Linear and quadratic trend surface formula get the same R

I want to analyze the relation of location and the population of a city. When I try linear and quadratic trend surface model, it strikes me that the two model has the same R^2:0.4590. Why does this happen .Data are in the attachment

My program is here:

% read data
fid = readtable('trend_surface.xlsx'); % 题目中的表1
POP = table2array(fid(1:50,2));        % 人口
x = table2array(fid(1:50,3));     % x坐标
y = table2array(fid(1:50,4));     % y坐标
% draw 3-D scatter
scatter3(x,y,POP,"blue",'+');
% Linear trend surface
X = [ones(50,1) x y];      % coeffecient matrix
Z = POP;                   % solution matrix
A = (X'*Z)\(X'*X);      % result
% examine the model
SS_D = sum((A(1)+A(2).*x+A(3).*y-Z).^2);
SS_R = sum((A(1)+A(2).*x+A(3).*y-mean(Z)).^2);
SS_T = SS_D+SS_R;
R_2 = SS_R/SS_T;   % R_2 = 0.4590
F = (SS_R/2)/(SS_D/47);   % F = 19.9399
% Quadratic trend surface
X_1 = [ones(50,1) x y x.*x x.*y y.*y];    % coeffecient matrix
Z_1 = POP;            % solution matrix
A_1 = (X_1'*Z)\(X_1'*X_1); % result
% examine the model
SS_D1 = sum((A_1(1)+A_1(2).*x+A_1(3).*y+A_1(4).*x.*x+A_1(5).*x.*y+A_1(6)*y.*y-Z_1).^2);
SS_R1 = sum((A_1(1)+A_1(2).*x+A_1(3).*y+A_1(4)*x.*x+A_1(5)*x.*y+A_1(6)*y.*y-mean(Z_1)).^2);
SS_T1 = SS_D1+SS_R1;
R_2_1 = SS_R1/SS_T1;   % R_2_1=0.4590
F_1 = (SS_R1/5)/(SS_D1/44);    % F_1=7.4668

[Mathieu NOE]

hello

without any change to your code , I get following results :

R_2 = 0.4486

F = 19.1212

R_2_1 = 0.5000

F_1 = 8.8000

Answer 1

[SOUMNATH PAUL]

Hi,

I also got the same results as @Mathieu NOE, without any change to your code.

R_2 = 0.4486

R_2_1 = 0.5000

But, it is fairly possible to encounter a situation where both linear and quadratic trend surface models yield the same result.

Here are some situations where the results can be same:

1) The dataset you are working with may have characteristics that make it equally well-suited for both linear and quadratic models.

2) With a limited sample size, the ability to detect differences in model performance may be constrained. A larger dataset might reveal distinctions between the linear and quadratic models.

3) The underlying relationship between location and population might genuinely be adequately described by a linear model. Adding quadratic terms could be unnecessary complexity if it does not significantly improve the model fit.

Hope it helps!

Regards,

Soumnath