Let's solve this step-by-step.

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1. Find the Sample Proportion

The sample proportion $\hat{p}$ is given by:

$\hat{p} = \frac{x}{n}$

Where:

• $x = 338$ (number of voters for the candidate)
• $n = 650$ (total number of likely voters)

$\hat{p} = \frac{338}{650} \approx 0.520$

Thus, the sample proportion is:

$\hat{p} = 0.520$

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2. Approximate the Margin of Error for a 95% Confidence Level

The formula for the margin of error (ME) for a 95% confidence level is:

$\text{ME} = z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $z = 1.96$ (critical value for 95% confidence)
• $\hat{p} = 0.520$ (sample proportion)
• $n = 650$ (sample size)

Plugging in the values:

$\text{ME} = 1.96 \times \sqrt{\frac{0.520 \times (1 - 0.520)}{650}}$

Let's calculate this.

$0.520 \times (1 - 0.520) = 0.520 \times 0.480 = 0.2496$

$\sqrt{\frac{0.2496}{650}} = \sqrt{0.000384} \approx 0.0196$

$\text{ME} = 1.96 \times 0.0196 \approx 0.0384$

Thus, the margin of error is approximately:

$\text{ME} = 0.038$

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3. Find the 95% Confidence Interval

The 95% confidence interval is given by:

$\hat{p} \pm \text{ME}$

$0.520 \pm 0.038$

Calculating the lower and upper bounds:

• Lower Bound: $0.520 - 0.038 = 0.482$
• Upper Bound: $0.520 + 0.038 = 0.558$

Convert these to percentages:

• $0.482 \times 100\% = 48.2\%$
• $0.558 \times 100\% = 55.8\%$

Thus, the 95% confidence interval is:

$(48.2\%, 55.8\%)$

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Final Answers

1. Sample Proportion: $0.520$
2. Margin of Error: $\pm 0.038$
3. 95% Confidence Interval: $(48.2\%, 55.8\%)$

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Do you need further clarification or details on any part of the solution?

Related Questions:

1. How do you interpret the confidence interval in terms of polling data?
2. What happens to the margin of error if the sample size increases?
3. What is the formula for the margin of error at a 99% confidence level?
4. How does the sample proportion affect the width of the confidence interval?
5. What assumptions are made when calculating the margin of error?

Tip:

When calculating a confidence interval, always ensure your sample size is large enough for the normal approximation to be valid (typically $n \hat{p} \geq 5$ and $n (1 - \hat{p}) \geq 5$).