To find the 95% confidence interval for the true fraction $p$ of the population intending to vote for candidate A, we can use the formula for the confidence interval for a proportion. Here’s a step-by-step explanation:

1. Sample Proportion: $\hat{p} = \frac{x}{n} = \frac{457}{1000} = 0.457$ where $x$ is the number of individuals intending to vote for candidate A, and $n$ is the total number of individuals polled.

2. Standard Error (SE): The standard error for the sample proportion is given by: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.457(1 - 0.457)}{1000}} = \sqrt{\frac{0.457 \times 0.543}{1000}} \approx \sqrt{0.000248} \approx 0.0157$

3. Z-Score for 95% Confidence: For a 95% confidence level, the critical z-value is approximately 1.96 (from standard normal distribution tables).

4. Margin of Error (ME): The margin of error can be calculated as: $ME = z \times SE = 1.96 \times 0.0157 \approx 0.0308$

5. Confidence Interval: Now, we can construct the confidence interval: $\hat{p} - ME \quad \text{to} \quad \hat{p} + ME$ Thus, the confidence interval is: $(0.457 - 0.0308, 0.457 + 0.0308) = (0.4262, 0.4878)$

So, the 95% confidence interval for the true fraction $p$ of the population intending to vote for candidate A is approximately $(0.426, 0.488)$.

Summary

The 95% confidence interval for $p$ is $(0.426, 0.488)$.

Would you like more details on any specific step, or do you have any questions? Here are some related questions for further exploration:

1. How would the confidence interval change if the sample size were larger?
2. What assumptions are made when using this confidence interval method?
3. How does the choice of confidence level affect the width of the interval?
4. Can you explain the law of large numbers in more detail?
5. What are the implications of the results if the true proportion is outside this interval?

Tip: Always consider the sample size and variability when interpreting confidence intervals, as larger samples generally provide more reliable estimates.