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To find the 90% confidence interval for the population proportion \( p \), we will use the formula for the confidence interval of a proportion:

\[
\hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
\]

### Step-by-step:

1. **Sample proportion**: 
   \[
   \hat{p} = \frac{105}{138} = 0.761
   \]
2. **Sample size**: 
   \( n = 138 \)
   
3. **Z-value for a 90% confidence interval**:
   The Z-value for a 90% confidence interval is 1.645 (from the Z-table).

4. **Standard error**:
   \[
   SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.761(1 - 0.761)}{138}} = \sqrt{\frac{0.761 \times 0.239}{138}} \approx \sqrt{0.001317} \approx 0.0363
   \]

5. **Margin of error**:
   \[
   ME = Z \times SE = 1.645 \times 0.0363 \approx 0.0597
   \]

6. **Confidence interval**:
   \[
   0.761 - 0.0597 < p < 0.761 + 0.0597
   \]
   \[
   0.701 < p < 0.821
   \]

Thus, the 90% confidence interval is \( 0.701 < p < 0.821 \).

Would you like further details or have any other questions?

### Additional related questions:
1. How does the sample size affect the width of the confidence interval?
2. What is the significance of the Z-value in the confidence interval formula?
3. How would the confidence interval change if the confidence level was 95% instead of 90%?
4. What is the impact of increasing the number of successes in the sample on the confidence interval?
5. Can the confidence interval include a proportion of 0 or 1?

**Tip**: Remember, increasing the confidence level widens the interval because you're more certain that the true proportion falls within that range.

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 138 with 105 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Calculate the 90% confidence interval for a population proportion based on a sample size of 138 with 105 successes. This problem involves statistical concepts like the confidence interval formula and the Central Limit Theorem. The solution shows how to compute the interval using standard error and margin of error.

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