To compute the 90% confidence interval for the mean IQ score, we need the following information from the problem:

• Sample mean ($\bar{x}$) = 105.1
• Population standard deviation ($\sigma$) = 10
• Sample size ($n$) = 73
• Critical value for 90% confidence (from the Z-distribution table) = 1.645 (already provided)

Formula for the confidence interval:

$CI = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$ Where:

• $\bar{x}$ = sample mean
• $Z_{\alpha/2}$ = critical value
• $\sigma$ = population standard deviation
• $n$ = sample size

Step 1: Calculate the margin of error (MOE):

$MOE = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 1.645 \times \frac{10}{\sqrt{73}}$ $MOE \approx 1.645 \times 1.169 = 1.922$

Step 2: Compute the confidence interval:

$CI = 105.1 \pm 1.922$ This gives the confidence interval: $CI \approx (103.18, 107.02)$

Thus, the 90% confidence interval for the mean IQ score is approximately $(103.18, 107.02)$.

Would you like further details or have any questions? Here are a few questions that might help expand on this:

1. How would the confidence interval change if the confidence level were 95%?
2. How does the sample size affect the width of the confidence interval?
3. Why is the population standard deviation used instead of the sample standard deviation?
4. What would happen if the sample size was much smaller, like 30?
5. How is the critical value determined for different confidence levels?

Tip: The margin of error shrinks as the sample size increases, making the confidence interval narrower.