186k views
5 votes
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. Using the data, construct the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.

User Alok Nayak
by
8.5k points

1 Answer

2 votes

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then


n = 1537, \pi = (353)/(1537) = 0.2297

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.2297 - 1.96\sqrt{(0.2297*0.7703)/(1537)} = 0.2087

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2297 + 1.96\sqrt{(0.2297*0.7703)/(1537)} = 0.2507

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

User Rampr
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.