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. (a) The results of a recent statewide test suggested that the proportion is 0.71. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.03?(b) Estimate the sample size needed if no estimate of  is available.

To construct a 90% confidence interval for the proportion of elementary school children in Colorado who are proficient in reading, we can use the formula for the margin of error in a proportion confidence interval:

\[ E = z \cdot \sqrt{\frac{p(1-p)}{n}} \]

where:
- \( E \) is the margin of error.
- \( z \) is the z-score corresponding to the confidence level.
- \( p \) is the estimated proportion.
- \( n \) is the sample size.

### Part (a): Using the estimate \( p = 0.71 \)

1. **Identify the z-score for a 90% confidence interval**:
   
   The z-score corresponding to a 90% confidence interval is approximately 1.645.

2. **Set up the margin of error equation**:

   Given \( E = 0.03 \), \( p = 0.71 \), and \( z = 1.645 \), we plug these values into the margin of error formula:
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.71(1-0.71)}{n}} \]

3. **Solve for \( n \)**:

   First, simplify the equation:
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.71 \cdot 0.29}{n}} \]
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.2059}{n}} \]

   Square both sides to eliminate the square root:
   
   \[ (0.03)^2 = (1.645)^2 \cdot \frac{0.2059}{n} \]
   
   \[ 0.0009 = 2.705025 \cdot \frac{0.2059}{n} \]

   Rearrange to solve for \( n \):
   
   \[ n = \frac{2.705025 \cdot 0.2059}{0.0009} \]
   
   \[ n = \frac{0.556731645}{0.0009} \]
   
   \[ n \approx 618.59 \]

   Since the sample size must be a whole number, round up to the next whole number:
   
   \[ n \approx 619 \]

### Part (b): No estimate available

When no estimate for \( p \) is available, we use \( p = 0.5 \) because this maximizes the sample size (providing the most conservative estimate).

1. **Set up the margin of error equation**:

   Given \( E = 0.03 \), \( p = 0.5 \), and \( z = 1.645 \):
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.5(1-0.5)}{n}} \]
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.25}{n}} \]

2. **Solve for \( n \)**:

   First, simplify the equation:
   
   \[ 0.03 = 1.645 \cdot \sqrt{\frac{0.25}{n}} \]
   
   Square both sides to eliminate the square root:
   
   \[ (0.03)^2 = (1.645)^2 \cdot \frac{0.25}{n} \]
   
   \[ 0.0009 = 2.705025 \cdot \frac{0.25}{n} \]

   Rearrange to solve for \( n \):
   
   \[ n = \frac{2.705025 \cdot 0.25}{0.0009} \]
   
   \[ n = \frac{0.67625625}{0.0009} \]
   
   \[ n \approx 751.4 \]

   Since the sample size must be a whole number, round up to the next whole number:
   
   \[ n \approx 752 \]

### Summary

- (a) The sample size needed using the estimate \( p = 0.71 \) is approximately \( 619 \).
- (b) The sample size needed with no estimate available is approximately \( 752 \).

Would you like more details or have any questions?

Here are 8 related questions to expand your understanding:

1. How does the margin of error change with the sample size?
2. Why do we use \( p = 0.5 \) when no estimate is available?
3. How does the confidence level affect the sample size?
4. What is the impact of the population size on the required sample size?
5. Can the sample size formula be applied to other confidence intervals (e.g., 95%, 99%)?
6. How would the sample size change if the margin of error is reduced to 0.02?
7. What assumptions are made when using this sample size formula?
8. How does the variability in the population affect the sample size?

**Tip:** Always round up the sample size to ensure the desired confidence level and margin of error are maintained.

Learn how to construct a 90% confidence interval for the proportion of elementary school children in Colorado proficient in reading. Calculate sample sizes with and without prior estimates using real-world data.

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