Question

(c) Suppose that 20% of voters are in favor of certain legislation. A large number n of voters are polled and a relative frequency estimate fA(n) for the above proportion is obtained. Use central limit theorem to estimate how many voters should be polled in order that the probability is at least .95 that fA(n) differs from 0.20 by less than 0.02

Answer 1

Answer:

1537 voters should be polled.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

20% of voters are in favor of certain legislation.

This means that $\pi = 0.2$

How many voters should be polled in order that the probability is at least .95 that fA(n) differs from 0.20 by less than 0.02

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.2*0.8}{n}}$

$0.02\sqrt{n} = 1.96\sqrt{0.2*0.98}$

$\sqrt{n} = \frac{1.96\sqrt{0.2*0.98}}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.2*0.98}}{0.02})^2$

$n = 1536.6$

Rounding up

1537 voters should be polled.