To calculate the margin of error (E) for the confidence interval, we can use the following formula:

$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $t_{\alpha/2}$ is the critical value from the t-distribution corresponding to the confidence level of 99%,
• $s = 0.00567 \, \text{lb}$ is the sample standard deviation,
• $n = 36$ is the sample size.

Step 1: Determine the critical t-value ($t_{\alpha/2}$)

For a 99% confidence level and $n-1 = 36-1 = 35$ degrees of freedom, we can find the critical t-value using a t-distribution table or a calculator. The critical value for a 99% confidence level with 35 degrees of freedom is approximately:

$t_{\alpha/2} \approx 2.724$

Step 2: Calculate the margin of error

Now, plug the values into the formula:

$E = 2.724 \times \frac{0.00567}{\sqrt{36}}$

Simplifying the expression:

$E = 2.724 \times \frac{0.00567}{6} = 2.724 \times 0.000945 = 0.002574 \, \text{lb}$

Thus, the margin of error is approximately:

$E = 0.00257 \, \text{lb}$

Would you like a detailed explanation of how this result was derived?

Here are 5 questions to expand on this topic:

1. How does the sample size affect the margin of error in confidence intervals?
2. What is the relationship between confidence level and the width of the confidence interval?
3. How is the critical t-value different from the z-value in confidence intervals?
4. Why do we use the t-distribution instead of the z-distribution for small sample sizes?
5. How can we interpret the margin of error in practical terms?

Tip: Always ensure that the degrees of freedom are correctly calculated when using the t-distribution, as it significantly impacts the critical value for small sample sizes.