Respuesta :
Answer:
The 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol is (0.2425, 0.3113).
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, p = \frac{180}{650} = 0.2769[/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}} = 0.2769 - 1.96\sqrt{\frac{0.2769*0.7231}{650}} = 0.2425[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2769 + 1.96\sqrt{\frac{0.2769*0.7231}{650}} = 0.3113[/tex]
The 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol is (0.2425, 0.3113).
