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Answer:
The 80% confidence interval for the proportion of all college students who drove a car the day before the survey was conducted is (0.479, 0.621).
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 80 college students showed that 44 had driven a car during the day before the survey was conducted.
This means that [tex]n = 80, \pi = \frac{44}{80} = 0.55[/tex]
80% confidence level
So [tex]\alpha = 0.2[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.2}{2} = 0.9[/tex], so [tex]Z = 1.28[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 - 1.28\sqrt{\frac{0.55*0.45}{80}} = 0.479[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 + 1.28\sqrt{\frac{0.55*0.45}{80}} = 0.621[/tex]
The 80% confidence interval for the proportion of all college students who drove a car the day before the survey was conducted is (0.479, 0.621).
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