Discussion
In a survey of finite population, the following method is applied to obtain a sample of size \(n\).
(i) In the 1st draw, the ith unit is selected with probability \(P_{i} \gt 0\), \(i = 1, \ldots, N\), \(\sum_{i=1}^{N} Pi = 1\) and \(P_{i} \neq \frac{1}{N}\) for some i.
(ii) The remaining \((n - 1)\) units are obtained by SRSWOR method from remaining N - 1 units of the population.
Let \(\overline{Y}\) denote the sample mean, then :
Answer: B
The key's answer is that there exists an unbiased estimator of the population mean and it is different from the sample mean.
This is the Midzuno-Sen scheme. Because the first unit is drawn with unequal probabilities \(P_{i}\), the sample mean \(\bar{y}\) gives units different weights from their population shares and is biased (A fails). But the scheme has known inclusion probabilities, so the Horvitz-Thompson estimator, and the ratio-type estimator \(\bar{y}/\big(\sum_{i \in s} P_{i} \cdot \frac{N}{n}\big)\) used with this scheme, are unbiased, which also rules out D.
The procedure draws from the whole population at each stage; it does not divide the population into strata, so C is wrong.
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