Respuesta :
Answer:
The 98% confidence interval for the true population proportion of all New York students who favor year round school is (0.3083, 0.4383).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
A survey of 300 students from a New York State School District reveals that 112 favor year round school with two week vacations every ten weeks.
This means that [tex]n = 300, \pi = \frac{112}{300} = 0.3733[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3733 - 2.327\sqrt{\frac{0.3733*0.6267}{300}} = 0.3083[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3733 + 2.327\sqrt{\frac{0.3733*0.6267}{300}} = 0.4383[/tex]
The 98% confidence interval for the true population proportion of all New York students who favor year round school is (0.3083, 0.4383).
