.7 -1.576 Yes Yes Increased grade results in greater scour, and subsequently less durability. Therefore, a negative value is expected, and the result is intuitive. MNTCE ($100’s per linear mile) 9.757 2.581 No Yes As more maintenance is completed, it is expected that the duration of the pipe-line system will increase. As a result, a positive coefficient is expected to represent the data, and the result is intuitive.

CIVE 3720 Civil Engineering Systems Name: Karen Vinal Date: February 18, 2024 HW 5: Statistical Models-Ch7 (Max score 80, 90 with bonus) Page 9 of 9 Data Set for Problem 4: PIPE ID AVFLOW MNTCE$ GRADE MATRL ACIDITY DURABILITY P 15916 11.031 2.25 0.07 0 0 14.55 P 16091 3.48 3 0.06 0 1 52.58 P 16099 3.14 4 0.06 0 1 4.68 P 16563 15.17 2.63 0.05 0 1 11.51 P 10001 10.943 2.5 0.07 0 1 9.35 P 10018 8.3 2.5 0.02 0 1 15.65 P 10019 7.882 3 0.06 0 0 13.7 P 10044 5.9 4.25 0.08 0 1 11.94 P 10045 10.45 2.86 0.05 0 1 8.36 P 10046 11.5 2.6 0.06 0 0 10.94 P 10049 3.86 3.5 0.06 0 1 38.24 P 10117 13.07 2 0 1 0 14.74 P 10119 4.9 2.5 0.07 0 1 10.38 P 10147 3.734 5.75 0.08 0 1 26.05 P 10199 6.68 3 0.06 0 1 14.63 P 10451 7.003 3 0.06 0 0 9.43 P 10455 3.279 4 0.08 0 0 17.57 P 10459 7.578 3 0.06 0 0 13.22 P 10507 4.78 3 0.08 0 1 11.09 P 10512 12.21 1.75 0.06 0 0 12.78 P 10513 0.18 1 0.06 0 0 51.67 P 10514 10.31 3.5 0.06 0 0 19.51 P 10519 10.58 2.19 0.06 0 0 11.44 P 10610 3.4 2.75 0.07 0 0 27.21 P 10660 9.147 5.2 0.06 0 0 116.52 P 10956 12.396 4.5 0.07 0 1 26.23 P 15099 3.74 3 0.06 0 0 38.73 P 15065 3.74 3 0.06 1 0 38.73 P 15160 8.05 2.6 0.05 0 1 11.17 P 15171 9.39 1.75 0.02 0 0 20.95 P 15597 0.6633 2.13 0.07 0 0 64.14 P 15600 9.22 2 0 0 1 19.73 P 15609 7.66 1.75 0.05 0 0 11.29 P 15610 14 2.5 0.02 0 1 26.68 P 15611 8.75 2.41 0.05 0 0 10.52 P 15618 5.57 2.18 0.05 0 1 12.87 P 15619 12.97 2.5 0.02 0 0 23.36 P 15911 4.5 3 0.06 0 1 21.46 P 15919 7.46 1.8 0.06 0 0 11.31

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