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A study conducted by an airline showed that a random sample of nine of its passengers disembarking at the Cleveland airport, took an average of 24.1 minutes to claim their luggage. From a previous survey it was willing to assume that time to claim luggage is normally distributed with a variance of 18 (min)2. A 95% confidence interval for the mean time to claim one's luggage has endpoints. Answer 24.1 ± 8.32 24.1 ± 3.92 24.1 ± 2.77 24.1 ± 3.26 24.1 ± 9.78
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95% CI for the MEAN is = mean +- 2 SE approx SE = standard error of the mean SE = std dev/sqrt(number in sample) SE = sqrt(18)/sqrt(9) = sqrt(2) 95%CI = mean + 2sqrt(2) approx. = 24.1 + 2.77 from options given
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