Question

find the margin of error and 95% confidence interval for the survey result described. according to a poll of 1374 people, about one - third (32%) of americans keep a dog for protection. margin of error = % (round to the nearest whole number as needed.)

Explanation:

Step1: Recall Margin of Error Formula

For a proportion, the margin of error (ME) for a 95% confidence interval is given by \( ME = z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( z_{\alpha/2} = 1.96 \) (for 95% confidence), \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.
Here, \( \hat{p} = 0.32 \), \( n = 1374 \).

Step2: Calculate the Standard Error

First, calculate \( \hat{p}(1 - \hat{p}) = 0.32 \times (1 - 0.32) = 0.32 \times 0.68 = 0.2176 \).
Then, the standard error (SE) is \( \sqrt{\frac{0.2176}{1374}} \).
\( \frac{0.2176}{1374} \approx 0.00015837 \), and \( \sqrt{0.00015837} \approx 0.01258 \).

Step3: Calculate Margin of Error

Multiply the standard error by \( z_{\alpha/2} = 1.96 \):
\( ME = 1.96 \times 0.01258 \approx 0.02465 \).
Convert to a percentage: \( 0.02465 \times 100 \approx 2.465\% \), round to the nearest whole number is 2%.

Answer:

2