To solve this problem, we need to determine the minimum sample size, $n$, required to estimate the mean completion time with a 99% confidence level and a margin of error of $E = 2.0$ minutes. Given information includes:

• Population standard deviation, $\sigma = 10.3$ minutes
• Confidence level, $99\%$
• Margin of error, $E = 2.0$ minutes

Step-by-Step Solution

1. Identify the Z-score for a 99% confidence level. For a 99% confidence level, the Z-score (critical value) associated with this level is approximately $z = 2.576$.

2. Apply the sample size formula for estimating a population mean. The formula to determine the sample size $n$ when estimating a mean is:

   $n = \left( \frac{z \cdot \sigma}{E} \right)^2$

   Plugging in the known values:

   $n = \left( \frac{2.576 \cdot 10.3}{2.0} \right)^2$

3. Calculate the value inside the parentheses first.

   $\frac{2.576 \cdot 10.3}{2.0} = \frac{26.5472}{2.0} = 13.2736$

4. Square the result.

   $n = (13.2736)^2 \approx 176.1689$

5. Round up to the nearest whole number. Since we need the minimum sample size, we round up to the next whole number.

   $n = 177$

Answer

The minimum sample size needed is 177.

Would you like further details on this calculation, or do you have any questions?

Here are 5 related questions for practice:

1. How would the required sample size change if the confidence level was reduced to 95%?
2. What is the impact of increasing the margin of error on the sample size?
3. How is the sample size calculation affected if the population standard deviation is not known?
4. If the margin of error is reduced to 1 minute, what would the new sample size be?
5. How does changing the sample size affect the confidence interval width?

Tip: Remember, increasing confidence levels or reducing margin of error both increase required sample sizes, reflecting more stringent conditions for precision.