Question

During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 75 of the residences had reduced their water consumption over that of the previous year. Find a 95% confidence interval for the proportion of residences that reduced their water consumption. Round the answers to three decimal places.

Answer 1

Answer:

The 95% confidence interval for the proportion of residences that reduced their water consumption is (0.665, 0.835).

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 $1 - \frac{\alpha}{2}$.

Sampled 100 residential water bills and found that 75 of the residences had reduced their water consumption over that of the previous year.

This means that $n = 100, \pi = \frac{75}{100} = 0.75$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.96\sqrt{\frac{0.75*0.25}{100}} = 0.665$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.96\sqrt{\frac{0.75*0.25}{100}} = 0.835$

The 95% confidence interval for the proportion of residences that reduced their water consumption is (0.665, 0.835).