To solve this problem, we need to compute the margin of error (M.E.) using the given formula for estimating a population proportion:

$\text{M.E.} = z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $z^*$ is the critical value for a given confidence level.
• $\hat{p}$ is the sample proportion (success rate).
• $n$ is the sample size.

Given Data:

• Sample size $n = 275$
• Success rate $\hat{p} = 78.2\% = 0.782$
• Confidence level = 99% (corresponding $z^*$ value = 2.576)

Step 1: Compute $\hat{p}(1 - \hat{p})$

$\hat{p}(1 - \hat{p}) = 0.782 \times (1 - 0.782) = 0.782 \times 0.218 = 0.170476$

Step 2: Plug into the formula

$\text{M.E.} = 2.576 \times \sqrt{\frac{0.170476}{275}}$

Step 3: Simplify

$\frac{0.170476}{275} \approx 0.000619$ $\sqrt{0.000619} \approx 0.02488$ $\text{M.E.} = 2.576 \times 0.02488 \approx 0.06406$

Step 4: Convert to percentage

$\text{M.E.} = 0.06406 \times 100 = 6.406\%$

Step 5: Round to one decimal place

$\text{M.E.} \approx 6.4\%$

Thus, the margin of error is 6.4%.

Would you like further details on any part of the calculation?

Here are five related questions:

1. How do we find the critical value for different confidence levels?
2. How does increasing the sample size affect the margin of error?
3. What is the relationship between confidence level and margin of error?
4. Can this method be applied to other types of population parameters?
5. How is the formula different when we don't know the population proportion?

Tip: As the confidence level increases, the critical value $z^*$ increases, which also increases the margin of error.