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.