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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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