Average of standard deviation makes sense?

[xeon123]

I've a set of tests, and for each test, they have different size.

I'll show an example.

Test1
size - Time(seconds)
100 - 10
100 - 23
100 - 17
200 - 37
200 - 42
200 - 47
300 - 53
300 - 53
300 - 53

For each test, I took the average.

Average1
size - average
100 - 16
200 - 42
300 - 53

And for each entry of the Test1, I took the standard deviation. Does't in make sense calculate the mean of the standard deviation?

Doing an average of standard deviation can proof that all examples ran at similar time? For example, having low average in the standard deviation means that all the 3 results of 100 were similar among themselves, the 3 results of 200 were similar results among themselves, and the 3 results of 300 were similar among themselves?

 

[Number Nine]

There are a few procedures used to test the hypothesis that the variances of two groups are equal, but they tend to be extremely unreliable (especially for your small N).

The standard deviation has a distribution, of course, and the distribution tends to be pretty normal for large N (which you don't have, so your distribution is going to be quite skewed). You're not going to be able to say much with confidence; the best thing to do may be to just point to the standard deviations of the three groups and note that they're pretty close to each other.

 

[xeon123]

I just showed here a small example. I've a set with N=100.

 

[mathman]

When combining data sets you should work with the variances and not the standard deviations.

 

[Bandit127]

When combining data sets you should work with the variances and not the standard deviations.

Where variance is SD².

So I presume that SD_(total) = √(SD₁² + SD₂²...SD_n²)

And SD_(mean) = SD_(total)/n

While I can't see a good use for SD_(mean), being able to understand SD_(total) is a powerful tool in process improvement. Often you can't reduce all variation in a process but you can sure tackle some of them individually and reduce the total that way.

 

[mathman]

Where variance is SD².

So I presume that SD_(total) = √(SD₁² + SD₂²...SD_n²)

And SD_(mean) = SD_(total)/n

While I can't see a good use for SD_(mean), being able to understand SD_(total) is a powerful tool in process improvement. Often you can't reduce all variation in a process but you can sure tackle some of them individually and reduce the total that way.

Do you have a further question?

 

[Bandit127]

Since you have not taken issue with it, I presume the workings are correct.

Should I take for granted that all the readers of this thread know what variance is and how to sum and take the average of standard deviations? If so my post is redundant.

If not you could have prevented me posting by explaining it yourself.