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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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

This means that $n = 42, \pi = \frac{26}{42} = 0.619$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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