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