Question

Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is particularly hazardous to infants under the age of 6 months. The maximum contaminant level (MCL) is the highest level of a contaminant the government allows in drinking water. For nitrates, the MCL is 10mg/L. The health department wants to know what proportion of wells in Madison Count that have nitrate levels above the MCL. A worker has been assigned to take a simple random sample of wells in the county, measure the nitrate levels, and assess compliance. What sample size should the health department obtain if the estimate is desired to be within 2% with 95% confidence if: (hint: there are two different methods)There is no prior information available?
A study conducted two years ago showed that approximately 7% of the wells in Madison County had nitrate levels exceeding the MCL?

Answer 1

Answer:

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

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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".  

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:

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

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 $ME =\pm 0.02$ 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)  

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$. And replacing into equation (b) the values from part a we got:

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

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.