A poll reported 60% support for a statewide election with a margin of error of 2.31 percentage points.

How many voters should be sampled for a 99% confidence interval, rounded to the nearest whole number?

(Use [tex]\( z^* = 2.576 \)[/tex] for 99% confidence.)

Example: 1234 voters



Answer :

To determine the number of voters that should be sampled for a 99% confidence interval with a margin of error of 2.31 percentage points when the support is reported to be 60%, we can follow these steps:

1. Identify the given values:
- Support percentage [tex]\( p = 0.60 \)[/tex] (since 60% is the support percentage)
- Margin of error [tex]\( E = 0.0231 \)[/tex] (since 2.31 percentage points is the margin of error)
- For a 99% confidence level, the z-score [tex]\( z^* = 2.576 \)[/tex]

2. Recall the formula for the margin of error:
The margin of error for a proportion is given by:
[tex]\[ E = z^* \cdot \sqrt{\frac{p(1-p)}{n}} \][/tex]
where [tex]\( n \)[/tex] is the sample size.

3. Rearrange the formula to solve for [tex]\( n \)[/tex]:
[tex]\[ n = \left( \frac{z^* \cdot \sqrt{p(1-p)}}{E} \right)^2 \][/tex]

4. Substitute the given values into the formula:
[tex]\[ n = \left( \frac{2.576 \cdot \sqrt{0.60 \cdot (1-0.60)}}{0.0231} \right)^2 \][/tex]

5. Calculate the intermediate values:
- [tex]\( p \cdot (1-p) = 0.60 \cdot 0.40 = 0.24 \)[/tex]
- [tex]\( \sqrt{0.24} = 0.4899 \)[/tex]
- [tex]\( \frac{2.576 \cdot 0.4899}{0.0231} = 54.6401 \)[/tex]

6. Square the result:
[tex]\[ n = 54.6401^2 = 2984.5509 \][/tex]

7. Round to the nearest whole number:
[tex]\[ n \approx 2985 \][/tex]

Therefore, to achieve a 99% confidence interval with a margin of error of 2.31 percentage points for a reported support percentage of 60%, you should sample approximately 2985 voters.

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