Asked by Ronny
on 20 Jul 2014

Hi everybody

I have an actually pretty simple problem which is driving me crazy right now. There are two sets of data: one for O2 and one for Heat. I made a linear regression in the plot of those two data sets which gives me an equation of the form O2 = a*Heat +b. So now I need to find the confidance interval of a. That for I need to find the standard deviation of a which I somehow just can't find out how to get it. Of course it would also work for me if there is a function that returns the confidance interval directly.

Cheers Ronny

Answer by Star Strider
on 20 Jul 2014

Edited by Star Strider
on 21 Jul 2014

Accepted Answer

With absolutely no humility at all I direct you to ** polyparci**. It calculates the confidence intervals for you for both parameters:

[p,S] = polyfit(Heat, O2, 1); CI = polyparci(p,S);

If you have two vectors, `Heat` and `O2`, and a linear fit is appropriate to your data, this code should work.

Ronny
on 21 Jul 2014

Thanks. That works.

Star Strider
on 21 Jul 2014

My pleasure!

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Answer by Shashank Prasanna
on 21 Jul 2014

Ronny, it is fairly easy to calculate in few lines of code, however it is easier to use functions such as fitlm to perform linear regression. fitlm gives you standard errors, tstats and goodness of fit statistics right out of the box:

http://www.mathworks.com/help/stats/fitlm.html

If you want to code it up yourself, its 5 or so lines of code, but I'll let you give it a shot first.

Ronny
on 21 Jul 2014

I already tried fitlm. Somehow it always gives me no intercept and a strange slope.

Shashank Prasanna
on 21 Jul 2014

What do you mean by **no intercept** and **strange slope** ?

You should get the same results no matter what approach you use to solve your least squares problem.

If you got something different, I'd be interested in knowing more about it.

fitlm by default computes the intercept for you, here is an example:

x = (1:10)' y = 10+5*x + randn(10,1) fitlm(x,y)

ans =

Linear regression model: y ~ 1 + x1

Estimated Coefficients: Estimate SE tStat pValue ________ _______ ______ __________

(Intercept) 10.164 0.85652 11.866 2.3351e-06 x1 5.0984 0.13804 36.934 3.1669e-10

Number of observations: 10, Error degrees of freedom: 8 Root Mean Squared Error: 1.25 R-squared: 0.994, Adjusted R-Squared 0.993 F-statistic vs. constant model: 1.36e+03, p-value = 3.17e-10

star
on 28 Jun 2016

these two methods (polyparci and fitlm) find the same trend values. But, the results of the confidence intervals are different in these two methods. Polyparci seems to be more optimistic. For a given set of data, polyparci results in confidence interval with 95% (3 sigma) between CI =

4.8911 7.1256 5.5913 11.4702

So, this means we have a trend value between 4.8911 and 5.5913 in 95% confidence interval. What we found from this result is that 1 sigma is 0.1167.

However, for the same data set fitlm results in SE

Estimate SE tStat pValue ________ _______ ______ __________

(Intercept) 9.2979 1.1682 7.9592 4.5304e-05 x1 5.2412 0.18827 27.838 2.9924e-09

this indicates 0.18827 for one sigma. So, the trend values are same. But, the sigma values of estimated trends are different.

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