t test ( not sure if my implementation is correct)

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Alexander Guillen
Alexander Guillen 2023 年 2 月 28 日
コメント済み: Alexander Guillen 2023 年 2 月 28 日
Suppose that I have two groups, N and R. Each group contains four columns which represent different subjects. For example (hypothetically):
N1 = [1:5]' .... N4 = [1:5]' so that N = { N1,N2,N3,N4}
R1 = [1:5]'.... R4 = [1:5]' so that R = {R1...R4}
Is it possible to enable a loop which which can perform the t-test so that I can avoid this manually:
h = ttest(N1,R1);
.
.
.
h = ttest(N1,R4);
---
h = ttest(N2,R1);
.
.
.
h = ttest(N2,R4); and so on
I understand it may be wrong my implementation of the t-test. But I would like to put into a loop what I just described. Any guidance and help will be appreciated.

採用された回答

David Hill
David Hill 2023 年 2 月 28 日
N=randi(100,5,4);%your matrix;
R=randi(100,5,4);%your matrix
h=1;
for n=1:4
for r=1:4
H(h)=ttest(N(:,n),R(:,r));
h=h+1;
end
end
H
H = 1×16
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
  2 件のコメント
Alexander Guillen
Alexander Guillen 2023 年 2 月 28 日
wow thank you
Alexander Guillen
Alexander Guillen 2023 年 2 月 28 日
By any chance, do you know how I can display the columns from N and R which h = 1?

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