To construct a 95% confidence interval for the true mean of all test scores, we use the formula for the confidence interval of the mean:

$\text{Confidence Interval} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right)$

Where:

• $\bar{x}$ is the sample mean,
• $z$ is the z-score corresponding to the desired confidence level,
• $\sigma$ is the population standard deviation,
• $n$ is the sample size.

Given:

• Sample mean ($\bar{x}$) = 60
• Population standard deviation ($\sigma$) = 12
• Sample size ($n$) = 70
• Confidence level = 95%

For a 95% confidence level, the corresponding z-score ($z$) is approximately 1.96.

Steps to Calculate:

1. Standard Error (SE):

$\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{70}}$

2. Margin of Error (ME):

$\text{ME} = z \times \text{SE} = 1.96 \times \frac{12}{\sqrt{70}}$

3. Confidence Interval:

$\text{Confidence Interval} = 60 \pm \text{ME}$

Let's compute these values.The 95% confidence interval for the true mean of all test scores is approximately $57.2$ to $62.8$.

Summary:

• Lower limit: 57.2
• Upper limit: 62.8

Would you like more details, or do you have any other questions?

Here are 8 related questions you might find interesting:

1. What is the significance of the standard deviation in calculating the confidence interval?
2. How does increasing the sample size affect the width of the confidence interval?
3. What would the confidence interval look like with a 99% confidence level?
4. What happens if the population standard deviation is unknown?
5. How would the confidence interval change if the sample mean was different?
6. Can you explain how to find the z-score for different confidence levels?
7. Why is the z-score for a 95% confidence level specifically 1.96?
8. What assumptions are made when calculating this confidence interval?

Tip: Confidence intervals provide a range of values within which the true population parameter is likely to fall, offering a way to quantify uncertainty in estimates.