To find a 95% confidence interval for the population proportion $p$ of passing scores, we can use the following formula:

$\hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $\hat{p}$ = sample proportion = $\frac{225}{300}$
• $n = 300$ is the sample size
• $Z$ is the Z-value for a 95% confidence level, which is 1.96.

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Step 1: Calculate the sample proportion $\hat{p}$

$\hat{p} = \frac{225}{300} = 0.750$

Step 2: Calculate the standard error (SE)

$SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.750(1 - 0.750)}{300}}$ $SE = \sqrt{\frac{0.750 \times 0.250}{300}} = \sqrt{\frac{0.1875}{300}} = \sqrt{0.000625} = 0.025$

Step 3: Calculate the margin of error (ME)

$ME = Z \times SE = 1.96 \times 0.025 = 0.049$

Step 4: Calculate the confidence interval

$\hat{p} \pm ME = 0.750 \pm 0.049$

• Lower limit: $0.750 - 0.049 = 0.701$
• Upper limit: $0.750 + 0.049 = 0.799$

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Final Answers:

• Lower limit: $0.70$
• Upper limit: $0.80$

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Would you like further clarifications or have any additional questions?

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Here are 5 related questions you could explore:

1. What happens to the confidence interval if the confidence level is increased to 99%?
2. How does sample size impact the width of the confidence interval?
3. What is the interpretation of a confidence interval in the context of a population proportion?
4. Can we assume normality when constructing confidence intervals for proportions?
5. How would the interval change if only 150 scores were passing instead of 225?

Tip: Confidence intervals provide a range of plausible values for a population parameter, but they are not guarantees!