How to find standard deviation of a linear regression?

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 1

[Star Strider]

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]

Thanks. That works.

[Star Strider]

My pleasure!

[Sophie]

this didn't work for me. it sent the error "Unrecognized function or variable 'polyparci'." what do i do?

[Torsten]

You have to download "polyparci" before you can use it:

https://uk.mathworks.com/matlabcentral/fileexchange/39126-polyparci

Answer 2

[Shashank Prasanna]

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.

https://en.wikipedia.org/wiki/Linear_regression

[Ronny]

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

[Shashank Prasanna]

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]

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.