from
Chi Square Test for Independence
by Kevin Wong
Chi Square Test for Independence: calculuate the chi square of a vector and the critical value.
|
| chi2test(data, num_int)
|
%?cs(data)
%Chi-Square Test =
% 490
%Critical Value =
% 552.074742731856
% n = 5000
% k = 500
% n/k = 10
%Since 490 < 552, We accept the null hypothesis H_0 and conclude
%that we don't have enough evidence at level alpha = 0.05 to say that
%the observation are not uniform.
function chi2test(data, num_int)
[m,n] = size(data);
E = m/num_int; %E=10
interval=zeros(num_int, 1);
for i=1:m
k = data(i,1);
interval(fi(num_int,k),1) = interval(fi(num_int,k),1) +1;
end
x = 0;
for j= 1:num_int
x=x+ (interval(j,1)-E)^2/E;
end
x
chi2inv(0.95,num_int-1)
function b = fi(i,n)
for j= 0:i
if n > (1/i)*(i-j)
b = i-j+1;
break
end
end
|
|
Contact us at files@mathworks.com