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A simple random sample of size $n=n_1 + n_2$ is drawn without replacement from a finite population of size $N$. Further a simple random sample of size $n_1$ is drawn without replacement from the first sample. Let $\bar y$ and $\bar {y_1}$ be the respective sample means.

Find V($\bar {y_1}$) and V ($\bar {y_2}$), where $\bar {y_2}$ is the mean of the remaining $n_2$ units in the first sample. Also find Cov($\bar {y_1}$,$\bar {y_2}$). Assume that $S^2$ is the population variance.


This is what I've been able to do thus far:

Note that $\bar {y_2}$ = $\frac{(n\bar y) - n_1\bar {y_1}}{n_2}$.

Also, V($\bar {y_1}$) = $E_1[V_2(\bar {y_1})] + V_1[E_2(\bar {y_1})]$. After simplification, this has turned out to be $(\frac1{n_1} - \frac1N)S^2$. Now am I on the right track? Also, some help in finding $V(\bar y_2)$ and the covariance would be appreciated.

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  • $\begingroup$ Hint: you already know $\text{V}(\bar{y}_2)$: the formula is in your question. $\endgroup$ – whuber May 7 '14 at 15:14
  • $\begingroup$ @whuber, kindly point it out meticulously. It is turning out to be quite complicated. $\endgroup$ – A.Chakraborty May 7 '14 at 17:40
  • $\begingroup$ Interchanging $1$ and $2$ in your question produces the answer for $\text{V}(\bar{y}_2)$, because the complement of a random sample without replacement is itself a random sample without replacement. This leaves the hard part, which is finding the covariance. $\endgroup$ – whuber May 7 '14 at 17:54

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