
#1
May2105, 12:04 PM

P: 48

Suppose you are conducting a hypothesis test to compare two sample means from independent samples, with the variance unknown, but you know it is the same for both populations. Then you use the pooled estimate of the variance given by [ (n1  1)s1^2 + (n21)s2^2 ] / (n1+n22)
I was just wondering why we use (n11) etc instead of using n1 and n2 and then dividing by n1 + n2? thanks 



#2
May2205, 11:45 AM

Sci Advisor
P: 2,751

In your example above I assume that s1^2 and s2^2 are based on the "N1" calculations. 



#3
May2305, 12:01 PM

Sci Advisor
P: 2,751

Here is the above in a bit more detail :
[tex]\[ s_n^2 = 1/n \sum (x_i\bar{x})^2 \][/tex] [tex]\[= 1/n \sum [ ( (x_i\mu)  (\bar{x}\mu) )^2 ]\][/tex] [tex]\[= 1/n \sum [(x_i\mu)^2  2 (x\mu)(\bar{x}\mu) + (\bar{x}\mu)^2 ] \][/tex] [tex]\[= 1/n \sum [(x_iu)^2)]  (\bar{x}u)^2 \][/tex] So, [tex]\[E[s_n^2] = 1/n \sum E[(x_iu)^2)]  E[(\bar{x}u)^2 ] \][/tex] [tex]\[= E[(xu)^2)]  E[(\bar{x}u)^2 ] \][/tex] [tex]\[= \sigma^2  \{\rm{term\ greater\ than\ or\ equal\ zero}\} \][/tex] This shows that the sample variance [tex]s_n^2[/tex] always underestimates the population variance [tex]\sigma^2[/tex]. You can further show (assuming all the samples are independant) that [tex]E[(\bar{x}u)^2 ] [/tex] is equal to [tex]$\sigma^2/n[/tex] and hence, [tex]\[s_n^2 = \sigma^2  \sigma^2/n = \frac{n1}{n} \sigma^2 \][/tex]. So not only is [tex]s_n^2[/tex] a biased estimator of [tex]\sigma^2[/tex] it is too small by a factor of precisely [tex](n1)/n[/tex]. Clearly using [tex](n1)[/tex] instead of [tex](n)[/tex] in the denominator fixes this and makes the expectation of this modified sample variance ([tex]E(s_{n1}^2)[/tex]) equal to the population variance ([tex]\sigma^2[/tex]). 


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