9.1 Question 17 In an opinion poll it is assumed that an unknown proportion p of the people are in favor of a proposed new law and a proportion 1 − p are against it. A sample of n people is taken to obtain their opinion. The proportion p ̄ in favor in the sample is taken as an estimate of p. Using the Central Limit Theorem, determine how large a sample will ensure that the estimate will, with probability .95, be correct to within .01.
Since the distribution of the standardized version of \(\overline{p}\) is approximated by the standard normal density, we know, for example, that 95% of its values will lie wtihin two standard deviations of its mean., and the same is true of \(\overline{p}\). So we have:
\[ P(p-2\sqrt{\frac{pq}{n}} < \overline{p} < p + 2\sqrt{\frac{pq}{n}}) \approx .954\]
The question stems asks that we find the sample size n required to be correct to be within 0.01 (margin of error). If we take the example in the book (on page 334), and assume that \(\overline{pq} \leq \frac{1}{4}\), no matter what the value of \(\overline{p}\) is, it is easy to show that if we chooses a value of n so that \(\frac{1}{\sqrt{n}} \leq 0.01\), we will be safe. If we solve for n:
\[\frac{1}{\sqrt{n}} \leq 0.01 => (\frac{1}{0.01})^2\leq n\]
Then,
\[n \geq 10000\]
So in order for the estimate will, with probability .95, be correct to within .01, n will need to be greater than or equal to 10000.