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Anum Ali

Error bar with CI 95 on bar graph

Anum Ali
さんによって質問されました 2019 年 11 月 15 日 4:51
最新アクティビティ Adam Danz
さんによって 編集されました 2019 年 11 月 15 日 13:31
Can anyone tell how to apply CI 95% error bars on grouped bar graph.

  6 件のコメント

Anum Ali
2019 年 11 月 15 日 5:31
For now I am testing on plot that can be replaced with bar() command.
Adam Danz
2019 年 11 月 15 日 5:39
Your method of computing CIs (using tinv) requires that your data form a normal distribution. You're also only computing 1-tail of the CI, is that intentional?
Anum Ali
2019 年 11 月 15 日 5:52
I require something like thisconfidence.png

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1 件の回答

Adam Danz
回答者: Adam Danz
2019 年 11 月 15 日 6:13
編集済み: Adam Danz
2019 年 11 月 15 日 6:14

Here's an anonymous function that computes the 95% CI based on the tinv method which requires that your data approximately form a normal distirbution. See this link for more information on this function.
% x is a vector, matrix, or any numeric array of data. NaNs are ignored.
% p is a the confident level (ie, 95 for 95% CI)
% The output is 1x2 vector showing the [lower,upper] interval values.
CIFcn = @(x,p)std(x(:),'omitnan')/sqrt(sum(~isnan(x(:)))) * tinv(abs([0,1]-(1-p/100)/2),sum(~isnan(x(:)))-1) + mean(x(:),'omitnan');
% Demo
% x = randn(100,1) + 5;
% p = 95;
% CI = CIFcn(x,p)
Here's a demo using your code
EE = [0.0363 0.0312 0.0274 0.0244 0.0220 0.0200 0.0183 0.0168 0.0155 0.0143];
CIFcn = @(x,p)std(x(:),'omitnan')/sqrt(sum(~isnan(x(:)))) * tinv(abs([0,1]-(1-p/100)/2),sum(~isnan(x(:)))-1) + mean(x(:),'omitnan');
CI = CIFcn(EE,96);
% Compute the distance of the upper and lower bounds
CIdist = abs(CI-mean(EE));
% plot
plot(1, mean(EE), 'bo')
hold on
errorbar(1, mean(EE), CIdist(1), CIdist(2))

  4 件のコメント

Adam Danz
2019 年 11 月 15 日 6:30
In my answer, EE is a vector of values. The center point is the mean of EE which you can clearly see in my code.
Confidence intervals generally show the range of possible mean values from a distribution. In my answer, I'm only dealing with 1 vector, 1 distribution, 1 mean value, 1 confidence interval.
Now you can apply that to your data but first, I recommend taking a few minutes to understand what's going on in my answer.
In your data, EE is probably the means from several distribtions. You'll want to provide the raw data in the CIFcn function in order to compute the CI.
Anum Ali
2019 年 11 月 15 日 6:50
Ya I get the concept of your solution but now I edited to all data still it gave the error
Error using errorbar (line 70)
X, Y, and error bars all must be the same length.
Error in ra30 (line 206)
errorbar(p_device, EE,CI)
because CI only had two values ? why CI has two values?
Adam Danz
2019 年 11 月 15 日 7:09
I couldn't possibly answer that without knowing what inputs you're providing.
I have no idea what your data look like. Are you provding the CIFcn() function a matrix? a vector? If you're providing a matrix and you'd like to compute the CIs for each column, you'll need to provide each column as input individually or rewrite the function.

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