Question

Construct a scatterplot and identify the mathematical model that best fits the data. Assume that the model is to be used only for the scope of the given data and consider only linear, quadratic, logarithmic, exponential, and power models. Use a calculator or computer to obtain the regression equation of the model that best fits the data. You may need to fit several models and compare the values of R².

x1 2 3 3 9 17 30 40 4 5

Group of answer choices

y = 2.96 x^1.628

y = -1.59 + 22.3 ln x

y = -8.70 + 9.50 x

y = 2.07 e^0.638x

Answer 1

Matlab code for regression line clear all close all All data points x=(1 2 3 4 5); y=[3 9 17 30 40); fequation for regression line p=polyfit(x,y,1); fprintf('Equation for regression line y=($f) *x+($f)\n',p(1),p(2)) yl -2.96*x. 1.628; fprintf('\n\tFor model y=2.96*x. 1.628, error =$f\n', norm(y1-y)) y2 =-1.59+22.3*log(x); fprintf('\n\tFor model y =-1.59 + 22.3 *ln(x), error =$f\n', norm(y2- y)) y3 = -8.70 + 9.50*x; fprintf('\n\tFor model y = -8.70 + 9.50 x, error =$f\n', norm(y 3-y)) y4 = 2.07*exp(0.638*x); fprintf('\n\tFor model y = 2.07*e*0.638*x, error =8f\n', norm(y 4-y)) fprintf('\nHence the best model is y=2.96**.*1.628\n') 8888888888888888% End of Code %%%%%%%%%%%%% Equation for regression line y=19.500000)*x+(-8.700000) For model y=2.96*x. *1.628, error =1.981261 For model y =-1.59 + 22.3 *In(x), error =10.612053 For model y = -8.70 + 9.50 x, error =4.037326 For model y = 2.07*e*0.638*x, error =11.382910 Hence the best model is y=2.96*x.-1.628 Published with MATLABRO R2018a

%Matlab code for regression line
clear all
close all

%All data points
x=[1 2 3 4 5];
y=[3 9 17 30 40];

%equation for regression line

p=polyfit(x,y,1);
fprintf('Equation for regression line y=(%f)*x+(%f)\n',p(1),p(2))

y1 =2.96*x.^1.628;
fprintf('\n\tFor model y=2.96*x.^1.628, error =%f\n',norm(y1-y))

y2 =-1.59+22.3*log(x);
fprintf('\n\tFor model y =-1.59 + 22.3 *ln(x), error =%f\n',norm(y2-y))

y3 = -8.70 + 9.50*x;
fprintf('\n\tFor model y = -8.70 + 9.50 x, error =%f\n',norm(y3-y))

y4 = 2.07*exp(0.638*x);
fprintf('\n\tFor model y = 2.07*e^0.638*x, error =%f\n',norm(y4-y))

fprintf('\nHence the best model is y=2.96*x.^1.628\n')

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