To calculate the margin of error (E), we use the formula for the margin of error in estimating a population proportion:

$E = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Step 1: Calculate the sample proportion ($\hat{p}$)

$\hat{p} = \frac{x}{n} = \frac{1672}{2388} \approx 0.7007$

Step 2: Find the critical value ($z$) for a 95% confidence level

For a 95% confidence level, the $z$-value is approximately 1.96 (from standard normal distribution tables).

Step 3: Calculate the margin of error

Substitute the values into the formula: $E = 1.96 \cdot \sqrt{\frac{0.7007(1 - 0.7007)}{2388}}$

First, calculate the variance: $0.7007 \cdot (1 - 0.7007) = 0.7007 \cdot 0.2993 \approx 0.2098$

Next, divide by the sample size: $\frac{0.2098}{2388} \approx 0.0000879$

Take the square root: $\sqrt{0.0000879} \approx 0.00938$

Multiply by the $z$-value: $E = 1.96 \cdot 0.00938 \approx 0.0184$

Final Answer:

The margin of error is approximately: $E \approx 0.0184 \, \text{or} \, 1.84\%$

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Would you like a breakdown of each step in more detail or have further questions?

Related Questions:

1. How is the critical $z$-value determined for different confidence levels?
2. What happens to the margin of error if the sample size increases?
3. How does the sample proportion ($\hat{p}$) influence the margin of error?
4. Can this formula be used for very small sample sizes? Why or why not?
5. How do confidence levels higher than 95% affect the margin of error?

Tip:

Always double-check whether the sample size is large enough for the approximation to hold true; $n \cdot \hat{p} \geq 5$ and $n \cdot (1 - \hat{p}) \geq 5$ are common guidelines.