The problem you're working on asks for the margin of error (M.E.) for estimating a population mean $\mu$, given a sample of size $n = 5$, a mean of 44.4, a standard deviation of 6.1, and a confidence level of 99.5%.

Margin of Error Formula:

The margin of error for a sample mean is given by:

$M.E. = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

where:

• $Z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the confidence level,
• $\sigma$ is the sample's standard deviation,
• $n$ is the sample size.

Step 1: Determine the Critical Value ($Z_{\alpha/2}$)

For a confidence level of 99.5%, the corresponding $Z_{\alpha/2}$ value is approximately 2.807.

Step 2: Apply the Formula

Now plug in the values:

$M.E. = 2.807 \times \frac{6.1}{\sqrt{5}} = 2.807 \times \frac{6.1}{2.236}$

$M.E. \approx 2.807 \times 2.728 \approx 7.66$

Step 3: Round to One Decimal Place

The margin of error is 7.7 when rounded to one decimal place.

If you have any more questions or need further details, let me know!

Here are five related questions you can think about:

1. What happens to the margin of error if the sample size increases?
2. How does the confidence level affect the margin of error?
3. What is the significance of using the critical value in the formula?
4. How would the margin of error change if the standard deviation was smaller?
5. How can we interpret the margin of error in practical terms?

Tip: Always remember that higher confidence levels result in larger margins of error due to wider confidence intervals!