Respuesta :
Answer:
Step 1: The estimate the proportion of tenth graders reading at or below the eighth grade level is 0.2.
Step 2: The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).
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].
Suppose a sample of 2845 tenth graders is drawn. Of the students sampled, 2276 read above the eighth grade level.
So 2845 - 2276 = 569 read below, and the estimate of the proportion of tenth graders reading at or below the eighth grade level is:
[tex]\pi = \frac{569}{2845} = 0.2[/tex], and the answer to step 1 is 0.2.
The sample size is [tex]n = 2845[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 2.575\sqrt{\frac{0.2*0.8}{2845}} = 0.181[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 2.575\sqrt{\frac{0.2*0.8}{2845}} = 0.219[/tex]
The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).
