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Homework #6 - 1800-1 1 3 Exercise 4.6'data.frame 16 obs of...

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1. Exercise 4.2 a)Welch Two Sample t-test data: age by group t = 8.0288, df = 16.626, p-value = 4.058e-07 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 17.55963 30.10704 sample estimates: mean in group 1 mean in group 3 49.75000 25.91667 b) Welch Two Sample t-test data: age by group t = 8.0288, df = 16.626, p-value = 4.058e-07 alternative hypothesis: true difference in means is not equal to 0 99 percent confidence interval: 15.20598 32.46069 sample estimates: mean in group 1 mean in group 3 49.75000 25.91667
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diff 800 1000 1200 1400 1600 1800 2. Exercise 4.4 'data.frame': 11 obs. of 3 variables: $ pre : num 5260 5470 5640 6180 6390 . .. $ post: num 3910 4220 3885 5160 5645 . .. $ diff: num 1350 1250 1755 1020 745 . ..
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qnorm diff 800 1000 1200 1400 1600
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Unformatted text preview: 1800-1 1 3. Exercise 4.6 'data.frame': 16 obs. of 3 variables: $ vas.active: int -167 -127 -58 -103 -35 -164 -3 25 -61 -45 . .. $ vas.plac : int -102 -39 32 28 16 -42 -27 -30 -47 8 . .. $ grp : int -65 -88 -90 -131 -51 -122 24 55 -14 -53 . .. grp 1.0 1.2 1.4 1.6 1.8 2.0 grp-100-50 50 It shifts it to left and makes the treatments negative. Therefore we would use the H1:MD<MDo hypothesis. And we cannot use a pair t test because the group variable had an effect on the other two variables. 4. Exercise 4.7 num [1:1000] 0.121 0.600 0.441 0.376 0.687 . .. qunif p 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 qnorm 0.0 0.2 0.4 0.6 0.8 1.0-2 2 A very few amount of the p-values are less than .005. About 5% should be less than .005....
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Homework #6 - 1800-1 1 3 Exercise 4.6'data.frame 16 obs of...

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