Respuesta :
Answer:
The margin of error associated with a 90% confidence interval that estimates the proportion of them that are involved in an after school activity is 0.0764.
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
A simple random sample of 49 8th graders at a large suburban middle school indicated that 88% of them are involved with some type of after school activity.
This means that [tex]n = 49, \pi = 0.88[/tex]
Margin of error:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.645\sqrt{\frac{0.88*0.12}{49}}[/tex]
[tex]M = 0.0764[/tex]
The margin of error associated with a 90% confidence interval that estimates the proportion of them that are involved in an after school activity is 0.0764.
