solpracfinalfall2011

solpracfinalfall2011 - Solutions to Practice Final;...

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Solutions to Practice Final; Statistics 371; Fall 2011; Professor Wardrop 1. First we note that in the collapsed table ˆ p 1 =0 . 34 is smaller than ˆ p 2 =0 . 356 .Fo r Simpson’s Paradox to occur, this inequality must be reversed in both component tables. In the frst component table, we need c/ 100 < 60 / 200 or c< 30 or c 29 . In the second component table, we need d/ 150 < 42 / 100 or d< 63 or d 62 . But we also need the component tables to be consistent with the collapsed table; i.e. we need c + d =89 . So, we have three conditions: c 29 , d 62 and c + d =89 .Cana l lthreecond i t ions be satisfed? Yes; there are three possible answers: ( c =29 ; d =60 ) ; ( c =28 ; d = 61) ;and ( c =27; d =62) . 2. We note that in the collapsed table ˆ p 1 = 0 . 362 is smaller than ˆ p 2 =0 . 42 .ForS imp- son’s Paradox to occur, this inequality must be reversed in both component tables. In the frst component table, we need c/ 35 < 21 / 75 or c< 9 . 8 or c 9 . In the second component table, we need d/ 65 < 26 / 55 or d< 30 . 7 or d 30 . But we also need the component tables to be consistent with the collapsed table; i.e. we need c + d =42 . So, we have three conditions: c 9 , d 30 and c + d =42 .C a na l lt h r e ec o n d i - tions be satisfed? No, because the largest c + d can be and still keep both reversals is 39. On the exam it would su±fce to say: It is impossible ±or c 9 , d 30 and c + d =42 ;Iw i l lassumeyouknowwhy . (You won’t have time ±or long essay an- swers on the exam.) 3. I will defne population 1 to be ±emales and population 2 to be males. Assume inde- pendent random samples ±rom these pop- ulations. Based on the data given, the 95% CI ±or p 1 - p 2 is (0 . 407 - 0 . 364) ± 1 . 96 ± 0 . 407(0 . 593) 450 + 0 . 364(0 . 636) 538 = 0 . 043 ± 1 . 96(0 . 0311) = 0 . 043 ± 0 . 061 = [ - 0 . 018 , 0 . 104] . 4. I will defne population 1 to be males and population 2 to be ±emales. Assume inde- pendent random samples ±rom these popu- lations. Based on the data given, the 95% CI ±or p 1 - p 2 is (0 . 257 - 0 . 173) ± 1 . 96 ± 0 . 257(0 . 743) 537 + 0 . 173(0 . 827) 446 = 0 . 084 ± 1 . 96(0 . 0260) = 0 . 084 ± 0 . 051 = [0 . 033 ,
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6. (a) The frequency of the interval is 5. (b) It cannot be a relative frequency his- togram because a relative frequency cannot exceed one. (c) The relative frequency of the interval is its area: 5(0 . 10) = 0 . 50 .Thu s ,i t s frequency is 0
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solpracfinalfall2011 - Solutions to Practice Final;...

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