Suppose we want to estimate the proportion p (or equivalently, the percentage 100p%) of the persons who intend to vote for a certain political party in the forthcoming general elections. For this reason, we plan to take a sample of size n of voters and ask them about their intention (assuming they

Question

Suppose we want to estimate the proportion p (or equivalently, the percentage 100p%) of the persons who intend to vote for a certain political party in the forthcoming general elections. For this reason, we plan to take a sample of size n of voters and ask them about their intention (assuming they are all going to vote), from the entire population of eligible voters. How large should the sample size n be so that the error we make in this estimate (sometimes in everyday language called the “statistical error”) does not exceed 1%?

Accepted Answer

This is a sampling problem. We have seen problems of this type in relation to the hypergeometric distribution earlier in Chapter 5. The problem here, as is very often in case of opinion polls, is to determine the sample size. If we select n persons from the population of eligible voters, the number, X, who will vote in favor of the party is a random variable which follows the binomial distribution with parameters n and p (neither of which is known at present). Then, [latex]\hat{p}=X/n[/latex] will be our estimate for p and we want the following condition to be satisfied:

[latex]|p-{\hat{p}}|\leq0.01.\qquad\qquad\qquad\qquad(7.15)[/latex]

We thus see that the problem here is stated in an ill-posed manner! No matter how large n is we cannot be sure that the above condition is met, unless we select the whole population. Otherwise it is even possible, for example, that no one in our sample votes in favor of the party in question. The best we can do to overcome this is to impose the condition that (7.15) holds with a high preassigned probability, such as 95%. Consequently, the problem now becomes that of finding n such that (7.15) holds with a probability of (at least) 0.95, i.e.

[latex]P(|p-{\hat{p}}|\leq0.01)=P(|n p-n{\hat{p}}|\leq0.01n)\geq0.95.[/latex]

But, [latex]X = n\hat{p}[/latex] has a b(n, p) distribution and since we expect n to be large (typically several hundreds of persons will be needed for opinion polls of this type), we approximate the distribution of X by the N(𝜇, 𝜎²) distribution, with

[latex]\mu=n p,\quad\sigma^{2}=n p(1-p).[/latex]

We thus obtain

[latex]P(|n p-n{\hat{p}}|\leq0.01n)=P(|\mu-X|\leq0.01n)=P(-0.01n\leq X-\mu\leq0.01n{\mathrm{)}}=P\left({\frac{-0.01n}{\sigma}}\leq{\frac{X-\mu}{\sigma}}\leq{\frac{0.01n}{\sigma}}\right).[/latex]

Setting for simplicity 𝛼 = 0.01n∕𝜎, we see that the last expression equals Φ(𝛼) − Φ(−𝛼). But this, in turn, is

[latex]\Phi(\alpha) − \Phi(−\alpha) = \Phi(\alpha)−[1 − \Phi(\alpha)] = 2\Phi(\alpha) − 1,[/latex]

in view of Proposition 7.4. So, the condition we need in order to determine the sample size n, becomes now 2Φ(𝛼) − 1 ≥ 0.95; that is, we need to find the value of 𝛼 such that

[latex]\Phi(\alpha)\geq{\frac{1+0.95}{2}}=0.975.[/latex]

From the tables of the standard normal distribution in Appendix B, we see that 𝛼 must be greater than or equal to 1.96. Recalling now the definition of 𝛼, we get

[latex]\alpha={\frac{0.01n}{\sigma}}={\frac{0.01n}{\sqrt{n p(1-p)}}}\geq1.96[/latex]

and, solving this for n, we derive that

[latex]0.01n\geq1.96{\sqrt{n p(1-p)}},[/latex]

or equivalently

[latex]\sqrt{n}\geq 1.96\frac{\sqrt{p(1-p)} }{0.01} =196\sqrt{p(1-p)} [/latex]

Note that the quantity [latex]\sqrt{p(1-p)}[/latex] is unknown since the population proportion of persons voting for that party, p, is not known. However, and since we only require a lower bound for n, we may use the fact that g(p) = p(1 − p) ≤ 1∕4 for all values of p ∈ [0, 1] (see the next figure).

Hence, for any p ∈ [0, 1], [latex]\sqrt{p(1-p)}[/latex] is at most 1∕2. This finally shows that a minimum sample size of

[latex]\left(196 \cdot \frac{1}{2} \right) ^2=9604[/latex]

is sufficient for our purposes.

Question 7.9: Suppose we want to estimate the proportion p (or equivalentl......

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