Chapter 7, Survey Sampling
Solution 9
Let $X$ denotes the population in the sample.
Denoting $X_i$ as the $i_{th}$ member of population $X$. Let $X_i = 1$ if the member votes for the preposition and $X_i = 0$ if the member votes against the preposition.
If $\bar X$ denotes the proportion of the population that votes for the preposition and then $1 - \bar X$ denotes the proportion that votes against the preposition.
Thus the difference between $\bar X$ and $1 - \bar X$, i.e. $2 \bar X - 1$, gives the estimated margin of victory in the sample population. The variance of this estimate is then $\mathrm{Var}(2 \bar X - 1) = 4 \mathrm{Var}(\bar X)$.
We know that $\mathrm{Var}(\bar X) = \frac { \bar p (1 - \bar p) } {n-1} = \frac { 0.55 \times 0.45 } { 1499 } = 0.000165$, ignoring the finite population correction.
Thus the variance of the estimated margin of victory is $4 \times 0.000165 = 0.00066$. And which gives the standard error = $\sqrt { 0.00066 } = 0.0256 \approx 0.026$.
The 95% confidence interval is $0.1 \pm 1.96 \times 0.026$, or $(0.04904,0.15096) \approx (0.05, 0.15)$.
Or in percentage, The interval is $(5, 15)$.
Note: Compared to last problem here we used the approximation for $\mathrm{Var}(\bar X) = s^2_{\bar p} = \frac {\bar p(1- \bar p)} {n-1}$, because population variance is not available.
$$\tag*{$\blacksquare$} $$