Standard deviation question!?

Question

is the following true?

(stdev(a)-stdev(b))/stdev(b) = stdev((a-b)/b)

...how do you know? standard deviation is a pretty complicated equation, it may not be transitive (i think that is the right word)

..eh, nevermind about the "transitive"

.........eh, i don't know, i seem skeptical of all of your answers

Answer 1

Yes due to simple arithmetic. For this purpose, stdev will be noted as "S"

1) [S(a)-S(b)]/[S(b)]
2) [S(a-b)]/[S(b)]
Therefore = S[(a-b)/b]

OR:

1) S[a/b]-S[b/b]
2) S[a/b]-S[1]
and the S[1] =1
Therefore, = S[a/b]-1

Answer 2

Now, I'm not too sure on this one, so don't take my word for it.

I'm not even sure how that works out. The standard deviation is a measure of deviance from the mean for a set of data. If you've got two different stddevs, they're most likely for different data sets, and what you have up there may not make sense (if it's right at all). If the data sets are supposed to be related, you may want to look in the correlation between the two.

I also may be speaking out bad parts of my body. Something about that just strikes me as wrong though. Maybe if you provide more background.

Answer 3

you cant apply arithematics to standard deviation...

in the above example a,b represents a colleciton of data... how can you do the subtraction a-b???

I mean to say how do you calculate stdev((a-b)/b)?? and what does that represent...?

So it is not possible

Answer 4

No.

As a first thought, you have to be working with variances, not standard deviations.

Answer 5

yes