Question

The state education commission wants to estimate the fraction of 10th grade students that have reading skills at or below the eighth grade level. And an earlier study, the population proportion was estimated to be 0.23. How large of a sample would be required in order to estimate the fraction of 10th graders reading at or below the eighth grade level at the 99% confidence level with an error of at most 0.02? Round answer up to the next integer

Answer 1

2938

Explanation:

Given:

Proportion (p) = 0.23

Margin of error (E) = 0.02

At 99% confidence level the z is:

`\begin{gathered} \alpha=1-99\% \\ \\ \alpha=1-0.99=0.01 \\ \\ \alpha\text{/2}=(0.01)/(2)=0.005 \\ \\ \text{ The corresponding z value is:} \\ \\ Z_{\alpha\text{/2}}=2.576 \end{gathered}`

We can determine the value of the sample size using the following formula:

`\begin{gathered} E=Z_{\alpha\text{/2}}\cdot\sqrt{(p\cdot(1-p))/(n)} \\ \\ \text{ We replacing:} \\ \\ 0.02=2.576\cdot\sqrt{(0.23\cdot(1-0.23))/(n)} \\ \\ \sqrt{(0.23\left(1-0.23\right))/(n)}=(0.02)/(2.576) \\ \\ (0.23\left(1-0.23\right))/(n)=\left((0.02)/(2.576)\right)^2 \\ \\ (1)/(n)=(\left((0.02)/(2.576)\right)^2)/(0.1771) \\ \\ n=(0.1771)/(\left((0.02)/(2.576)\right)^2) \\ \\ n=2937.99\approx2938 \end{gathered}`

The sample size is 2938