Reading proficiency: An educator wants to construct a 99% confidence interval for the proportion of elementary school children in Colorado who are proficient in reading. (a) The results of a recent statewide test suggested that the proportion is 0.63. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.06?(b) Estimate the sample size needed if no estimate of  is available.

To construct a 99% confidence interval for the proportion of elementary school children in Colorado who are proficient in reading with a margin of error (E) of 0.06, we need to determine the required sample size. Let's break this problem into two parts: one with an estimated proportion and one without.

### (a) Sample Size with an Estimated Proportion

Given:
- Estimated proportion (\(\hat{p}\)) = 0.63
- Margin of error (E) = 0.06
- Confidence level = 99%

For a 99% confidence level, the Z-score (Z) corresponding to the critical value is approximately 2.576.

The formula for the margin of error for a proportion is:

\[ E = Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Rearranging this formula to solve for the sample size \( n \):

\[ n = \left( \frac{Z^2 \hat{p} (1 - \hat{p})}{E^2} \right) \]

Substitute the given values into the formula:

\[ n = \left( \frac{2.576^2 \times 0.63 \times (1 - 0.63)}{0.06^2} \right) \]

Calculating the values:

\[ n = \left( \frac{2.576^2 \times 0.63 \times 0.37}{0.06^2} \right) \]
\[ n = \left( \frac{6.635776 \times 0.63 \times 0.37}{0.0036} \right) \]
\[ n = \left( \frac{1.541830592}{0.0036} \right) \]
\[ n \approx 428.287 \]

Since the sample size must be a whole number, we round up to the next whole number:

\[ n \approx 429 \]

### (b) Sample Size without an Estimated Proportion

If no estimate of the proportion (\(\hat{p}\)) is available, we use 0.5 for the proportion because it maximizes the sample size, making our calculation conservative and ensuring the margin of error is within the specified limit.

Given:
- \(\hat{p}\) = 0.5
- Margin of error (E) = 0.06
- Confidence level = 99%

Using the same formula for the margin of error and solving for \( n \):

\[ n = \left( \frac{Z^2 \hat{p} (1 - \hat{p})}{E^2} \right) \]

Substitute the values:

\[ n = \left( \frac{2.576^2 \times 0.5 \times (1 - 0.5)}{0.06^2} \right) \]

Calculating the values:

\[ n = \left( \frac{2.576^2 \times 0.5 \times 0.5}{0.06^2} \right) \]
\[ n = \left( \frac{6.635776 \times 0.25}{0.0036} \right) \]
\[ n = \left( \frac{1.658944}{0.0036} \right) \]
\[ n \approx 460.818 \]

Again, rounding up to the next whole number:

\[ n \approx 461 \]

### Summary
(a) The required sample size with an estimated proportion of 0.63 is approximately 429.

(b) The required sample size without an estimated proportion is approximately 461.

Would you like further details or have any questions about this process?

### Additional Questions
1. What is the impact of the confidence level on the sample size?
2. How does the margin of error affect the required sample size?
3. Why do we use 0.5 as the estimate when no prior proportion is available?
4. What are some other methods to estimate the sample size for proportions?
5. How would the sample size change if we used a 95% confidence interval instead?
6. Can the sample size formula be applied to other confidence levels and margins of error?
7. What is the importance of rounding up the sample size?
8. How does the variability in the population affect the sample size calculation?

### Tip
Always use the most conservative estimate (usually 0.5) when no prior information is available to ensure that the sample size is large enough to meet the desired margin of error.

Learn how to construct a 99% confidence interval for the proportion of elementary school children proficient in reading in Colorado using statistical methods. Includes calculations for sample size with and without an estimated proportion.

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