Question 1186596
Here's how to calculate the margin of error for this poll:

1. **Identify the key values:**

*   Sample size (n) = 580
*   Sample proportion (p̂) = 0.22 (22% expressed as a decimal)
*   Confidence level = 95%

2. **Find the critical z-score:**

*   For a 95% confidence level, the alpha (α) is 1 - 0.95 = 0.05.
*   Since we are dealing with a two-tailed confidence interval, we divide alpha by 2: 0.05 / 2 = 0.025
*   The z-score corresponding to 0.025 in each tail (or 0.975 in the center) is 1.96. You can find this using a z-table or calculator.

3. **Calculate the standard error:**

*   Standard Error (SE) = sqrt[ (p̂ * (1 - p̂)) / n ]
*   SE = sqrt[ (0.22 * (1 - 0.22)) / 580 ]
*   SE = sqrt[ (0.22 * 0.78) / 580 ]
*   SE = sqrt(0.1716 / 580)
*   SE ≈ sqrt(0.00029586)
*   SE ≈ 0.0172

4. **Calculate the margin of error:**

*   Margin of Error (ME) = z * SE
*   ME = 1.96 * 0.0172
*   ME ≈ 0.0337

5. **Round to three decimals:**

*   ME ≈ 0.034

Therefore, the margin of error of this poll, at the 95% confidence level, is approximately 0.034.