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.