To determine the approximate 90% confidence interval for the proportion of students needing external exams, we can use the formula for the confidence interval of a proportion.

Given:

• Proportion ($\hat{p}$) = 64% = 0.64
• Sample size ($n$) = 100
• Confidence level = 90%

First, we need to find the critical value ($z$) for a 90% confidence interval. For a 90% confidence level, the critical value ($z$) is approximately 1.645 (from the standard normal distribution table).

The formula for the confidence interval is: $\hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Let's calculate this step by step:

1. Calculate the standard error (SE): $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.64 \times (1 - 0.64)}{100}} = \sqrt{\frac{0.64 \times 0.36}{100}} = \sqrt{0.2304 / 100} = \sqrt{0.002304} \approx 0.048$

2. Calculate the margin of error (ME): $ME = z \times SE = 1.645 \times 0.048 \approx 0.079$

3. Calculate the confidence interval: $\text{Lower limit} = \hat{p} - ME = 0.64 - 0.079 = 0.561$ $\text{Upper limit} = \hat{p} + ME = 0.64 + 0.079 = 0.719$

Therefore, the approximate 90% confidence interval for the proportion of students that need external exams is $[0.561, 0.719]$.

Would you like any further details or have any questions?

Here are 5 numbered questions you might consider next:

1. How do you calculate the critical value for different confidence levels?
2. What is the standard error in the context of confidence intervals?
3. How do you interpret the confidence interval in a real-world context?
4. What assumptions are made when calculating confidence intervals for proportions?
5. How does increasing the sample size affect the width of the confidence interval?

Tip: When determining the confidence interval, the choice of confidence level (e.g., 90%, 95%) affects the critical value and, consequently, the width of the interval. A higher confidence level results in a wider interval.