Final Answer:
the 95% confidence interval for the mean score, μ, of all students taking the test is c. 23.5<μ<25.7. Thus the correct option is C.
Step-by-step explanation:
The formula for a 95% confidence interval for the mean (\(\mu\)) can be calculated using the formula:
\[ \bar{x} \pm Z \left(\frac{s}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean,
- \(s\) is the standard deviation,
- \(n\) is the sample size,
- \(Z\) is the Z-score corresponding to the desired confidence level.
Given that \(\bar{x} = 24.6\), \(s = 3.3\), \(n = 30\), and for a 95% confidence interval, the Z-score is approximately 1.96 (commonly used for a normal distribution):
\[ 24.6 \pm 1.96 \left(\frac{3.3}{\sqrt{30}}\right) \]
Now, calculate the margin of error:
\[ 1.96 \times \left(\frac{3.3}{\sqrt{30}}\right) \approx 1.96 \times 0.6 \approx 1.176 \]
Finally, calculate the confidence interval:
\[ 24.6 - 1.176 \text{ to } 24.6 + 1.176 \]
This results in the confidence interval: \(23.424 < \mu < 25.776\), which can be rounded to \(23.5 < \mu < 25.8\). Therefore, option c is the correct 95% confidence interval.