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The text in the image is a bit blurred, but here is what I can extract:

"The mean score for all students, p(hat), is estimated as 0.60. The sampling distribution is approximately normal with a standard deviation of 0.03. Construct and interpret a 95% confidence interval for p, the true proportion of students who agree."

This question involves constructing a confidence interval for a population proportion \(p\), based on the sample proportion \( \hat{p} = 0.60 \) and the standard deviation of the sampling distribution, which is given as \( 0.03 \).

### Step-by-step solution:

1. **Determine the z-value for a 95% confidence interval:**
   For a 95% confidence interval, the critical value (z*) corresponds to approximately 1.96, based on the standard normal distribution.

2. **Compute the margin of error (ME):**
   \[
   ME = z^* \times \text{standard deviation}
   = 1.96 \times 0.03 = 0.0588
   \]

3. **Construct the confidence interval:**
   The confidence interval is given by:
   \[
   \hat{p} \pm ME
   = 0.60 \pm 0.0588
   \]

   This gives:
   \[
   (0.60 - 0.0588, 0.60 + 0.0588)
   = (0.5412, 0.6588)
   \]

4. **Interpretation:**
   We are 95% confident that the true proportion of students who agree lies between 54.12% and 65.88%.

Would you like further details on any step? 

Here are 5 related questions for further understanding:
1. What is the importance of using a 95% confidence level in this context?
2. How do we choose the z-value for different confidence levels?
3. Why is the sampling distribution assumed to be normal?
4. How would the confidence interval change with a larger sample size?
5. What does the margin of error represent in practical terms?

**Tip:** Always ensure that the sample size is large enough for the Central Limit Theorem to hold, ensuring that the sampling distribution of the sample proportion is approximately normal.

The mean score for all students, p(hat), is estimated as 0.60. The sampling distribution is approximately normal with a standard deviation of 0.03. Construct and interpret a 95% confidence interval for p, the true proportion of students who agree.

Learn how to construct a 95% confidence interval for the true proportion of students who agree, using a sample proportion of 0.60 and a standard deviation of 0.03. Understand how to calculate the margin of error and apply the normal distribution.

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