Comparing Sample Means with Pooled Variance Estimate

[adamg]

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 + (n2-1)s2^2 ] / (n1+n2-2)

I was just wondering why we use (n1-1) etc instead of using n1 and n2 and then dividing by n1 + n2?

thanks

 

[uart]

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 + (n2-1)s2^2 ] / (n1+n2-2)

I was just wondering why we use (n1-1) etc instead of using n1 and n2 and then dividing by n1 + n2?

thanks

When you calculate the sample variance using "(sum of squared difference from mean)/N" then it turns out that this gives a biased estimate of the population variance (and it's square-root a biased estimate of the population standard deviation). Replacing "N" with "N-1" gives an unbiased estimate of the population variance and standard deviation so it's usually preferred. Unfortunately there is often a bit of ambiguity whenever sample var and sd are discussed as there doesn't seem to be a universal standard of whether to use "N" or "N-1" in the definition.

In your example above I assume that s1^2 and s2^2 are based on the "N-1" calculations.

 

[uart]

Here is the above in a bit more detail :

$$\[ s_n^2 = 1/n \sum (x_i-\bar{x})^2 \]$$

$$\[= 1/n \sum [ ( (x_i-\mu) - (\bar{x}-\mu) )^2 ]\]$$

$$\[= 1/n \sum [(x_i-\mu)^2 - 2 (x-\mu)(\bar{x}-\mu) + (\bar{x}-\mu)^2 ] \]$$

$$\[= 1/n \sum [(x_i-u)^2)] - (\bar{x}-u)^2 \]$$

So,

$$\[E[s_n^2] = 1/n \sum E[(x_i-u)^2)] - E[(\bar{x}-u)^2 ] \]$$

$$\[= E[(x-u)^2)] - E[(\bar{x}-u)^2 ] \]$$

$$\[= \sigma^2 - \{\rm{term\ greater\ than\ or\ equal\ zero}\} \]$$

This shows that the sample variance $$s_n^2$$ always under-estimates the population variance $$\sigma^2$$.

You can further show (assuming all the samples are independent) that $$E[(\bar{x}-u)^2 ]$$ is equal to $$\sigma^2/n$$ and hence,

$$\[s_n^2 = \sigma^2 - \sigma^2/n = \frac{n-1}{n} \sigma^2 \]$$.

So not only is $$s_n^2$$ a biased estimator of $$\sigma^2$$ it is too small by a factor of precisely $$(n-1)/n$$. Clearly using $$(n-1)$$ instead of $$(n)$$ in the denominator fixes this and makes the expectation of this modified sample variance ($$E(s_{n-1}^2)$$) equal to the population variance ($$\sigma^2$$).