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.