SOLUTION: In a recent poll, 580 people were asked if they liked dogs, and 22% said they did. Find the margin of error of this poll, at the 95% confidence level. Give your answer to three

Algebra ->  Statistics  -> Confidence-intervals -> SOLUTION: In a recent poll, 580 people were asked if they liked dogs, and 22% said they did. Find the margin of error of this poll, at the 95% confidence level. Give your answer to three       Log On


   



Question 1186596: In a recent poll, 580 people were asked if they liked dogs, and 22% said they did. Find the margin of error of this poll, at the 95% confidence level.
Give your answer to three decimals

Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
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.