Let $N = 12$. For purposes of this problem, assume that all population values are known. We have,
$$i = {\{1,2,3,4,5,6,7,8,9,10,11,12\}}$$
$$y_i = {\{1,2,5,1,2,5,1,2,5,1,2,5\}}$$
Consider all systematic samples of size $n = 4$.
a. Find the inclusion probability $\pi_i$ for each unit $i$
b. Find the sampling distribution of $\hat\tau$ = $12\bar{y}$
c. Using the sampling distribution, find $E(\hat\tau)$ and $V(\hat\tau)$
d. Is $\hat\tau$ an unbiased estimator of $\tau$, the population total? Why or why not?
e. Using the following formula, calculate the variance of $\hat\tau$ for a simple random sample of size $n = 4$:
$$V(\hat\tau) = N^2\left(\frac{N-n}{N-1}\right) {\sigma^2\over n}$$
f. Which sampling plan (systematic or SRS) do you think is better? Why?
Attempted Solution:
a. The skip number is $N\over{n}$ = $12\over{4}$ = $3$. Thus $\pi_i$ = $1\over{3}$ for each unit $i$.
b. $\bar{y}$ = $\frac{4(1+2+5)}{12}$=$32\over{12}$ = $2.67$ so $\hat\tau$ = $12(2.67)$ = $32$
c. I suppose I should start by considering the different systematic samples. ${\{1,4,7,10}\},{\{2,5,8,11}\}, {\{3,6,9,12}\}$. These are the indices so for the $y_i$'s we have ${\{1,1,1,1}\},{\{2,2,2,2}\}, {\{5,5,5,5}\}$. The values of $\bar{y}$ for each of these are $1,2,$ and $5$, respectively, and so the values of $\hat\tau$ for each of these would be $4,8,$ and $20$, respectively. Thus $E(\hat\tau)$ = $\frac{4+8+20}{3}$ = $10.7$. Since the values for $\hat\tau^2$ would be $16,64,$ and $400$, we get $E(\hat\tau^2)$ = $\frac{16+64+400}{3}$ = $160$. Thus,
$V(\hat\tau)$ = $E(\hat\tau^2)$ - $E(\hat\tau)^2$ = $160 - 10.7^2$ = $46.2$
d. I'm not sure how to determine this but I don't think it would be considering the size of the variance.
e. $V(\hat\tau)$ = ${N^2}$$(\frac{N-n}{N-1})$$\sigma^{2}\over{n}$ = ${12^2}$$(\frac{12-4}{12-1})$$1.7^{2}\over{4}$ = $75.7$
f. I think knowing if I did c and d correctly (I don't think I did) would help me answer this.