Question

Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want to be at least 95% confident that the error is at most 0.04. How large a sample will we need if:
a) We have no idea what the true proportion might be
b) We know that the true proportion is approximately 0.12?

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

And rounded up we have that n=601

b)  $n=\frac{0.12(1-0.12)}{(\frac{0.04}{1.96})^2}=253.546$  

And rounded up we have that n=254

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 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 confidence interval for the proportion is given by the following formula:  

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

Part a

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 $ME =\pm 0.04$ 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)  

Since we don't know prior estimation for the population proportion we can use $\hat p=0.5$. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25$  

And rounded up we have that n=601

Part b

For this case we use $\hat p =0.12$ and if we solve for n we got:

$n=\frac{0.12(1-0.12)}{(\frac{0.04}{1.96})^2}=253.546$  

And rounded up we have that n=254