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:

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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

This means that [tex]n = 150, \pi = \frac{48}{150} = 0.32[/tex]

95% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{150}} = 0.2453[/tex]

The upper limit of this interval is:

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

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

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Salary Data
77600
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90700
97200
90700
101800
78700
81300
84200
97600
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89400
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78700
84600
87700
103400
83800
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94700
69200
95400
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68800

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