How to calculate mean square error

2 次查看（过去 30 天）
Adriana
Adriana 2015-3-13
编辑： Guillaume 2015-3-13
I am supposed to find the mean square error of a least square fit that I did. I cannot find where the bug in my code is. My error is huge- should be ~ 1e-4
%Hw 7
clear all
close all
y3=[1960;
1965;
1970;
1975;
1980;
1985;
1990;
1995];
y=[0;5;10;15;20;25;30;35];
w=[26.4;
31.9;
38.6;
46.7;
56.5;
68.3;
82.7;
100.0];
p=polyfit(y,w,1);
%Compute in 5 years and 10 years
k=2000;
future=((k-1995)+35);
Eq=2.0669*future+20.2167;
%After 40 years
Eq1=2.0669*40+20.2167;
%After 45 years
Eq2=2.0669*45+20.2167;
save w_2000_deg1.dat Eq1 -ascii
save w_2005_deg1.dat Eq2 -ascii
%Find error
y1=[0;5;10;15;20;25;30;35;40;45];
w1=[26.4;31.9;38.6;46.7;56.5;68.3;82.7;100.0;102.8927;113.2272];
n=10;
c=1;
e=0;
for i=1:n
e(i)=((2.066*y1(i,c)+20.2167)-w1(i,c))^2;
end
r=sum(e);
e_ans=(1/10)*r;
save e_deg1.dat e_ans -ascii
%Part b
s=log(w);
y4=y3-1959;
pp=polyfit(y4,s,1);
%model=0.0381*t+3.2347;
%c=e^alpha a=beta
g1=exp(3.2728)*exp(0.0381*40);
g2=exp(3.2728)*exp(0.0381*45);
save w_2000_exp.dat g1 -ascii
save w_2005_exp.dat g2 -ascii
n=10;
c=1;
h=0;
for i=1:n
h(i)=((exp(3.2728)*exp(0.0381*y1(i,c)))-w1(i,c))^2;
end
r=sum(h);
h_ans=(1/10)*r;
save e_exp.dat h_ans -ascii
Any input is appreciated!!

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Interpolation 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by