Respuesta :
Answer:
The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).
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:
972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so [tex]n = 972, \pi = \frac{272}{972} = 0.28[/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} = 0.925[/tex], so [tex]Z = 1.44[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301[/tex]
The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).
