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. Step 2 of 2 : Suppose a sample of 1291 tenth graders is drawn. Of the students sampled, 1098 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. Round your answers to three decimal places.

Answer 1

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).

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:

1291 tenth graders, 1098 read above the eighth grade level.

1291 - 1098 = 193 read at or below this level.

We want the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.

So
`n = 1291, \pi = (193)/(1291) = 0.149`

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)} = 0.149 - 1.96\sqrt{(0.149*0.851)/(1291)} = 0.13`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.149 - 1.96\sqrt{(0.149*0.851)/(1291)} = 0.168`

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.13, 0.168).