Homework Assignment 8: EGM 3344 Problems and Solutions

HOMEWORK ASSIGNMENT 8 EGM 3344 Problems from Chapra book chapters 14, 15 Questions answered: 14.1, 14.5, 14.8, 14.11, 14.18, 15.5, 15.6, 15.5 14.1 Given the data 0.90 1.42 1.30 1.55 1.63 1.32 1.35 1.47 1.95 1.66 1.96 1.47 1.92 1.35 1.05 1.85 1.74 1.65 1.78 1.71 2.29 1.82 2.06 2.14 1.27 (a) the mean μ μ = ∑ 𝑎?? ?𝑎???? ÷ number of values = 40.61 25 = 1 .6244 (b) median 𝑀??𝑖𝑎? = ?𝑖???? ?𝑎??? = 1.65 ( c) mode 𝑀??? = ???? ?????? ?𝑎??? = 1 .35 𝑎?? 1.47 The data is multimodal. (d) range ?𝑎??? = ℎ𝑖?ℎ??? − ?????? ?𝑎??? = 2.29 − 0.9 = 1 .39 (e) standard deviation 𝜎 𝜎 = √?𝑎?𝑖𝑎??? ?𝑎?𝑖𝑎??? = 1 ? ∑(? 𝑖 − ??𝑎?) = 0.115184 𝑛 𝑖=1 𝜎 = 0 .339388 (f) variance ?𝑎?𝑖𝑎??? = 1 ? ∑ (? 𝑖 − ??𝑎?) = 0 .115184 𝑛 𝑖=1

2 (g) coefficient of variation. ?????𝑖?𝑖??? ?? ?𝑎?𝑖𝑎?𝑖?? = 𝜎 𝜇 × 100% = 0.339388 1.6244 × 100% = 20 .89

3 14.5 Use least-squares regression to fit a straight line Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. Then repeat the problem, but regress x versus y — that is, switch the variables. Interpret your results. SEE MATLAB CODE question145 % Given data x = [0 2 4 6 9 11 12 15 17 19]; y = [5 6 7 6 9 8 8 10 12 12]; % Perform y vs. x regression and calculate statistics [slope_y_vs_x, intercept_y_vs_x, SEE_y_vs_x, r_y_vs_x] = performRegression(x, y); % Perform x vs. y [slope_x_vs_y, intercept_x_vs_y, SEE_x_vs_y, r_x_vs_y] = performRegression(y, x); % results for y vs. x disp( 'y vs. x:' ); fprintf( 'y = %.5fx + %.5f\n' , slope_y_vs_x, intercept_y_vs_x); fprintf( 'SEE: %.4f\n' , SEE_y_vs_x); fprintf( 'r: %.4f\n' , r_y_vs_x); % results for x vs. y disp( 'x vs. y:' ); fprintf( 'x = %.5fy + %.5f\n' , slope_x_vs_y, intercept_x_vs_y); fprintf( 'SEE: %.4f\n' , SEE_x_vs_y); fprintf( 'r: %.4f\n' , r_x_vs_y); % Function to perform regression and calculate statistics function [slope, intercept, SEE, r] = performRegression(x, y) coeff = polyfit(x, y, 1); slope = coeff(1); intercept = coeff(2); y_pred = slope * x + intercept; residuals = y - y_pred; SSR = sum(residuals.^2); SEE = sqrt(SSR / (length(x) - 2)); r = corrcoef(x, y); r = r(1, 2); end >> question145 y vs. x: y = 0.35915x + 4.88812 SEE: 0.8511 r: 0.9449 x vs. y:

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4 x = 2.48614y + -11.13494 SEE: 2.2393 r: 0.9449

5 14.8 Beyond the examples in Fig. 14.13, there are other models that can be linearized using transformations. For example, 4 4 x x y e   = Linearize this model and use it to estimate a4 and b4 based on the following data. Develop a plot of your fit along with the data. Code: % Given Data x = [0.1 0.2 0.4 0.6 0.9 1.3 1.5 1.7 1.8]; y = [0.75 1.25 1.45 1.25 0.85 0.55 0.35 0.28 0.18]; % Linearized Variables X and Y X = x; Y = log(y./x); % ./ is elementwise division operator p = polyfit(X,Y,1); A = p(2); B = p(1); alpha4 = exp(A); beta4 = B; % print results fprintf( 'Coefficients of fit :- \n\t alpha4 = %.4f \n\t beta4 = %.4f\n' ,alpha4,beta4) % Ploting x_plot = linspace(0.1,1.8); y_plot = alpha4*x_plot.*exp(beta4*x_plot); plot(x,y, '.' ,x_plot,y_plot, '-' ) legend( 'Data Points' , 'Fitted Curve' , 'location' , 'Best' ) ylabel( 'y' ); Output: Coefficients of fit : alpha4 = 9.6618 beta4 = -2.4733 Graph:

