Question

Suppose a sample of size 400 yields pˆ = .5. You'd like to construct a confidence interval with a margin of error only half as great as the one produced by this sample. What's the minimum sample size necessary to accomplish this?a. 400b. 800c. 1,600d. 1,200e. 2,400

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600$  

c. 1600

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

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

Solution to the problem

In order to solve this problem we need to assume a confidence level. Let's assume that is 95%

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 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:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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)  

First we need to find the margin of error from the original sample given by:

$ME=1.96\sqrt{\frac{0.5 (1-0.5)}{400}}=0.049$

And on this case we have that $ME =\pm 0.049/2=0.0245$ 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)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600$  

c. 1600