Quiz6Sols

# Quiz6Sols - z use sample SD = 120 SEsum = sqrt(50*120=848.5...

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Quiz 6 Solutions 1) a.The average math SAT score of all 3,500 seniors at Sunshine High School is estimated as 710, and this estimate is likely to be off by 2.82 or so. SD = 60, SE_sum = sqrt(400)*60 = 1200, SE_avg = SE_sum/400 = 1200/400 = 3. SE_w/o = SE_w/ * cf, where cf = sqrt((3500-400)/(3500-1)) = sqrt(0.886). SE_w/o = 3*cf = 2.82 b. 710 +/- 1.4*2.82 (1) c.The 84% confidence interval is (about right) because the probability histogram for the sample average (follows) the normal curve. d. i.False. (No CI for statistic.) ii.False. (No chance with parameter.) iii.False. (SE is for sample average, not for spread of actual scores.) 2) null: a) the difference can be explained by chance b) English grad students have same average GRE as other Berkeley grad students alternative: a) can't be explained by chance b) English grad students have a higher average GRE
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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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