6

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7 14.11 Determine an equation to predict metabolism rate as a function of mass based on the following data. Use it to predict the metabolism rate of a 200-kg tiger. See MATLAB code ‘ question1411 ’ Code: % data mass = [400; 70; 45; 3; 0.3; 0.16]; % Mass in kg metabolism = [270; 82; 50; 4.8; 1.45; 0.97]; % Metabolism rate in calories per day coeff = polyfit(mass, metabolism, 1); slope = coeff(1); intercept = coeff(2); % Predict tiger mass_tiger = 200; metabolism_tiger = slope * mass_tiger + intercept; % Display the linear regression equation and the predicted metabolism rate for the tiger fprintf( 'Metabolism Rate = %.2f * Mass + %.2f\n' , slope, intercept); fprintf( 'Predicted Metabolism Rate for a 200-kg tiger: %.2f calories per day\n' , metabolism_tiger); output: question1411 Metabolism Rate = 0.66 * Mass + 11.15 Predicted Metabolism Rate for a 200-kg tiger: 143.21 calories per day

8 14.18 The following data show the relationship between the viscosity of SAE 70 oil and temperature. After taking the log of data, use linear regression to find the equation of the line that best fits the data and the r 2 value. See MATLAB code ‘ question1418 ’ Code: % data temperature = [26.67, 93.33, 148.89, 315.56]; viscosity = [1.35, 0.085, 0.012, 0.00075]; % natural log of the data log_temperature = log(temperature); log_viscosity = log(viscosity); % Fit a straight line to transformed data coeff = polyfit(log_temperature, log_viscosity, 1); slope = coeff(1); intercept = coeff(2); % Calculate the predicted values of log(viscosity) log_viscosity_pred = polyval(coeff, log_temperature); % Calculate the sum of squared residuals SSR = sum((log_viscosity - log_viscosity_pred).^2); % Calculate the total sum of squares SST = sum((log_viscosity - mean(log_viscosity)).^2); % Calculate the R-squared value r_squared = 1 - SSR / SST; % Display the linear regression equation and the R-squared value fprintf( 'Linear Regression Equation (log-transformed data): log(viscosity) = %.4f * log(temperature) + %.4f\n' , slope, intercept); fprintf( 'R-squared value: %.4f\n' , r_squared); Output: Linear Regression Equation (log-transformed data): log(viscosity) = -3.0134 * log(temperature) + 10.5492 R-squared value: 0.9757

9 15.3 fit a cubic polynomial to the following data: Along with the coefficients, determine r 2 and S y/x . See MATLAB code ‘ question1530 ’ Code: % data x = [3, 4, 5, 7, 8, 9, 11, 12]; y = [1.6, 3.6, 4.4, 3.4, 2.2, 2.8, 3.8, 4.6]; % Fit a cubic polynomial coefficients = polyfit(x, y, 3); %coefficients cubic_poly = polyval(coefficients, x); % Calculate SSR and SST SSR = sum((y - cubic_poly).^2); SST = sum((y - mean(y)).^2); %R-squared r_squared = 1 - SSR / SST; disp( 'Coefficients of the cubic polynomial:' ); disp(coefficients); fprintf( 'R^2 (coefficient of determination): %.3f\n' , r_squared); output: Coefficients of the cubic polynomial: 0.0467 -1.0412 7.1438 -11.4887 R^2 (coefficient of determination): 0.829

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10 15.6 Use multiple linear regression to derive a predictive equation for dissolved oxygen concentration as a function of temperature and chloride based on the data from Table P15.5. Use the equation to estimate the concentration of dissolved oxygen for a chloride concentration of 15 g/L at T = 12 °C. Note that the true value is 9.09 mg/L. Compute the percent relative error for your prediction. Explain possible causes for the discrepancy. Code: % given data T = [0 5 10 15 20 25 30]; C = [0 10 20]; o = [14.6 12.8 11.3 10.1 9.09 8.26 7.56; 12.9 11.3 10.1 9.03 8.17 7.46 6.85; 11.4 10.3 8.96 8.08 7.35 6.73 6.20]; % Reshape data for regression T = repmat(T, 3, 1); C = kron(C, ones(1, 7)); % Perform multiple linear regression X = [ones(21, 1), T(:), C(:)]; coeff = X \ o(:); % Extract coefficients a0 = coeff(1); a1 = coeff(2); a2 = coeff(3); % Estimate the concentration of dissolved oxygen for C = 15 g/L and T = 12°C C_pred = 15; T_pred = 12; o_pred = a0 + a1 * T_pred + a2 * C_pred; % Compute the percent relative error true_value = 9.09; percent_error = abs(o_pred - true_value) / true_value * 100; % Display the results fprintf( 'Estimated Dissolved Oxygen Concentration: %.2f mg/L\n' , o_pred); fprintf( 'Percent Relative Error: %.2f%%\n' , percent_error);

11 output: Estimated Dissolved Oxygen Concentration: 9.26 mg/L Percent Relative Error: 1.83%

12 15.15 Given the data use least-squares regression to fit (a) a straight line (b) a power equation

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13 (c) a saturation-growth-rate equation (d) a parabola.

14 Plot the data along with all the curves. Is any one of the curves superior? If so, justify. The saturation-growth-rate equation is superior - Best fit to data points - highest R^2 value

HW8

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