Respuesta :
Answer:
A 95% confidence interval estimate for the population proportion is (0.6570, 0.7096).
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 interval [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 survey of 1,200 airline travelers, 820 responded that the airline fee that is most unreasonable is additional charges to redeem points/miles. This means that [tex]n = 1200, \pi = \frac{820}{1200} = 0.6833[/tex]
95% confidence interval
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}} = 0.6833 - 1.96\sqrt{\frac{0.6833*0.3167}{1200}} = 0.6570[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6833 + 1.96\sqrt{\frac{0.6833*0.3167}{1200}} = 0.7096[/tex]
A 95% confidence interval estimate for the population proportion is (0.6570, 0.7096).
