pvalue for rate of change

I have a matrix (aa) 5000 x 12 matrix containing temperature for 5000 (1x1) grids for 12 years. I estimated the rate of change in temperature/year over each grid by estimating the slope in temperature through 12 years using the polyfit function. How to get a p-value for the rate of change of temperature in each grid which will give an idea if the rate of change estimated is significant? Attached is the code for estimating the rate of change of temperature for each grid.

y=1:12; 

kq= zeros(5000,2);
X = mat2cell(aa,1*ones(5000,1),12);
for i=1:5000;
u(i,:)= aa(i,:);
kq(i,:)= polyfit(y,u(i,:),1);
end

Answer 1

[Star Strider]

The File Exchange contribution polyparci (link) will give you the confidence intervals on the parameters.

[Sourangsu Chowdhury]

Thank you for taking efforts to build the function @Star Strider

This returns an error Struct contents reference from a non-struct array object.

Error in polyparci (line 36) COVB = (PolyS.R'*PolyS.R)\eye(size(PolyS.R)) * PolyS.normr^2/PolyS.df;

y=1:12; 

kq= zeros(5000,2);
X = mat2cell(aa,1*ones(5000,1),12);
for i=1:5000;
u(i,:)= aa(i,:);
kq(i,:)= polyfit(y,u(i,:),1);
cx(i)=polyparci(kq(i,1),kq(i,2));
end

[Star Strider]

My pleasure.

The polyparci function requires that you return the second ‘S’ output of polyfit to pass to it as well as the parameter vector.

Here, that would be:

      [kq{i},S{i}] = polyfit(y,u(i,:),1);
      cx{i} = polyparci(kq{i},S{i});

I am not certain what your loop and variables do, so I am guessing. I would also save the coefficients and the ‘S’ output as cell arrays, and then pass them as such to polyparci. Note that I also saved the polyparci output as a cell array.

I have never used cell arrays with polyparci. Nevertheless, this should work.

[Sourangsu Chowdhury]

Thank you for the reply, however, an error appears * Cell contents assignment to a non-cell array object.*

for i=1:5000;

    u(i,:)= aa(i,:);
          [kq{i},S{i}] = polyfit(y,u(i,:),1);
           cx{i} = polyparci(kq{i},S{i});
    end

[Star Strider]

I cannot reproduce the result you got. However, this works for me when I test it with random data:

for i=1:5000
    u(i,:)= aa(i,:);
    [kq(i,:),S] = polyfit(y,u(i,:),12);
    cx{i} = polyparci(kq(i,:),S);
end

Also, a degree 12 polynomial is not likely to correctly model your data. I would choose as small an order as possible, preferably not greater than about 3.

[Star Strider]

No worries.

The code I just posted should solve that problem. Note that I save ‘cx’ as a cell array.

[Sourangsu Chowdhury]

thanks this works!

[Star Strider]

As always, my pleasure!

You have also given me the idea of calculating and optionally returning the p-values in polyparci. I will consider updating it.

[Sourangsu Chowdhury]

Thank you so much, you may consider https://www.bmj.com/content/343/bmj.d2304 for updating the function for providing the p values as a function of the CIs returned as of now

[Star Strider]

The polyparci function calculates the t-statistic internally from the stated (or default) ‘alpha’ value. I will have to consider calculating the p-values given the paramater variances.

Thank you for the BMJ reference. I am a physician and Diplomate of the American Board of Internal Medicine, so I am familiar with it, although I do not subscribe to it.