To find the margin of error $E$ for a population proportion $p$, given a confidence level and sample size, we can use the formula:

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

where:

• $z$ is the z-score corresponding to the desired confidence level,
• $p$ is the sample proportion (success rate),
• $n$ is the sample size.

Step-by-Step Solution

1. Identify values from the question:

   • Confidence level: 98%
   • Sample size $n = 800$
   • Sample proportion $p = 0.40$

2. Find the z-score for a 98% confidence level:

   • For a 98% confidence level, the z-score is approximately $z = 2.3263$.

3. Plug values into the formula:

   $E = 2.3263 \times \sqrt{\frac{0.40 \times (1 - 0.40)}{800}}$

4. Calculate step-by-step:

   • Calculate $p(1 - p) = 0.40 \times 0.60 = 0.24$.
   • Calculate $\frac{p(1 - p)}{n} = \frac{0.24}{800} = 0.0003$.
   • Take the square root: $\sqrt{0.0003} \approx 0.0173$.
   • Multiply by the z-score: $2.3263 \times 0.0173 \approx 0.0404$.

Answer:

The margin of error $E$ is approximately 0.0404.

Correct Option:

The correct answer is:

• 0.0404

Would you like a more detailed breakdown or have any questions?

Related Questions:

1. What is the margin of error for a 95% confidence level with the same sample?
2. How would the margin of error change if the sample size increased?
3. What happens to the margin of error if the success rate is closer to 50%?
4. How can we interpret the margin of error in a confidence interval?
5. How do we determine the z-score for other confidence levels?

Tip: Remember that a higher confidence level or smaller sample size will increase the margin of error, making the interval wider.