Linear Regression and Spring Testing Analysis - CE 344 Assignment

Page 1 of 7 CIVIL ENGINEERING ANALYSIS I- CE 344 Assignment No. 6 Due: November 7, 2023 – 12:15PM 50 points max I had to solve these on excel and move my numbers in over to the word doc 1. (12 points) Given the following data pairs (x, y): (1, 1.24), (2, 5.23), (3, 7.24), (4, 7.60), (5, 9.97), (6, 14.31), (7, 13.99), (8, 14.88), (9, 18.04), (10, 20.70) (a) Determine the linear regression equation (4 points) (b) Calculate the correlation coefficient (4 points) (c) Is the coefficient of determination strong? (2 points) (d) Obtain a prediction for x = 4.5 (2 points) The computation of parameters: x y 1 1.24 2 5.23 3 7.24 4 7.6 5 9.97 6 14.31 7 13.99 8 14.88 9 18.04 10 20.7 mean 5.50 11.320 variance 9.167 36.987 covariance 18.140 b^ 1.979 a^ 0.436 correlation 0.985 Explanation: 1

Page 2 of 7 Used excel to find the mean , average(array), var.s(array) to find sample variance, covariance.s(array1,array2) to find the sample covariance, and correl(array1,array2) for the correlation between x and y The estimated regression model is y^=0.436+1.979×x The correlation coefficient is 0.985 The correlation coefficient is close to 1 and it is positive, so, The linear relationship between the X and Y is high and positively correlated The coefficient of determination R2 = 0.9852 = 0.970 for x=4.5, the value of y is; y= 0.436+1.979 x 4.5 = 9.342 The predicted value of y for x=4.5 is 9.342 a) The estimated regression model is y = 0.436 + 1.979 x x b)The correlation coefficient is r=0.985 c) Yes, The coefficient of determination is strong and it is 0.970 d) The predicted value of y for x = 4.5 is 9.342 2. (28 points; 4 points for each question ) A new spring is being tested. Twenty-five weights, x , are hung, and the spring length, y , is measured for each. The weights 2

Page 3 of 7 are measured in lbs, and the length is measured in inches. The following summary statistics are recorded: _ _ 25 _ x = 2.40 y = 12.18  ( x i – x ) 2 = 52.00 i =1 25 _ 25 _ _  ( y i – y ) 2 = 498.96  ( x i – x ) ( y i – y ) = 160.27 i =1 i =1 Let  o represent the length of the spring at rest, and let  1 represent the increase in length caused by a load of 1.0 lb.   (a) Compute the least-squares estimates  o and  1 . (b) Compute the error variance estimate s 2 . The error variance estimate s2 is 21.69 (c) Determine the 95% confidence intervals for  o and  1 . The 95% confidence interval for ^0 � is approximately (2.9745, 6.5913), and the 95% confidence interval for ^1 � is approximately (1.7457, 4.4185). 3

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Page 4 of 7 (d) The manufacturer of the spring claims that the spring constant is no more than 3.0 in/lb. Do the data provide sufficient evidence for you to conclude that the claim is false? The data do not provide sufficient evidence to conclude that the manufacturer's claim is false. (e) It is also claimed that the unloaded length of the spring is at least 5.5 inches. Do the data provide sufficient evidence for you to conclude that the claim is false? The data do not provide sufficient evidence to conclude that the claim of the unloaded length being at least 5.5 inches is false. 4

Page 5 of 7 (f) Determine a 99% confidence interval for the length of the spring under a load of 1.5 lbs. The 99% confidence interval for the length of the spring under a load of 1.5 lbs is approximately (6.3238 inches, 12.488 inches). (g) Determine a 99% prediction interval for the measured length of the spring under a load of 1.5 lbs. The 99% prediction interval for the measured length of the spring under a load of 1.5 lbs is approximately (-4.0252 inches, 22.837 5

Page 6 of 7 inches). This interval accounts for the uncertainty in predicting an individual measurement. 3. (10 points) A transportation engineer studied the association between Hours of Delay and Miles Traveled on a principal arterial street. The linear regression statistics are shown below. 6

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Page 7 of 7 (a) Measurements show that 1,531 miles were traveled on the principal arterial street. Can the engineer use the information provided to estimate the delay (4 points) ? Delay = -664.6465 + 1.5443783 * 1,531= 1697.1579 (b) Based on the findings of the regression analysis, how would you characterize the fit of linear line to the data (excellent, good, weak, poor) and why (3 points). Based on the findings of the regression analysis, the fit of the linear line to the data can be characterized as good. This is because the R-squared value (RSquare) is 0.759, which indicates that approximately 75.9% of the variability in the delay can be explained by the miles traveled on the principal arterial street. (c) Explain in your own words what is the purpose of calculating the t ratio and P-value (Prob >│t│) for the intercept, β 0 , and the miles traveled variable, β 1 (3 points) . For the intercept (Bo), the t-ratio is calculated by dividing the estimated intercept value by its standard error. A low t-ratio and a low p-value indicate that the intercept is significantly different from zero, suggesting that it is a statistically significant component of the model. For the miles traveled variable (B), the t-ratio is calculated in a similar way. A high t-ratio and a low p-value indicate that the miles traveled variable is statistically significant and has a significant effect on the dependent variable (delay). the t-ratio and p-value help determine whether the coefficients in the regression model are statistically significant or not. If the p-value is less than a chosen significance level (commonly 0.05), it suggests that the coefficient is significant. 7

CE 344 Homework Assignment No. 6-2023

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