Error Accumulation - Combined Result without going to Partial Derivitaves

[Statisticale]

Hi,

I have an area of forest that has been stratified into 2, & I wish to calculate the +/- 3 standard deviation (sigma) distribution of tree heights in the combined area. The weighting of the first area is 0.73, & the second is 0.27. Means heights are 8.3 & 5.5 meters, with standard deviations of 1.2 & 0.9 meters.

If we make the assumption that co-variance can be ignored, can the combined result be calculated without going to partial derivitaves? If so, how?

Best regards.

[Darius]

Sorry to ask for more info..., what are you searching?. That you can calculate something doesn't mean you should.

If you need to inform a data value (standard deviation for all set), maybe could be better to obtain the weighted mean for stdev or much better could be to calculate the correlation between height and variation and don't show a value but it's equation.

IMHO, if you want to do SPC with such data, you can transform the data to Z values (a Zed chart), so the limits will be between 3 and -3 sigma, and you can chart both kind of products (tree-heights) as one set of data.
http://elsmar.com/Forums/showthread....ht=zed+chart#4

[Statistical Steven]

In Reply to Parent Post by Statisticale

Hi,

I have an area of forest that has been stratified into 2, & I wish to calculate the +/- 3 standard deviation (sigma) distribution of tree heights in the combined area. The weighting of the first area is 0.73, & the second is 0.27. Means heights are 8.3 & 5.5 meters, with standard deviations of 1.2 & 0.9 meters.

If we make the assumption that co-variance can be ignored, can the combined result be calculated without going to partial derivitaves? If so, how?

Best regards.

If you assume independence, you can use the property of variances to solve the problem.

Var (X + Y) = Var(X) + Vary (Y)
Var (0.73X) = (0.73^2)Var(X)
Var (0.27Y) = (0.27^2)Var(Y)
Var (0.73X + 0.27Y) = (0.73^2)Var(X) + (0.27^2)Var(Y)

The calculation of the means is straightforward.

Overall Mean = ((n1*0.73*mean1) + (n2*0.27*mean2))/(n1+n2)

Does that make sense. Of course the assumnption of independence is CRITICAL here.