When I do the standard deviation for the 2 arrays by hand I get a different answer than Matlab. Ive had 2 colleagues check and we all got the same answer; is there a problem with my code?

year1_10 = [1460,1240,1230,1270,861,1355,612,822,1370,1380];

year11_20 = [810,735,259,1290,1125,528,622,468,664,717];

disp('The means of A and B are:')

mean(year1_10)

mean(year11_20)

w = 1;

disp('The standard deviation of A and B are:')

std2(year1_10)

std2(year11_20)

[Star Strider]

What results do you and your colleagues get?

[Amir Dagazau]

we get:

273.47 and 286.78

whereas matlab returns:

288.2657 and 302.3713

Answer 1

[John D'Errico]

How can we possibly know if there is a problem with your code? You fail to show how you computed the standard deviation. std2 is not part of MATLAB itself, so did you write the code?

My guess is you divided by the number of elements in the array, instead of n-1.

year1_10 = [1460,1240,1230,1270,861,1355,612,822,1370,1380];
std(year1_10) % the default
ans =
          288.265695342181
std(year1_10,1)
ans =
          273.472850572045

The default method is the one that is usually appropriate. My guess is you used the alternative, as that is the common mistake made by people when they think there is an error.

[Amir Dagazau]

this is the code i used:

year1_10 = [1460,1240,1230,1270,861,1355,612,822,1370,1380];

year11_20 = [810,735,259,1290,1125,528,622,468,664,717];

disp('The means of A and B are:')

mean(year1_10)

mean(year11_20)

disp('The standard deviation of A and B are:')

std(year1_10)

std(year11_20)

I added std2 because that was a suggestion i found online, when i removed the 2 i still got the same answer mentioned above; not the one that you managed to get

[John D'Errico]

As I said, you are essentially using the wrong formula. There is a difference.

If you have measured data, and you want to estimate the standard deviation of the set of numbers you have measured, then the default for std is corrrect. It produces an unbiased estimator of the standard deviation. Odds are that does not make a lot of sense to you, but to appreciate the difference, a beginning class in statistics is probably necessary.

The numbers that you computed are the result of dividing by n, not n-1. You say that you got 273.47.

std(year1_10,1)
ans =
          273.472850572045

which is exactly the result we get from std, IF we use the non-default option.

Here are the two formulas, written out as MATLAB code:

X = year1_10;
sqrt(sum((X - mean(X)).^2/numel(X)))
ans =
          273.472850572045

sqrt(sum((X - mean(X)).^2/(numel(X)-1)))
ans =
          288.265695342181

You can see the difference. When I divide by numel(X) in the first case, I get what you did. However, that is the wrong way to compute a standard deviation, UNLESS you are computing the standaard deviation of a complete population of numbers. So if the set of numbers we have here are not measured, but represent all the values in that population we will ever see? Only then should you use the formula that you used. Otherwise, you will produce a biased estimate of the standard deviation. Again, take a class in statistics. Or do some reading here:

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

[Steven Lord]

As John indicated, I think the difference in normalization is responsible for the difference in answers. From the description of the w input argument on the documentation page for the std function:

"S = std(A,w) specifies a weighting scheme for any of the previous syntaxes. When w = 0 (default), S is normalized by N-1. When w = 1, S is normalized by the number of observations, N."

Which normalization factor are you using in your hand calculations? If you're using N, specify 1 as the second input to std. If you're using N-1, either specify just one input argument or specify 0 as the second input to std. Based on John's results, you're using N.

[Amir Dagazau]

yes thank you both, I've amended my calculation. thanks again for the assistance.

[Image Analyst]

Please Accept this answer to give John reputation points for helping you, and to let others know the problem is already solved. Thanks in advance.