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Answer :
Final Answer:
The 95% confidence interval for the population mean proficiency score, based on the given sample of eighteen subjects with a mean of [tex]\(_\bar{X} = 492.3\)[/tex] and a standard deviation of (s = 37.6), is (481.1, 503.5).
Explanation:
To construct the 95% confidence interval for the population mean [tex](\(_\mu\)),[/tex] we can use the formula:
[tex]\[\bar{X} \pm Z \left(\frac{s}{\sqrt{n}}\right)\][/tex]
where [tex]\(\bar{X}\)[/tex] is the sample mean, (s) is the sample standard deviation, (n) is the sample size, and (Z) is the critical value from the standard normal distribution corresponding to the desired confidence level. For a 95% confidence interval, (Z) is approximately 1.96.
Given that [tex]\(_\bar{X} = 492.3\), \(_s = 37.6\)[/tex], and the sample size (n) is 18, we can substitute these values into the formula:
[tex]\[492.3 \pm 1.96 \left(\frac{37.6}{\sqrt{18}}\right)\][/tex]
After performing the calculations, we get the 95% confidence interval (481.1, 503.5). This interval suggests that we can be 95% confident that the true population mean proficiency score falls within this range based on the sample data.
In conclusion, this confidence interval provides a range of values within which we can reasonably estimate the population mean. The margin of error is influenced by both the sample size and the variability of the data, and the 95% confidence level indicates the level of certainty associated with the interval.
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