Respuesta :
Answer:
The 85% confidence interval for the population proportion of people who black out at G forces greater than 66 is (0.261, 0.319)
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:
[tex]n = 518, p = \frac{150}{518} = 0.29[/tex]
85% confidence level
So [tex]\alpha = 0.15[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.15}{2} = 0.9250[/tex], so [tex]z = 1.44[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.29 - 1.44\sqrt{\frac{0.29*0.71}{518}} = 0.261[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.29 + 1.44\sqrt{\frac{0.29*0.71}{518}} = 0.319[/tex]
The 85% confidence interval for the population proportion of people who black out at G forces greater than 66 is (0.261, 0.319)
