Let's break this down step by step:

1. Sample Proportion (p):

The sample proportion $p$ represents the fraction of voters who are against the school bond measures. It is calculated as:

$p = \frac{\text{Number of voters against the measure}}{\text{Total number of voters in the sample}} = \frac{424}{1000} = 0.424$

So, the sample proportion $p$ is 0.424.

2. Margin of Error (m) for the 90% Confidence Level:

The margin of error for a sample proportion is given by the formula:

$m = Z_{\alpha/2} \times \sqrt{\frac{p(1-p)}{n}}$

Where:

• $Z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level (for 90% confidence, $Z_{\alpha/2} \approx 1.645$),
• $p$ is the sample proportion,
• $n$ is the sample size (1000 voters in this case).

Now, let's calculate:

1. $p(1 - p) = 0.424 \times (1 - 0.424) = 0.424 \times 0.576 = 0.244224$
2. $\frac{p(1 - p)}{n} = \frac{0.244224}{1000} = 0.000244224$
3. $\sqrt{0.000244224} = 0.01563$
4. $m = 1.645 \times 0.01563 = 0.0257$

Thus, the margin of error $m$ is approximately 0.0257 or 2.57%.

Summary:

• Sample Proportion $p$: 0.424
• Margin of Error $m$ at 90% Confidence Level: 0.0257 or 2.57%

Would you like to see further details on any of these steps, or have any other questions?

---

Related questions:

1. How does the margin of error change with different confidence levels (e.g., 95%)?
2. What would the sample proportion be if 476 voters were against the bond measure instead?
3. How would increasing the sample size affect the margin of error?
4. How do you calculate the critical value for a 95% or 99% confidence interval?
5. What happens to the margin of error if the sample proportion $p$ is very close to 0 or 1?

Tip: The margin of error decreases as the sample size increases, making your estimate more precise.