Answer :
Final Answer:
The 95% confidence interval for the population mean proficiency score, based on a sample mean of 492.3 and a standard deviation of 37.6 for 18 subjects, is (478.7, 505.9).
Explanation:
To establish a 95% confidence interval for the population mean (\(\mu\)), we use the formula:
[tex]\[ \text{Confidence Interval} = \bar{X} \pm Z \left( \frac{s}{\sqrt{n}} \right) \][/tex]
Where:
- [tex]\(\bar{X}\)[/tex] is the sample mean (492.3),
- s is the standard deviation (37.6),
- n is the sample size (18),
- Z is the Z-score corresponding to the desired confidence level.
For a 95% confidence interval, the Z-score is approximately 1.96. Now, substitute the values into the formula:
[tex]\[ \text{Confidence Interval} = 492.3 \pm 1.96 \left( \frac{37.6}{\sqrt{18}} \right) \][/tex]
Calculate the margin of error:
[tex]\[ 1.96 \times \frac{37.6}{\sqrt{18}} \approx 13.6 \][/tex]
Finally, construct the confidence interval:
[tex]\[ 492.3 - 13.6 \approx 478.7 \][/tex]
[tex]\[ 492.3 + 13.6 \approx 505.9 \][/tex]
Thus, the 95% confidence interval for the population mean proficiency score is (478.7, 505.9), indicating that we can be 95% confident that the true population mean falls within this range.