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?

Answer 1

The sample size is
`n = 2401`

Explanation:

From the question we are told that

The margin of error is
`E = 0.02`

Given that the confidence level is 95% then the level of significance can be mathematically evaluated as

`\alpha = 100 - 95`

`\alpha = 5\%`

`\alpha = 0.05`

Next we would obtain the critical value of
`(\alpha )/(2)` from the z-table , the values is

`Z_{(\alpha )/(2) } = Z_{(0.05 )/(2) } = 1.96`

The reason we are obtaining critical value of
`(\alpha )/(2)` instead of
`\alpha` is because

`\alpha` represents the area under the normal curve where the confidence level interval (
`1-\alpha` ) did not cover which include both the left and right tail while

`(\alpha )/(2)` is just the area of one tail which what we required to calculate the sample size

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

Generally the sample size is mathematically evaluated as

`n = [ \frac{Z_{(\alpha )/(2) }}{E} ]^2 * \r p (1- \r p)`

Where
`\r p` is the proportion of sample taken which we will assume to be
`\r p = 0.5`

substituting values

`n = [( 1.96)/(0.02) ]^2 *( 0.5 (1- 0.5)`

`n = 2401`