Asked by sophia
3. Prove (not show) that following two methods, long and short respectively, to compute the population variance yield the same result.
? (x i - ?) 2 / n, where i = 1, n
? (x i 2 - (? x i) 2) / n, where i = 1, n
? (x i - ?) 2 / n, where i = 1, n
? (x i 2 - (? x i) 2) / n, where i = 1, n
Answers
Answered by
MathGuru
The "short" method:
∑(x - µ)^2/N
...where ∑ means to sum or add up, µ = population mean, and x = each piece of individual data.
The numerator in the "short" method ∑(x - µ)^2 is called the Sum of Squares, which can also be written as:
∑x^2 - (∑x)^2/N
Now we have this using ∑x^2 - (∑x)^2/N for the numerator of the "short" method to derive the "long" method:
[∑x^2 - (∑x)^2/N]/N
Taking each term and multiplying it by 1/N, we have:
∑x^2/N - (∑x)^2/N^2
Since the second term is the same thing as µ^2 (you can check this using actual data), we end up with this:
∑x^2/N - µ^2
Using either the "short" or "long" method will yield the same result.
I hope this is what you were asking.
∑(x - µ)^2/N
...where ∑ means to sum or add up, µ = population mean, and x = each piece of individual data.
The numerator in the "short" method ∑(x - µ)^2 is called the Sum of Squares, which can also be written as:
∑x^2 - (∑x)^2/N
Now we have this using ∑x^2 - (∑x)^2/N for the numerator of the "short" method to derive the "long" method:
[∑x^2 - (∑x)^2/N]/N
Taking each term and multiplying it by 1/N, we have:
∑x^2/N - (∑x)^2/N^2
Since the second term is the same thing as µ^2 (you can check this using actual data), we end up with this:
∑x^2/N - µ^2
Using either the "short" or "long" method will yield the same result.
I hope this is what you were asking.
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