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### 27-4soln

Course: MAT 108, Fall 2009
School: SUNY Albany
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Solution. :4. SE of sum of 1971 sample scores = 14 x 40 = 560 SE of average of 1971 sample scores = (560/1600) x 100% = 14/40 = 7/20 = 35/100 = .35 SE of sum of 1975 sample scores = same as 1971's = .35 (The box SD is bootstraped by the sample SD in both cases) SE of the difference = the square root (.35)^2 of + (.35)^2 = .49 z = ( 1975...

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Solution. :4. SE of sum of 1971 sample scores = 14 x 40 = 560 SE of average of 1971 sample scores = (560/1600) x 100% = 14/40 = 7/20 = 35/100 = .35 SE of sum of 1975 sample scores = same as 1971's = .35 (The box SD is bootstraped by the sample SD in both cases) SE of the difference = the square root (.35)^2 of + (.35)^2 = .49 z = ( 1975 ave - 1971 ave - 0 )/ SE of difference = (68.5 - 67.2)/.49 = 1.2/.49 = 2.6 The area under the normal curve from -2.6 to 2.6 is 99.07%, so the area to the right of 2.6 is .93/2 = .465%. .465/100 = 4.65/1000 = 5/1000 (approx) = 1/200
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:4. Solution. SE of sum of 1971 sample scores = 14 x 40 = 560 SE of average of 1971 sample scores = (560/1600) x 100% = 14/40 = 7/20 = 35/100 = .35 SE of sum of 1975 sample scores = same as 1971's = .35 (Th
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