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Answer :
Final answer:
The 95% confidence interval for the population mean based on a sample mean of 492.3 and a standard deviation of 37.6 with 18 subjects is approximately (472.62, 511.98).
Explanation:
To construct a 95% confidence interval for the population mean when the sample mean is 492.3 and the standard deviation is 37.6 with a sample size of 18, you would typically use the t-distribution because the sample size is small (n < 30) and the population standard deviation is unknown.
Looking up the t-value with 17 degrees of freedom (n-1) for a 95% confidence interval, we get a value of approximately 2.110. The formula for the confidence interval is:
CI = ± t * (s / √{n})
Where CI is the confidence interval, t is the t-value for the desired confidence level, s is the sample standard deviation, and n is the sample size. Plugging in the values:
CI = ± (2.110) * (37.6 / √{18})
We find the margin of error to be:
Margin of Error = 2.110 * (37.6 / √{18}) ≈ 19.68
Thus, the 95% confidence interval is:
(492.3 - 19.68, 492.3 + 19.68) ≈ (472.62, 511.98)
We estimate with 95 percent confidence that the true population mean for all such proficiency tests is between 472.62 and 511.98.
Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean for this type of exam score.
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