To calculate the margin of error for a poll, we can use the formula for the margin of error (ME) in a proportion, which is:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( z \) is the z-score corresponding to the desired confidence level (typically 1.96 for a 95% confidence level),
- \( p \) is the proportion of respondents who support a candidate (in this case, 0.12 for candidate A),
- \( n \) is the total number of respondents (in this case, 450).
Step 1: Determine variables
- \( p = 0.12 \)
- \( n = 450 \)
- \( z \) for a 95% confidence level = 1.96
Step 2: Calculate the standard error (SE)
\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.12 \times (1 - 0.12)}{450}} = \sqrt{\frac{0.12 \times 0.88}{450}} \]
Calculating \( 0.12 \times 0.88 \):
\[ 0.12 \times 0.88 = 0.1056 \]
Now substitute into the equation:
\[ SE = \sqrt{\frac{0.1056}{450}} = \sqrt{0.000235778} \approx 0.0154 \]
Step 3: Calculate the margin of error (ME)
\[ ME = z \times SE = 1.96 \times 0.0154 \approx 0.0302 \]
Step 4: Convert the margin of error to percentage
To convert the margin of error to a percentage, multiply by 100:
\[ ME \approx 0.0302 \times 100 = 3.02% \]
Step 5: Round to the nearest tenth of a percent
Rounding 3.02% to the nearest tenth of a percent gives us 3.0%.
Thus, the margin of error for this poll is approximately 3.0%.