Question

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct a 95% condence interval for the proportion of water specimens that contain detectable levels of lead

Answer 1

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead.

This means that
`n = 42, \pi = (26)/(42) = 0.619`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.619 - 1.96\sqrt{(0.619*0.381)/(42)} = 0.472`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.619 + 1.96\sqrt{(0.619*0.381)/(42)} = 0.766`

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).