Question

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.

Answer 1

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).