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Unformatted text preview: Introduction to confidence intervals – Examples (with solutions) 1. The temperature in degrees F of 20 individuals was measured: 101.8 98.3 95.9 97.5 99.4 97.3 98.2 103.8 101.0 97.2 100.2 100.3 96.8 96.5 97.9 99.4 97.8 93.5 99.7 99.2 The observed sample mean is ¯ x = 98 . 585 degrees F and the observed sample standard deviation is s = 2 . 277181 degrees F. (a) Compute a 90% confidence interval for μ , the population mean temperature (i.e. the mean of the distribution used to model temperature of individuals in a large population). State all assumptions made and interpret the interval. Solution: We must assume that the temperature measurements x 1 ,...,x 20 are a realization of a random sample from N ( μ,σ ). A confidence level of 90% corresponds to α = 1 . 90 = 0 . 10, thus t 1 α/ 2 ,n 1 = t . 95 , 19 = qt (0 . 95 , 19) = 1 . 729133. Plugging in the known quantities to the formula, ¯ x ± t 1 α/ 2 ,n 1 s √ n = 98 . 585 ± 1 . 729133 2 . 277181 √ 20 = (97 . 70454 , 99 . 46546) Our interpretation is that we are 90% confident that μ , the population distribution mean temperature, is between 97.70 degrees and 99.47 degrees. (b) If three of the individuals who’s temperature was measured are a father and his two sons, explain why it may be unreasonable to assume that x 1 ,...,x 20 are a realization of a random sample from a distribution (e.g. from N ( μ,σ )). Solution: The three temperature measurements (suppose they are the first three we measured): x 1 ,x 2 ,x 3 are a realization of X 1 ,X 2 ,X 3 which may be dependent random variables, since a father and his two sons may have strongly associated characteristics, such as temperature (e.g. knowledge that the father’s temperature is less 95 degrees F may impact the uncertainty about the yettobe measured sons’ temperatures). This would imply that x 1 ,...,x 20 are not a realization of a random sample, since X 1 ,X 2 ,X 3 would be dependent....
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 Spring '08
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 Standard Deviation, Confidence Level

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