0
$\begingroup$

Find the variance of $S^2_p$ under the conditions; $\bar{x_1}, \bar{x_2}, s_1, s_2$ are the means and standard deviations of independent random samples of sizes $n_1$ and $n_2$ from normal populations with equal variances.

The answer to this is $\frac {\sigma^4}{n_1 +n_2- \lambda}$. I am confused as to how to acquire this answer.

$\endgroup$
0
$\begingroup$

The formula for the pooled variance given two samples $x_1^1,\dots, x_1^{n_1}$ and $x_2^1,\dots, x_2^{n_2}$ of sizes $n_1$ and $n_2$ respectively is given by $$S_p = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2}$$ where $s_1^2$ and $s_2^2$ are the sample variances of the respective samples.

You want $$Var(S_p) = \left(\frac{n_1-1}{n_1+n_2}\right)^2 Var(s_1^2) + \left(\frac{n_2-1}{n_1+n_2}\right)^2 Var(s_2^2)$$ where $$s_i^2 = \frac{1}{n_i-1} \sum_{j=1}^{n_i} (x_i^j - \overline{x}_i)^2 \mbox{ for } i=1,2.$$

You have that your samples are drawn from a normal distribution with equal variances, that is $$x_i^1,\dots, x_i^{n_i} \sim N(\mu_i,\sigma^2 ) \mbox{ for } i=1,2$$ and hence $$\overline{x}_i \sim N(\mu_i, \frac{\sigma^2}{n_i}) \mbox{ for } i=1,2.$$

Using the above information compute now $Var(s_i^2)$ for $i=1$ and $i=2$ under the assumptions above and plug it in the formula for $Var(S_p)$.

Good luck!

P.S.: By the way, it's common within statistics that capital letters denote random variables, for instance $X\sim N(\mu,\sigma^2)$ and small letters denote realizations of these random variables, that is $x$ is an observation from $X$ means that $x$ is a given value drawn from $X\sim N(\mu,\sigma^2)$. I tell you because here we have not been consistent with this notation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.