High School

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Eighteen subjects are randomly selected and given proficiency tests. The mean of this group is 492.3, and the standard deviation is 37.6. Construct the 95% confidence interval for the population mean.

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.

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