Error bar with CI 95 on bar graph
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Hi,
Can anyone tell how to apply CI 95% error bars on grouped bar graph.
Thanks
6 件のコメント
回答 (1 件)
Adam Danz
2019 年 11 月 15 日
編集済み: Adam Danz
2020 年 12 月 9 日
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))
ylim([0, .05])
grid on
4 件のコメント
Adam Danz
2019 年 11 月 15 日
編集済み: Adam Danz
2019 年 11 月 15 日
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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