Chapter 7, Survey Sampling
Solution 11
(a)
In simple random sampling, a sample of a given size is chosen from all possible samples with same size from the population. Thus the total samples are $\, {4 \choose 2} = 6 \,$.
(b)
We have sample mean equals to:
Thus the expected value of the sample mean is $\, E(\bar X) = \frac 1 n \mathrm{E}(\sum_{i=1}^n X_i) = \frac 1 2 \sum_{i=1}^2 \mathrm{E} X_i \,$
In the given samples $\, (x_1, x_2), (x_2, x_3), (x_3, x_4), (x_1, x_4) \,$, we can see that every element of the population is present in equal number of samples(i.e. 2). (Note that in a sample, order does not matter $\, (x_1, x_2) \,$ and $\, (x_2, x_1) \,$ are same samples.)
Thus every $\, i^{th} \,$ member of a random sample has an equal chance to be any element $\, x_1, x_2, x_3 \,$ or $\, x_4 \,$ of the population. In other words, every element $\, x_j \,$ has an equal probability to become a part of a random sample. Thus probability $\, P(X_i = x_j) \,$ for any $\, i^{th} \in \{1,2\} \,$ member of sample and any $\, j^{th} \in \{1,2,3,4\} \,$ element of the population, is $\, \frac 1 N = \frac 1 4 \,$. It follows that $\, \mathrm{E}(X_i) = \frac 1 4 \sum_{j=1}^4 x_j \,$.
( Note: Initially, I made a rookie mistake here by computing the probability $\, = \frac 2 4 = \frac 1 2 \,$, thinking that each member appears twice in total 4 samples. This is wrong because this probability is for whether a specific element is a member of a sample or not but the probability that we require is the probability of selection of an element in a random sample. Another way to see that $\, \frac 1 2 \,$ is wrong is that the sum of probabilities of choosing each of the members should equal 1 but in this case it becomes $\, 4 \times \frac 1 2 = 2 \,$, which is clearly wrong!)
Thus $\, \mathrm{E}(\bar X) = \frac 1 2 \sum_{i=1}^2 \mathrm{E} X_i \,$
But this expression is the population mean $\, \bar x \,$.
It follows that $\, \mathrm{E}(\bar X) = \bar x \,$. Thus the sample mean is unbiased.
$$\tag*{$\blacksquare$} $$