To solve this problem, we need to construct a 99% confidence interval for the mean score based on the information provided.

Given dataTo solve this problem, we need to construct a 99% confidence interval for the mean score based on the information provided.

Given

• Standard deviation ($\sigma$) = 10
• Sample size ($n$) = 60
• Sample mean ($\bar{x}$) = 70
• Confidence level = 99%

Steps to construct the confidence interval:

1. Determine the Z-score for a 99% confidence interval. The Z-score for a 99% confidence level (which leaves 0.5% in each tail of the normal distribution) is approximately 2.576.

2. Calculate the standard error of the mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{60}}$

3. Compute the margin of error (ME): $\text{ME} = Z \times \text{SEM} = 2.576 \times \frac{10}{\sqrt{60}}$

4. Determine the confidence interval:

   • Lower limit = $\bar{x} - \text{ME}$
   • Upper limit = $\bar{x} + \text{ME}$

Let's calculate each step.The 99% confidence interval for the true mean of all test scores is:

• Lower limit: 66.7
• Upper limit: 73.3

Would you like a detailed explanation of each calculation step, or do you have any other questions?

Here are some related questions to deepen your understanding:

1. How would the confidence interval change if we used a 95% confidence level instead?
2. What impact does increasing the sample size have on the width of the confidence interval?
3. Why do we use the Z-score and not the T-score in this case?
4. How would a larger standard deviation affect the confidence interval?
5. Can we construct a confidence interval if the standard deviation is unknown?

Tip: When constructing confidence intervals, remember that higher confidence levels yield wider intervals, as they cover a larger range to ensure the mean lies within the interval with greater certainty.