### Chapter 7, Survey Sampling

#### Solution 29

#### (a)

Let $\, t \,$ denote the probability of answerting yes if Question 2 was asked. Similar to the problem-28, we have:

$$
\,
\begin{align*}
r = P(yes) \\
&= P(yes \,\vert\, \text{Q1 was asked})P(\text{Q1 was asked}) + P(yes \,\vert\, \text{Q2 was asked})P(\text{Q2 was asked}) \\
&= qp + t(1-p) \\
\end{align*}
\,
$$

Thus we can estimate r, by $\, R = Qp + t(1-p) \,$(Note that $\, t \,$ and $\, p \,$ and constant for an experiment. Thus $\, Q = \frac {R-t(1-p)} p \,$.

#### (b)

$$
\,
\begin{align*}
\Exp Q \\
&= \Exp\Prn{\frac {R-t(1-p)} p} \\
&= \frac 1 {p^2} \Exp(R-t(1-p)) \\
&= \frac 1 {p^2} (\Exp(R)-\Exp(t(1-p))) \\
&= \frac 1 {p^2} (r-t(1-p)) \\
&= q
\end{align*}
\,
$$

Thus $\, \Exp(Q) = q \,$. It follows that $\, Q \,$ is unbiased in estimating $\, q \,$.

#### (c)

$$
\,
\begin{align*}
\Var(Q) \\
&= \Var(\frac {R-t(1-p)} p) \\
&= \frac 1 {p^2} \Var(R-t(1-p)) \\
&= \frac 1 {p^2} \Var(R) && \text{$t$ and $p$ are constants} \\
&= \frac {r(1-r)} {np^2} && \text{Using Problem 28, part-d}
\end{align*}
\,
$$

$$\tag*{$\blacksquare$} $$