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

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

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)`.

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 = (75)/(100) = 0.75`

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)} = 0.75 - 1.96\sqrt{(0.75*0.25)/(100)} = 0.665`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.75 + 1.96\sqrt{(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).