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. In an earlier study, the population proportion was estimated to be 0.21. How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

Answer 1

Answer:

We need a sample size of at least 383.

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 $1 - \frac{\alpha}{2}$.

The margin of error is:

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

85% confidence level

So $\alpha = 0.15$, z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$, so $Z = 1.44$.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with $M = 0.03, \pi = 0.21$

Then

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

$0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}$

$n = 382.23$

Rounding up

We need a sample size of at least 383.