Reading proficiency: An educator wants to construct a $90 \%$ confidence interval for the proportion of elementary school children in Colorado who are proficient in reading. Part 1 of 3 (a) The results of a recent statewide test suggested that the proportion is 0.66 . Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.04 ? A sample of 380 elementary school children is needed to obtain a $90 \%$ confidence interval with a margin of error of 0.04 . Part: $1 / 3$ Part 2 of 3 (b) Estimate the sample size needed if no estimate of $p$ is available. A sampte of $\square$ etementary school children is needed to obtain a $90 \%$ confidence interval with a margin of error of 0.04 .

Step 1 ：An educator wants to construct a $90 \%$ confidence interval for the proportion of elementary school children in Colorado who are proficient in reading.

Step 2 ：The results of a recent statewide test suggested that the proportion is 0.66 .

Step 3 ：The educator wants the confidence interval to have a margin of error of 0.04 .

Step 4 ：To calculate the sample size needed, we use the formula \(n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}\), where \(Z\) is the z-score, \(p\) is the proportion, and \(E\) is the margin of error.

Step 5 ：Substituting the given values into the formula, we get \(n = \frac{(1.645)^2 \cdot 0.66 \cdot (1-0.66)}{(0.04)^2}\).

Step 6 ：Calculating the above expression, we find that \(n = 380\).

Step 7 ：Since we can't have a fraction of a person, we round up to the nearest whole number.

Step 8 ：Final Answer: The sample size needed so that the confidence interval will have a margin of error of 0.04 is \(\boxed{380}\).