Are Two Formulas for Finding Standard Deviation the Same?
Date: 10/29/2007 at 12:20:39 From: Ashley Subject: shortcut formula for standard deviation My statistics text states that a shortcut formula for calculating sample standard deviation is to take the square root of [(the sum of x^2) minus (the sum of x)^2/n all divided by n-1]. Is this really a shortcut formula for calculating S.D.? I am confused to how this equation is equal to the standard deviation formula that I was initially taught, which is the square root of [the sum of (x minus x bar)^2 divided by n-1]. Can you explain how these formulas are equal to one another? I have been trying to factor the original equation [i.e. (x-x bar)(x- x bar), but multiplying that out still does not seem to show a way to obtain the shortcut method from the original formula.
Date: 10/29/2007 at 13:15:58 From: Doctor Peterson Subject: Re: shortcut formula for standard deviation Hi, Ashley. This is hard to type, so I'll use "m" for "x bar", and a few other notations that I hope will be clear: Definition: sqrt(SUM[(x - m)^2] / (n-1)) where m = SUM[x] / n Desired formula: sqrt((SUM[x^2] - SUM[x]^2)/n / (n-1)) Now let's do what you started to do, and see if we can manipulate the definition to look like the shortcut: Std Dev = sqrt(SUM[(x - m)^2] / (n-1)) = sqrt(SUM[(x - m)(x - m)] / (n-1)) = sqrt(SUM[x^2 - 2mx + m^2] / (n-1)) = sqrt((SUM[x^2] - SUM[2mx] + SUM[m^2]) / (n-1)) = sqrt((SUM[x^2] - 2m * SUM[x] + n * m^2) / (n-1)) Are you okay so far? The next to last step rearranges the sum. The last factors the constant 2m out of the second sum, and rewrites the third sum, which was the sum of n copies of a constant, m^2, making a total of n * m^2. Now let's use the definition of x bar (my m): = sqrt((SUM[x^2] - 2 SUM[x]/n * SUM[x] + n * (SUM[x]/n)^2) / (n-1)) = sqrt((SUM[x^2] - 2 SUM[x]^2 / n + SUM[x]^2 / n) / (n-1)) = sqrt((SUM[x^2] - SUM[x]^2 / n) / (n-1)) There you have it! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 10/29/2007 at 16:38:57 From: Ashley Subject: Thank you (shortcut formula for standard deviation) Dr. Peterson- Thanks for your help! I had been working on that for hours, but I guess I just needed to do a few more steps of factoring out. I really appreciate it! :) ~Ashley
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