Question 1178703
Let's construct the 90% confidence interval for the population proportion.

**1. Define the Variables:**

* **Sample size (n):** 82
* **Number of adults with health insurance (x):** 66
* **Sample proportion (p̂):** x / n = 66 / 82 ≈ 0.8049
* **Confidence level:** 90%

**2. Find the Z-score:**

* For a 90% confidence interval, the z-score (zα/2) is 1.645 (from a z-table or calculator).

**3. Calculate the Standard Error (SE):**

* SE = √[p̂(1 - p̂) / n]
* SE = √[0.8049(1 - 0.8049) / 82]
* SE = √[0.8049(0.1951) / 82]
* SE = √[0.157036 / 82]
* SE = √0.001915073
* SE ≈ 0.04376

**4. Calculate the Margin of Error (ME):**

* ME = zα/2 * SE
* ME = 1.645 * 0.04376
* ME ≈ 0.07198

**5. Calculate the Confidence Interval:**

* Lower Bound: p̂ - ME = 0.8049 - 0.07198 ≈ 0.73292
* Upper Bound: p̂ + ME = 0.8049 + 0.07198 ≈ 0.87688

**6. Express the Confidence Interval:**

* The 90% confidence interval for the true proportion of adults in the town who have health insurance is approximately (0.7329, 0.8769).

**In percentage form:**

* (73.29%, 87.69%)