
Reading proficiency: An educator wants to construct a 99.8% 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 0.61. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.04?

To construct a 99.8% confidence interval for the proportion of elementary school children in Colorado who are proficient in reading with a margin of error (E) of 0.04, we need to determine the required sample size.

### Given:
- Estimated proportion (\(\hat{p}\)) = 0.61
- Margin of error (E) = 0.04
- Confidence level = 99.8%

### Steps:

1. **Determine the Z-score** for a 99.8% confidence level.

For a 99.8% confidence level, the Z-score (Z) corresponding to the critical value is approximately 3.090.

2. **Use the formula** for the margin of error for a proportion and solve for the sample size \( n \):

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

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:

\[ n = \left( \frac{3.090^2 \times 0.61 \times (1 - 0.61)}{0.04^2} \right) \]

Calculating the values:

\[ n = \left( \frac{3.090^2 \times 0.61 \times 0.39}{0.04^2} \right) \]
\[ n = \left( \frac{9.5481 \times 0.61 \times 0.39}{0.0016} \right) \]
\[ n = \left( \frac{2.27383379}{0.0016} \right) \]
\[ n \approx 1421.15 \]

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

\[ n \approx 1422 \]

### Summary:
The required sample size with an estimated proportion of 0.61, a 99.8% confidence level, and a margin of error of 0.04 is approximately 1422.

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

### Additional Questions
1. How does increasing the confidence level affect the required sample size?
2. What would the sample size be if the margin of error were 0.02 instead of 0.04?
3. How does the variability in the estimated proportion affect the sample size calculation?
4. What are the implications of using a higher confidence level on the interpretation of the results?
5. How does rounding the sample size up ensure the margin of error is met?
6. Can this method be used for other confidence levels, such as 95% or 99%?
7. Why is it important to have a precise estimate of the proportion when calculating sample size?
8. How would the sample size change if the estimated proportion were 0.5 instead of 0.61?

### Tip
When dealing with high confidence levels, be prepared for larger sample sizes. Higher confidence requires more data to ensure that the interval estimate is accurate.

Learn how to construct a 99.8% confidence interval for the proportion of elementary school children in Colorado proficient in reading. Calculate the required sample size with a margin of error of 0.04 using the estimated proportion of 0.61. Detailed step-by-step solution included.

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