Question

1. Verizon Wireless would like to estimate the proportion of households that use cell phones for their phone service without a land line. A random sample of 150 households was selected and 48 relied strictly on cell phones for their service. Based on the sample, construct a 95% confidence interval for the true proportion of households which rely strictly on cell phones for their phone service.

Answer 1

The 95% confidence interval for the true proportion of households which rely strictly on cell phones for their phone serviceis (0.2453, 0.3947).

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)`.

A random sample of 150 households was selected and 48 relied strictly on cell phones for their service.

This means that
`n = 150, \pi = (48)/(150) = 0.32`

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)} = 0.32 - 1.96\sqrt{(0.32*0.68)/(150)} = 0.2453`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.32 + 1.96\sqrt{(0.32*0.68)/(150)} = 0.3947`

The 95% confidence interval for the true proportion of households which rely strictly on cell phones for their phone serviceis (0.2453, 0.3947).