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Unformatted text preview: z: use sample SD = 120 SEsum = sqrt(50)*120=848.5 SEavg = 848.5/50 = 17 cf=0.71, SEavg w/out rep = 17*0.71 = 12 z=(585-550)/12 = 2.9 p-value: 0.002% conclusion: a) reject null at 1% level b) doesn't look like chance, difference looks real, or something similar c)looks like English grad students have higher average than others Use the sample SD since SD of box is unknown; box is just English grad students. 3) null: a) the difference can be explained by chance b) all bags have an average of 24 cookies alternative: a) can't be explained by chance b) bags have average lower than 24 t: SD = 2.74 SD+ 3.06 SEsum = sqrt(5)*3.06 = 6.85 SEavg = 6.85/5 = 1.37 t=(21-24)/1.37 = 2.19 p-value: df=4, p<5% conclusion: a) reject null at 5% level b) doesn't seem to be (or look like) chance c) bags look like they have smaller average than 24...
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- Fall '08