Question

After the 2008 elections, it is desired to estimate the proportion of Florida voters who now regret that they did not vote.

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$  

And rounded we got 1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$. And the critical value would be given by:  

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$  

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)  

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimates proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:  

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$  

And rounded we got 1681