well I get that the first term in the numerator of my answer is zero since it's the sum of xi and xbar but what about the denominator? isn't it supposed to stay as is? Is it a typo or should the denominator change from
the sum of (xi minus xbar)^2 to
the sum of (xi minus xbar^2)
So I've been trying to simplify the following for entirely too long: $\displaystyle \frac{\sigma^2}{n}+\bar{x}^2\frac{\sigma^2}{\sum (x_i -\bar{x}^)^2}$
becomes $\displaystyle \frac{\sigma^2}{n}+\frac{{}\sigma^2\frac{1}{n}\sum x_i^2}{\sum (x_i - \bar{x})^2}$
but WHY did xbar^2 become 1/n times the sum of x_i squared? shouldn't the 1/n be squared there as well? shouldn't $\displaystyle \bar{x}^2$ be equal to $\displaystyle (\frac{1}{n}\sum(x_i))^2$ which would square the$\displaystyle \frac{1}{n} $as well?
Also this is supposed to simplify to
$\displaystyle \frac{\sigma^2 \sum x_i^2}{\sum (x_i -\bar{x}^)^2}$
but I can't get their either. I specifically would like to know this:
If we let sigma^2 =A
N=B
1/n sum of xi squared = C
and sum of (xi-xbar)^2 = D
then for the simplification we have
$\displaystyle \frac{A}{B} + \frac{AC}{D}$ which is supposed to simplify to$\displaystyle \frac{AC}{D}$
except that$\displaystyle \frac{A}{B} + \frac{AC}{D}$ simplifies to
$\displaystyle \frac{A}{B}+\frac{AC}{D} = \frac{AD}{BD}+\frac{ACB}{DB} = \frac{AD+ABC}{DB}=A\frac{D+BC}{DB}$
as far as I can see and that does not compute with the correct answer. which is supposed to be$\displaystyle \frac{AC}{D}$=$\displaystyle \frac{\sigma^2\frac{1}{n}\sum x_i^2}{\sum (x_i-\bar{x})^2}} $
If anyone can just tell me what I'm supposed to do here it would be great.
Or is sigma^2/n = to zero for some reason?
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