Consider a population of nine taxpayers. Suppose five of the taxpayers are Democrats and the rest Republican.
Consider all SRS of size $n = 3$.
a. Find the sampling distribution of $\hat{p}$, the proportion of Democrats in the sample.
b. Using the sampling distribution, find $E(\hat{p})$ and $V(\hat{p})$.
c. Is $\hat{p}$ an unbiased estimator of $p$, the population proportion? Why or why not?
d. Verify that $V(\hat{p})$ computed using the sampling distribution equals:
${N-n}\over{N-1}$$\frac{\sigma^2}{n}$ = ${N-n}\over{N-1}$$\frac{p(1-p)}{n}$
Attempted Solution:
a. $5\over9$
b. The probability of choosing $0, 1, 2$, and $3$ democrats from a sample of three are
${\binom{5}{0}\binom{4}{3}}\over{9 \choose {3}}$, ${\binom{5}{1}\binom{4}{2}}\over{9 \choose {3}}$, ${\binom{5}{2}\binom{4}{1}}\over{9 \choose {3}}$, ${\binom{5}{3}\binom{4}{0}}\over{9 \choose {3}}$
The value of $\hat{p}$ that we would associate each of these with would be $0$, $1\over3$, $2\over3$, and $3\over3$.
Then, multiplying each $p$ with its respective $\hat{p}$ and summing, we get
${\binom{5}{0}\binom{4}{3}}\over{9 \choose {3}}$$(0)$ + ${\binom{5}{1}\binom{4}{2}}\over{9 \choose {3}}$$(\frac{1}{3})$ + ${\binom{5}{2}\binom{4}{1}}\over{9 \choose {3}}$$(\frac{2}{3})$ + ${\binom{5}{3}\binom{4}{0}}\over{9 \choose {3}}$$(\frac{3}{3})$ = $5\over9$ which is the same value that I got in part a.
$V(\hat{p})$ = ${{\binom{5}{0}\binom{4}{3}}\over{9 \choose {3}}}^2$$(0)$ + ${{\binom{5}{1}\binom{4}{2}}\over{9 \choose {3}}}^2$$(\frac{1}{3})$ + ${{\binom{5}{2}\binom{4}{1}}\over{9 \choose {3}}}^2$$(\frac{2}{3})$ + ${{\binom{5}{3}\binom{4}{0}}\over{9 \choose {3}}}^2$$(\frac{3}{3})$ - $(\frac{5}{9})^2$ = $-0.1$
c. Since $\hat{p}$ = $p$ = $5\over9$, then $\hat{p}$ is an unbiased estimator of $p$.
d. Need to find $V(\hat{p})$ to answer this but I have that
${N-n}\over{N-1}$$\frac{p(1-p)}{n}$ = ${9-3}\over{9-1}$$\frac{\frac{5}{9}\frac{4}{9}}{3}$
(a) To "find the sampling distribution" actually means to do something you've done in part (b), which is to list out the values that $\hat p$ can be along with their associated probabilities.
(b) To calculate $V(\hat p)$, first recall that for any random variable $X$, we have $V(X) = E[X^2] - (E[X])^2$. You've already done the hard part, which is finding the distribution of $\hat p$; now, compute $E[\hat p^2]$ as $$\frac{\binom 5 0 \binom 4 3}{\binom 9 3} (0)^2 + \frac{\binom 5 1 \binom 4 2}{\binom 9 3} (\frac 1 3)^2 + \dots$$ and then subtract $(5/9)^2$ from this.
(c) What you mean is that $E[\hat p] = p$. Remember that $p$ is a fixed number, which in this case is $\frac 5 9$, and that $\hat p$ is a random variable because it depends on the sample. Sometimes it will be $0$, sometimes it will be $2/3$, etc. -- but on average, it is indeed $5/9$.
(d) You should now be in a position to verify this.