- #1

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I'm not looking for something very accurate here. Something quick and dirty will do.

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- Thread starter wahaj
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- #1

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I'm not looking for something very accurate here. Something quick and dirty will do.

- #2

Simon Bridge

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ie. each of the 5 sets will have a mean ... from that you can find the mean of the means, and a standard deviation of the means ... which will let you compare the means.

- #3

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This is probably the simplest way to go about doing this. Thanks for the help

- #4

jim mcnamara

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Since we do not have much to go on, let's start with ANOVA - analysis of variance, and an assumption that the data fit a normal distribution.

Analysis of Variance (

It will tell you if subset A and subset B are from the same population. Or not. It is hard to tell what you did because you said 'settings' which is plural. Your experimental design and th answers you need impact how you analyze data. Big time.

If you are just playing, then @Simon Bridge gave you the best answer.

- #5

EnumaElish

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The answer partially depends on degrees of freedom your samples have. If each sample is large you can perform a separate matched-pairs t test on each sample. If their results point to the same conclusion then you're home. But suppose 3 point in one direction and 2 in the other. Then you may conclude "time" and "location" matter. If you don't have sufficient DF then my advice is to estimate a regression model with "sample fixed effects."

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- #6

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- #7

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Mathematically, if ##X_i## has a normal distribution ##N(\mu,\sigma^2)##. Then the mean ##\overline{X}## has distribution ##N(\mu, \sigma^2/n)##.

So if we take the means of ##5## means, then this will certainly be an unbiased estimator, no problem there. But the variance of the mean of ##5## means is

[tex]\frac{\sigma^2}{25}\sum_{i=1}^{5} \frac{1}{n_i}[/tex]

where ##n_i## are the sample sizes of the ##5## groups.

Compare this with the variance if we just put all the groups together:

[tex]\frac{\sigma^2}{\sum_{i=1}^{5} n_i}[/tex]

It is easy to see that this variance is always smaller with equality if ##n_1=...=n_5##.

So in order to get more precise estimates, it is better to just put all the observations together in one big set (and perhaps introduce a blocking factor) than taking a mean of means.

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- #9

chiro

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It might help to understand what property you are trying to assess.

Means, medians and things like variances have a very good interpretation but it may not necessarily be what you are looking for.

To get this answer you have to understand what questions you are trying to answer that are completely independent of statistics and then formulate the appropriate test statistics, collection techniques, and processing of information to facilitate this answer.

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