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

Answer:

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

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

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

This means that $n = 150, \pi = \frac{48}{150} = 0.32$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{150}} = 0.2453$

The upper limit of this interval is:

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

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