In a randomly selected group of 650 automobile deaths, 180 were alcohol related. Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.

a. [.243, .311]
b. [.262, .292]
c. [.259, .294]
d. [.263, .291]

Answer :

Answer:

a. [.243, .311]

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].

For this problem, we have that:

In a randomly selected group of 650 automobile deaths, 180 were alcohol related. This means that [tex]n = 650, \pi = \frac{180}{650} = 0.277[/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}} = 0.277 - 1.96\sqrt{\frac{0.277*0.723}{650}} = 0.243[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.277 + 1.96\sqrt{\frac{0.277*0.723}{650}}{119}} = 0.311[/tex]

So the correct answer is:

a. [.243, .311]

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