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To solve this, let's break down the information given:

1. **Problem Statement**:
   - A sample of 1000 voters was selected.
   - Out of these, 153 were against the school bond measure.
   - We need to find the sample proportion \( \hat{p} \) and calculate the margin of error \( E \) for a 90% confidence interval.

2. **Sample Proportion (\( \hat{p} \))**:
   \[
   \hat{p} = \frac{153}{1000} = 0.153
   \]

3. **Confidence Interval**:
   - For a 90% confidence level, the \( z \)-value (from the table) is \( z = 1.645 \).

4. **Margin of Error (\( E \))**:
   The margin of error for a population proportion is calculated using:
   \[
   E = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
   \]
   where:
   - \( z = 1.645 \)
   - \( \hat{p} = 0.153 \)
   - \( n = 1000 \)

   Plugging in these values:
   \[
   E = 1.645 \cdot \sqrt{\frac{0.153 \times (1 - 0.153)}{1000}}
   \]
   Calculating inside the square root:
   \[
   E = 1.645 \cdot \sqrt{\frac{0.153 \times 0.847}{1000}}
   \]
   \[
   E = 1.645 \cdot \sqrt{0.000129591}
   \]
   \[
   E = 1.645 \cdot 0.01138
   \]
   \[
   E \approx 0.0187
   \]

So, the margin of error \( E \) for a 90% confidence interval is approximately **0.0187**.

Would you like more details on any part of this calculation?

Here are some additional questions related to this topic:
1. How would the confidence interval change if the confidence level were increased to 95%?
2. What effect does increasing the sample size have on the margin of error?
3. How can the sample proportion \( \hat{p} \) help in determining population trends?
4. Why is it essential to use the correct \( z \)-value for each confidence level?
5. How would the calculation differ if we were dealing with a population mean rather than a proportion?

**Tip**: Always ensure the sample size is large enough when using normal approximation for proportions; generally, both \( n\hat{p} \) and \( n(1 - \hat{p}) \) should be greater than 5.

In a poll of 1000 randomly selected voters in a local election, 153 voters were against the school bond measure. What is the sample proportion, and what is the error in estimation for 90% confidence?

Learn how to calculate the sample proportion and margin of error for a 90% confidence interval in a poll of 1000 voters, with 153 voters against a school bond measure. This problem covers statistics concepts, including confidence intervals, sample proportions, and margin of error calculation.

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