Question 1178703: Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
Of 82 adults selected randomly from one town, 66 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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%)
|
|
|
