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

Answer:

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

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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 $n = 1291, \pi = \frac{193}{1291} = 0.149$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 - 1.96\sqrt{\frac{0.149*0.851}{1291}} = 0.13$

The upper limit of this interval is:

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

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