Question

Suppose that a researcher is designing a survey to estimate the proportion of adults in your state who oppose a proposed law that requires all automobile passengers to wear a seat belt. Which sample size will provide a margin of error of about 2%?

Answer 1

Answer:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

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".  

The population proportion have the following distribution  

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

Solution to the problem

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)  

If solve n from equation (a) we got:  

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

The margin of error desired for this case is $ME= \pm 0.02$ equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is $\hat p =0.5$

In order to find the critical value we need to take in count that we are finding the margin of error for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:  

$z_{\alpha/2}=\pm 1.96$  

Now we have all the values needed and if we replace into equation (b) we got:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.