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Let's go through each part of this problem step-by-step.

### Given Data
- Sample size, \( n = 500 \)
- Number who planned to vote for a candidate, \( x = 275 \)
- Sample proportion, \( \hat{p} = \frac{x}{n} = \frac{275}{500} = 0.55 \)
  
The table provides \( z \)-values for different confidence levels.

### Part (a): Error in Estimation \( E \) with 95% Confidence
The error \( E \) is calculated as:
\[
E = z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
For 95% confidence, \( z_{\alpha/2} = 1.960 \).

Substitute values:
\[
E = 1.960 \cdot \sqrt{\frac{0.55 \cdot (1 - 0.55)}{500}}
\]
\[
= 1.960 \cdot \sqrt{\frac{0.55 \cdot 0.45}{500}}
\]
\[
= 1.960 \cdot \sqrt{\frac{0.2475}{500}}
\]
\[
= 1.960 \cdot \sqrt{0.000495}
\]
\[
= 1.960 \cdot 0.02225
\]
\[
\approx 0.0436
\]

Thus, the error \( E \approx 0.0436 \).

### Part (b): 95% Confidence Interval for \( p \)
The 95% confidence interval for the true proportion \( p \) is given by:
\[
\hat{p} - E \leq p \leq \hat{p} + E
\]

Substitute values:
\[
0.55 - 0.0436 \leq p \leq 0.55 + 0.0436
\]
\[
0.5064 \leq p \leq 0.5936
\]

So, the 95% confidence interval is approximately:
\[
0.5064 \leq p \leq 0.5936
\]

### Part (c): Sample Size \( n_{90} \) for 90% Confidence with Same Error \( E \)
To find the sample size required for a 90% confidence level with the same error \( E \), we use:
\[
n = \left( \frac{z_{\alpha/2}^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2} \right)
\]
For 90% confidence, \( z_{\alpha/2} = 1.645 \).

Substitute values with \( E = 0.0436 \) and \( \hat{p} = 0.55 \):
\[
n = \frac{(1.645)^2 \cdot 0.55 \cdot 0.45}{(0.0436)^2}
\]
\[
= \frac{2.706025 \cdot 0.2475}{0.00190176}
\]
\[
\approx \frac{0.6697}{0.00190176}
\]
\[
\approx 352
\]

Thus, for the same error \( E \) with 90% confidence, a sample size of approximately \( n = 352 \) voters is required.

---

### Summary
- **(a)** Error \( E \approx 0.0436 \)
- **(b)** 95% Confidence Interval: \( 0.5064 \leq p \leq 0.5936 \)
- **(c)** Required sample size for 90% confidence: \( n \approx 352 \)

Would you like further clarification or details on any part? 

### Related Questions:
1. How would the confidence interval change if the sample size increased?
2. What if the proportion \( \hat{p} \) was closer to 0.5; how would that affect the margin of error?
3. Why is \( z_{\alpha/2} \) different for each confidence level?
4. How would you calculate the confidence interval for a different confidence level, like 99%?
5. What assumptions must be met for this method to provide accurate confidence intervals?

### Tip
When calculating confidence intervals, using a larger sample size will generally reduce the margin of error, giving a narrower interval.

In a poll of 500 randomly selected voters, 275 planned to vote for a certain candidate. (a) With 95% confidence, what is the error in estimation? (b) What is the 95% confidence interval for voter proportion? (c) For the same error in estimation, how many voters should be sampled for 90% confidence?

Learn how to calculate the confidence interval and required sample size for a voter poll. This problem involves determining the margin of error and confidence interval at 95% confidence and adjusting sample size for 90% confidence, using a sample of 500 voters with 275 in favor of a candidate.

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