STAT 571 Final: STAT 571 UW Madison Fall 03Final Solutions
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Brief solutions: this is a case of paired samples. use t = (a) for testing h0 : d = 0 versus ha : d 6= 0, Since n = 5, d = 8. 5/5 = 1. 7, and sd = p(22. 56 5 1. 72)/4 = 2. 03 = Compare with t4, the p-value = 2 p (t4 . There is weak evidence of change in bacteria count. /2 = 0. 025, = 1 0. 95 = 0. 05, z0. 025 = 1. 96, z0. 05 = 1. 645, a 0 = 1. 5, we have n = (1. 96+1. 645)2 (z /2+z )2 ( a 0)2 . (1. 5)2. 3. (a) let p1 denote the preference of bs1 for young leaves. Then 1 p1 is the the preference of bs1 for old leaves. 0. 5 versus h0 : p1 6= 0. 5 and the test statistic is. Since n1 = 60, p1 = 38/60 = Compare with standard normal distribution, the p-value = 2 p (z 2. 06) < 0. 05.
