How to correctly do a Student Test for atmoshperic data

I've been having some troubles to understand how to do a ttest over my data.

I have an OLR netcdf with dimensions 180 x 51 x 14245, i.e., lon x lat x time. I filtered this data with a bandpass of 20 to 70 days. After that, I select an specified number of times to analyze my phenomena. That way, I select 30 events with that phenomena, with 31 times each event, having dimensions 180 x 51 x 31 x 30. Finally, I calculated the mean over the number of events to obtain only 180 x 51 x 30.

I want to see which data has significative differences compared with the original OLR file (unfiltered), to plot only the statistically significant values with pcolor.

My code was this one:

for i=1:180
    for j=1:51
        k = olr_event(i,j,:); % olr_event is the composite olr data I mentioned before
        l = olr_orig(i,j,:); % olr_orig is the original olr data with 14245 times
        prueba = ttest2(k,l);
    end
end

The problem is that the whole area rejects the null hypothesis, meaning that each grid point has statistical differences, which I dont really expect.

Am I doing it correctly?

Thank you!

Answer 1

[Star Strider]

I’m not certain what you’re doing. I would use ttest2 on the filtered and unfiltered events themselves (since that is how ttest2 works), not test one event against the mean of the other.

Testing the effect of filtering on the signals themselves —

t = (0 : 150);
x = randn(size(t))

x = 1×151

0.2578 -0.2180 0.6520 0.0698 0.5735 0.9295 0.4737 -0.2826 0.2126 0.8254 -0.7138 1.4316 -0.2453 0.1224 1.5273 -1.4869 -0.2666 0.9059 0.6516 1.6866 -1.1261 -0.7402 0.5401 -0.5479 0.0190 0.2516 0.0470 1.8034 0.3905 -0.3171

y = randn(size(t))+0.1

y = 1×151

0.4744 2.1065 -0.9683 -0.0745 0.8046 0.3834 -0.1840 -0.1694 0.5952 -0.9526 1.1424 0.7380 0.6034 -0.2772 -0.7986 0.5767 0.1454 0.0328 0.8433 -0.2866 -2.2497 0.4187 0.2863 1.3744 0.2786 0.8586 -0.9505 -1.0958 0.8295 1.4762

yf = bandpass(y, [0.30 0.31], 1, 'ImpulseResponse','iir');

[h1,p1] = ttest2(x, y)                                          % Unfiltered

h1 = 0

p1 = 0.1913

[h2,p2] = ttest2(x, yf)                                         % Filtered

h2 = 1

p2 = 0.0024

It is certainly possible that the filter removes enough of the signal energy to result in a statistically significant difference.

.

[Luis Jesús Olvera Lazcano]

Thank you! I checked and indeed the filter does most of the job to remove the signal of the main time series

As always thanks for your help

[Star Strider]

As always, my pleasure!