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A random sample of 80 college students showed that 44 had driven a car during the day before the survey was conducted. Suppose that we are interested in forming a 80 percent confidence interval for the proportion of all college students who drove a car the day before the survey was conducted.

Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places.

User Jdbs
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7 votes

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).

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 80 college students showed that 44 had driven a car during the day before the survey was conducted.

This means that
n = 80, \pi = (44)/(80) = 0.55

80% confidence level

So
\alpha = 0.2, z is the value of Z that has a pvalue of
1 - (0.2)/(2) = 0.9, so
Z = 1.28.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.55 - 1.28\sqrt{(0.55*0.45)/(80)} = 0.479

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.55 + 1.28\sqrt{(0.55*0.45)/(80)} = 0.621

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).

User Ian Warwick
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