Cannot solve simple MATH problem

[zealot]

Hello!
I have a very simple math calculation which does not SUM up!!!! No high math involved - only averages....
Can yo please check the attached sheet? You will see cells which are not equal but THEY SHOULD BE EQUAL...or have I missed something?
Thank you!)
Dave

[Richard Buttrey]

Why do you believe they should be equal?

Yes you have missed something. You can't, or to be more precise you shouldn't average averages which seems to be what you are trying to do.

However explain what your data represents and tell us what statistic or effect you are trying to demonstrate.

Incidentally please update your location. It is often useful when we need to consider dates, times or regional list separators

[xlnitwit]

Hi,

Why do you think they should be equal?

[zealot]

thank you because the PRODUCT OF TWO AVERAGES OF TWP SETS OF DATA should be equal to the AVERAGE OF THE PRODUCT OF EACH DATA POINTS IN THESE SUBSETS
OR NOT?

[davsth]

No. I dont think they should

if you had
A,B

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averaging averages is virtually always bad practice, unless the groups are the same size. I would echo Richard's views

[zealot]

now I CHANGED THE PERCENTAGES TO THE INTEGERS and it works!
now sure why?????

[davsth]

fluke. put some other integers in and it wont work

[zealot]

and yes I am not averaging averages,,,,

I am comparing the product of averages AND average of the products...

[Richard Buttrey]

thank you because the PRODUCT OF TWO AVERAGES OF TWP SETS OF DATA should be equal to the AVERAGE OF THE PRODUCT OF EACH DATA POINTS IN THESE SUBSETS
OR NOT?

Not! As Dave has demonstrated with his example.
In essence you are not taking into account any weighting of the data points.

But tell us what you data represents/means. Pure numbers /%s are not much help if we don't know the context or indeed what you are trying to demonstrate.

[zealot]

fluke. put some other integers in and it wont work

yes this is mystery.....is this some kind of a mathematical paradox? btw I found those working integers by chance..

[zealot]

those are parameters in a complex model.....which calculates the relative strengths of the sentiment storms affecting the stock markets...

[Richard Buttrey]

fluke. put some other integers in and it wont work

or even swap two of the column A numbers around. The average doesn't change but your 'results' do. i.e you are ignoring the weighting factor as I said.

...and don't forget to add your location...

[zealot]

This is the formula:

((a+b+c)/3)*((d+e+f)/3) <> ((a*d)+(b*e)+(c*f))/3

[Richard Buttrey]

Yes. That formula is correct, i.e. the two calculations are NOT equal

- except in the special case that either a=b=c,c OR d=e=f

Your point being what exactly?


..Location????

[zealot]

this is the working set which is working but a=b=c this is not true...

1.00 2.00
3.00 2.00
2.00 3.00

My location is undefined for now...

[zealot]

someone has mathcad? - can someone SYMBOLICALLY solve the above equations quick and see where the results differ?

[Richard Buttrey]

this is the working set which is working but a=b=c this is not true...

1.00 2.00
3.00 2.00
2.00 3.00

My location is undefined for now...

You've been given all the reasons why your assumptions are wrong.
For the last time and then I'm out. Your example is NOT working as you believe. Swap 1 & 3 around and you'll see a different result although the average remains the same.

Why is this so difficult for you to understand and why aren't you accepting what we're saying?

And why do you think you are a special case and shouldn't declare your location?

[xlnitwit]

In order for the two equations to be equal

(a+b+c)/3*(d+e+f)/3 = ((a*d)+(b*e)+(c*f))/3

(a+b+c)*(d+e+f)/9 = ((a*d)+(b*e)+(c*f))/3

(a+b+c)*(d+e+f)/3 = ((a*d)+(b*e)+(c*f))

((a*d)+(b*e)+(c*f)+(a*e)+(a*f)+(b*d)+(b*f)+(c*d)+(c*e))/3 = ((a*d)+(b*e)+(c*f))

((ad)+(be)+(cf)+(ae)+(af)+(bd)+(bf)+(cd)+(ce)) = (3(a*d)+3(b*e)+(c*f))

ad+be+cf+ae+af+bd+bf+cd+ce = 3ad+3be+3cf

2ad+2be+2cf-ae-af-bd-bf-cd-ce = 0

Only if the last line is true will the two equations be equal, which will certainly not necessarily be the case. I imagine there is a more generic approach to this but it has been too long since I did this kind of thing!

[zealot]

thank you xlnitwit- this is spectacular!)