To calculate the margin of error for a poll, you can use the formula for the margin of error (ME) at a 95% confidence level, which is given by:
\[ ME = \frac{Z \cdot \sqrt{p(1-p)}}{\sqrt{n}} \]
where:
- \( Z \) is the Z-score (approximately 1.96 for a 95% confidence level),
- \( p \) is the proportion of respondents supporting candidate A (0.38 in this case),
- \( n \) is the total number of respondents (1,325).
Now we can plug in the values:
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Calculate \( p(1-p) \): \[ p(1-p) = 0.38 \times (1 - 0.38) = 0.38 \times 0.62 = 0.2356 \]
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Take the square root: \[ \sqrt{p(1-p)} = \sqrt{0.2356} \approx 0.4853 \]
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Calculate the margin of error: \[ ME = \frac{1.96 \cdot 0.4853}{\sqrt{1325}} \]
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Calculate \( \sqrt{1325} \): \[ \sqrt{1325} \approx 36.4 \]
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Now plug it back into the ME formula: \[ ME = \frac{1.96 \cdot 0.4853}{36.4} \approx \frac{0.9511}{36.4} \approx 0.0261 \]
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Convert to percentage: \[ ME \approx 0.0261 \times 100 \approx 2.61% \]
Finally, rounding to the nearest tenth of a percent: \[ \text{Margin of error} \approx 2.6% \]
Thus, the margin of error for this poll is 2.6%.