Proving formula for deviation scores, using sums, equation

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pc678

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Hi. I'm taking a stats class and the teacher gave us a question regarding deviation scores x-xbar=0

The question the teacher gave us is to mathematically prove this formula. I don't quite understand how to do this. She said that the sums of scores and equation for the mean are part of this??????

Can anyone help me????

--pc
 
I'm not sure about this notation ...

Hi Mac, er, PC:

Is xbar the mean of x? x is a set of numbers?

I'm not sure I understand the equation, and I don't remember the definition of the phrase "derivation scores"..

~ Mark :|

My edit: Okay, I get it now.

[ ? {x[sub:bbcplpgf]n[/sub:bbcplpgf]- xbar} for n = 1 thru numberOFmembers{x} ] = 0

I will think about a proof while some of the CRAY processors who monitor this board post their approach...
 
One strategy for a proof ...

Hi PC:

Excuse me for not remembering to come back to this; the dogs kept me walking for 3 hours, and I forgot.

See if the following gives you any ideas. (Note: I switched the symbol name from x to a, so that I can use the a-bar character.)

CASE WITH 2 SCORES:

a = {a[sub:fdmk5p1e]1[/sub:fdmk5p1e], a[sub:fdmk5p1e]2[/sub:fdmk5p1e]}

? = (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2

(a[sub:fdmk5p1e]1[/sub:fdmk5p1e] — ?) + (a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — ?) = 0

a[sub:fdmk5p1e]1[/sub:fdmk5p1e] — (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2 + a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2 = 0

a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2 — (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2 = 0

a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — 2 * (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e])/2 = 0

a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — (a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e]) = 0

a[sub:fdmk5p1e]1[/sub:fdmk5p1e] + a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — a[sub:fdmk5p1e]1[/sub:fdmk5p1e] — a[sub:fdmk5p1e]2[/sub:fdmk5p1e] = 0

(a[sub:fdmk5p1e]1[/sub:fdmk5p1e] — a[sub:fdmk5p1e]1[/sub:fdmk5p1e]) + (a[sub:fdmk5p1e]2[/sub:fdmk5p1e] — a[sub:fdmk5p1e]2[/sub:fdmk5p1e]) = 0

0 + 0 = 0

The same pattern will happen with a set of 3 scores, 4 scores, 5 scores, ... n scores.

Try to write a proof with notation that covers any number of scores in set X.

Let us know if you have more questions. PLEASE show your work so far, and try to explain why you cannot continue.

Cheers,

~ Mark :)
 
Re: Proving deviation scores

\(\displaystyle mean = \mu = \frac{\sum{x}}{n}\)

\(\displaystyle \sum {(x - \, \mu)} = (\sum {x} )\, - \, n \cdot \mu \, = \sum {x} \, - \, n \cdot \frac{\sum{x}}{n} \, = \, \sum {x} \, - \, \sum {x} = 0\)
 
Re: Proving deviation scores

This is great guys. I guess I'm trying to understand verbally and conceptually why this proof works. Can someone explain the answers in english??
 
Do you understand "average"?

Consider the four numbers 6, 4, 7, and 3. Their average is 5.

Let's say you have four tubes, and the first tube contains a stack of 6 tennis balls. The next tube contains a stack of 4 tennis balls. The next tube has 7 tennis balls. The last tube has 3.

[attachment=0:3t6oc8tv]average.JPG[/attachment:3t6oc8tv]

The red line shows the average. (Double-click to expand image.)

The first tube contains one ball over the average. Let's call this an "overage".

The next tube contains one ball less than the average. Let's call this an "underage".

Likewise, the tube with 7 balls has an overage of 2, and the last tube has an underage of 2.

In order for all tubes to contain the same amount of balls, the overages would need to be shifted to fill in the underages.

This is what it means to average a set of numbers.

Is it clear that all of the underages are filled in by all of the overages in an average?

In other words, they cancel each other out when added together.

This is why the sum in your original equation is zero.

I've shown in green what happens when you subtract the average from each number in the set.

If the number has an overage, then subtracting the average will yield a positive amount.

When the average is subtracted from a number with an underage, then the result is negative.

The sum of all the overages is equal to the sum of all the underages.

I showed you one strategy hint of how to show the proof algebraically.

Subhotosh gave you a concise way of expressing it using Sigma notation.

Is it clear now why your equation equals zero?

Do you see how the mathematical statements reveal it?

~ Mark :)
 

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Hey. Sorry I keep bugging you. I've been looking over Khan's proof more. I think I understand for the most part. I know how to plug in the scores and get 0.

I think my problem is that I don't know enough algebra. Is there a way to show more of the work on Khan's proof? I've been looking through some old algebra tutor books I have and can't figure it out.
 
pc678 said:
Is there a way to show more of the work on Khan's proof?
Click to expand...

Yes, there is.

I already posted it in red, blue, and black for the case where n = 2.

Did you find information in those books about Sigma notation?

There's a ton of information ... here's one page from a Google search ...

Try writing it all out (i.e., rewrite the steps without Sigma notation), and see if that helps. If you need further explanation of the notation, then please be more specific.

Cheers,

~ Mark :)
 
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