To calculate the maximal margin of error for the 99% confidence interval, we can use the formula for the margin of error in a confidence interval when the population standard deviation is known:

$\text{Margin of Error} = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

Where:

• $Z_{\alpha/2}$ is the Z-value corresponding to the confidence level (99% in this case),
• $\sigma$ is the population standard deviation,
• $n$ is the sample size.

Step 1: Find the Z-value

For a 99% confidence level, the Z-value corresponds to the critical value that leaves 0.5% of the probability in each tail (because 1% is split between the two tails of the normal distribution). The Z-value for 99% confidence is approximately 2.576.

Step 2: Plug the values into the formula

• Population standard deviation $\sigma = 8.8$ ounces,
• Sample size $n = 25$.

$\text{Margin of Error} = 2.576 \times \frac{8.8}{\sqrt{25}}$ $\text{Margin of Error} = 2.576 \times \frac{8.8}{5}$ $\text{Margin of Error} = 2.576 \times 1.76$ $\text{Margin of Error} \approx 4.53$

Final Answer:

The maximal margin of error is approximately 4.53 ounces.

Would you like more details on how this formula works or any further clarifications?

Additional Questions:

1. How do you determine the Z-value for different confidence levels?
2. What happens to the margin of error if the sample size increases?
3. How does the margin of error change when you use a sample standard deviation instead of the population standard deviation?
4. How would the margin of error be affected if the population standard deviation were larger?
5. What is the purpose of a confidence interval in statistics?

Tip:

When working with confidence intervals, it's always important to check if you know the population standard deviation. If not, you'll need to use the t-distribution instead of the normal distribution.