Combining Standard Deviations Watch

EmuEmu
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Hello ,

I'm new to the forum, and I was wondering if any of you could help me.

I am analysing some wind speed data (recorded using a cup anemometer). The log gives me the mean wind speed over a minute and the standard deviation of the wind speed over the minute (I think the data collection rate was 1Hz).

What I really need is the mean wind speed and the standard deviation of the wind speed over the course of 10 minutes. I know I can just take the mean of the means for the wind speed. But how do I calculate the standard deviation over 10 minutes :confused: ?

Thank you!
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DavyS
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If you can assume they are normally distributed, you can use the pooling formula (the one for two variables I put below, it's not that hard to extrapolate for ten sets of data.)

s_p^2 = \frac{(n_x-1)s_x^2 + (n_y-1)s_y^2}{(n_x-1)+(n_y-1)}

where n_x is the number of readings for the first minute, and s_x^2 is the variance for the first minute, and similarly for y

You'll notice that this is just the weighted average of the two sets of data. Since you took them at regular intervals, you can assume that n_y=n_x and you get:

s_p^2=\frac{s_x^2+s_y^2}{2}

Note, however, this form of the formula only works for if n_x=n_y, otherwise you'll need to work it out using the more complicated weighted average formula above.
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ukdragon37
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I would do this :o: :

Let X_i be the mean for each minute. Using the laws of expectation and variance, E(\sum{X_i}})=\sum{E(X_i)}=\sum{  \mu_i} and V(\sum{X_i}})=\sum{V(X_i)}=\sum{  \sigma^2_i}.
Dividing by n gives E(\frac{\sum{X_i}}{n}})=\frac{\s  um{E(X_i)}}{n}= \frac{\sum{\mu_i}}{n}
and V(\frac{\sum{X_i}}{n}})= \frac{\sum{V(X_i)}}{n^2}=\frac{\  sum{\sigma^2_i}}{n^2}.
SD=\sqrt{V(\frac{\sum{X_i}}{n})}  =\frac{\sqrt{\sum{\sigma^2_i}}}{  n}.
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EmuEmu
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Thank you so much for the replies - I really appreciate your time!

However I'm wondering if that is the whole picture.

Eg. consider two 10 minute periods:

The first has all the 1 minute means the same, but all the 1 minute SDs large.
The second has all the 1 minute means varying widely, but all the 1 minute SDs small.

Then the formulas above would say the 10 minute SD would be larger for the first 10 minute period than the second. Whereas actually the 10 minute SD should be larger for the second 10 minute period than the first.

I guess I'm asking if the answer should include a term for the 1 minutes means????

Sorry - not very good at expressing myself - hope that makes sense!

And thank you again - more thoughts appreciated!
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ukdragon37
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(Original post by EmuEmu)
Thank you so much for the replies - I really appreciate your time!

However I'm wondering if that is the whole picture.

Eg. consider two 10 minute periods:

The first has all the 1 minute means the same, but all the 1 minute SDs large.
The second has all the 1 minute means varying widely, but all the 1 minute SDs small.

Then the formulas above would say the 10 minute SD would be larger for the first 10 minute period than the second. Whereas actually the 10 minute SD should be larger for the second 10 minute period than the first.

I guess I'm asking if the answer should include a term for the 1 minutes means????

Sorry - not very good at expressing myself - hope that makes sense!

And thank you again - more thoughts appreciated!
I thought you were asking for the SD of the 10 minute period based on the combined SDs of all the 1 minute periods. But if you are just working out the SD of the 1 minute means in the 10 minutes why can't you use the standard formula \sigma=\sqrt{\frac{\sum{\left(x-\bar{x}\right)^2}}{n}}?
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EmuEmu
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Because I am looking for the standard deviation of the 1 second data - and the standard deviation of the 1 minute data will be smaller than that. Can I combine the standard deviation of the 1 minute data with the combined standard deviations of the 1 minutes periods?
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ukdragon37
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(Original post by EmuEmu)
Because I am looking for the standard deviation of the 1 second data - and the standard deviation of the 1 minute data will be smaller than that. Can I combine the standard deviation of the 1 minute data with the combined standard deviations of the 1 minutes periods?
They are SDs for completely different things. Why would the standard deviation become smaller? Combining ten inaccurate things do not make it any more accurate (and I don't think this is sampling).
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EmuEmu
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I'm not looking for an accurate measurement (if you see what I mean) - it's just that the 10 minute standard deviation is "standard" measurement and can tell you something empirically about the turbulence of the wind (which is what I'm interested in).

I have set up an Excel spreadsheet with random example numbers and found that I can approximate the 10 minute standard deviation by a simple equation combining the standard deviations and the means. However the answers vary so I fear it's not something for which a precise answer is possible. I will use my approximation and try to make sure the datalogger is set up better next time!!

Thank you everyone!!
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1721
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(Original post by ukdragon37)
I would do this :o: :

Let X_i be the mean for each minute. Using the laws of expectation and variance, E(\sum{X_i}})=\sum{E(X_i)}=\sum{  \mu_i} and V(\sum{X_i}})=\sum{V(X_i)}=\sum{  \sigma^2_i}.
Dividing by n gives E(\frac{\sum{X_i}}{n}})=\frac{\s  um{E(X_i)}}{n}= \frac{\sum{\mu_i}}{n}
and V(\frac{\sum{X_i}}{n}})= \frac{\sum{V(X_i)}}{n^2}=\frac{\  sum{\sigma^2_i}}{n^2}.
SD=\sqrt{V(\frac{\sum{X_i}}{n})}  =\frac{\sqrt{\sum{\sigma^2_i}}}{  n}.
what is sigma?
i was under the impression sigma was standard deviation in which case your formula makes no sense
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ukdragon37
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(Original post by 1721)
what is sigma?
i was under the impression sigma was standard deviation in which case your formula makes no sense
Sigma=SD and Sigma^2=Variance=V(X)

Why doesn't it make sense? :confused:
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1721
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(Original post by ukdragon37)
Sigma=SD and Sigma^2=Variance=V(X)

Why doesn't it make sense? :confused:
so sigma= (root varience)/n
???
i guess the fact sigma is on both sides surely you want to make sigma the subject and if there is one on the other side doesnt really help
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ukdragon37
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(Original post by 1721)
so sigma= (root varience)/n
???
i guess the fact sigma is on both sides surely you want to make sigma the subject and if there is one on the other side doesnt really help
No :p: The SD refers to to the overall standard deviation combined from all the 1-minute periods. Notice that the variances in the equation has an index subscript. That's because it's the individual variance from each of the 1-minute periods.

The SD and the sigma_i are standard deviations of different things.
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sarina_nl
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Sigma means 'The sum of'. So if for example I want to find the mean of a set of 20 numbers, I would write.....Sigma (the sum of) x/20

x bar (mean)=∑x/n
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ncdave4life
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Here's how to do what you want:
http://stats.stackexchange.com/a/43936/16987
or http://www.burtonsys.com/climate/com...eviations.html

Sorry this is a few years late. Maybe it'll help someone else.
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