Unformatted text preview: (60, 0.025) 4.8 2.00 1.116 75 S S ME t t n n= = 9 = . Hence A 95% confidence interval for the population mean, μ, is ( 29 ( 29 (3.8 1.116) 2.684 , 4.916 Estimate ME e = = However, your book accepts using the standard normal distribution, considering n = 74 as large. Then, /2 4.8 1.96 1.094 74 S ME z n α = = = and hence, A 95% confidence interval for the population mean, μ, is ( 29 ( 29 (3.8 1.094) 2.706 , 4.894 Estimate ME e = = d. A surveyor claims that the mean error is between 3.2 and 4.4 cm. At what level of confidence can this statement be made? [This is part c in the text.] We need to follow the above steps in reverse order and write (3.2,4.4) CI = hence /2 /2 4.8 (4.4 3.2) / 2 0.6 0.558 74 ME z z == = = . Then /2 0.6 1.07 0.558 z = = and P(Z > 1.07) = P(Z <  1.07) = 0.1423 = α/2. Therefore α = 0.2846 and 1 – α = 0.7154. So the CI is correct at 71.54% level of significance....
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 Spring '08
 Kyung
 Statistics, Normal Distribution, Ellipsoid Heights, O. Baykal

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