Sampling Question

Question

Answer:

Sample 9604 people.

I don't even know to start this question, and how did the textbook get an exact number with no information on the total population?

Answer 1

Using a normal population (suitable for large $n$ as contemplated here), a 95% confidence interval for the proportion $p$ of the population for the Candidate is of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$$ where $\hat p = X/n,$ the sample proportion for the Candidate, $\hat E = \sqrt{\frac{\hat p(1-\hat p)}{n}}$ is the estimated standard error of the estimate $\hat p$, the standard error of $\hat p$ is $E = \sqrt{\frac{p(1-p)}{n}},$ and the margin error of the confidence interval is $M = 1.96\hat E.$

You want to find $n$ large enough that $M = .01.$ Because we do not have data it is best to use $M \approx 1.96E.$ Because we do not know $p$ we consider the worst-case where $E$ is as large as possible, and that is for $p = 1/2$. [The largest possible value of $p(1-p) = 1/4$ occurs when $p = 1/2.]$

Thus we want to solve $1.96\sqrt{1/4n} = .01$ for $n.$ The solution is $n = 9604.$

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Notes: (1) A slight modification of this argument is widely used to plan polls and to give rough margins of sampling error in reported results of polls. If we use $1.96 \approx 2,$ then we get $M \approx 1/\sqrt{n}$ for a 95% confidence interval. Thus a poll that uses random sampling with replacement of $n = 2500$ subjects should have a margin of error of at most $\pm 2\%.$ And a poll with $n = 1100$ respondents might have a margin of error for estimating $p$ with $M = 3\%$

(2) It is important to understand that the formulas used here apply only to error that results from random sampling. If subjects are not randomly selected from the population (e.g., people in favor of the Candidate are more likely to be available), if people with some opinions are more likely to answer than others, or if people do not tell pollsters what they really believe, then these are not really probability issues and no probability formula can account for the resulting biases and errors.

(3) If $p$ is very close to $0$ or $1$ then the formula discussed above may overstate the required sample size.