ssrs 2012 barcode font Nonparametric Tests in Software

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CHAPTER 10 Nonparametric Tests
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Step 1. Combining all 18 sample values in an array from the smallest to the largest gives us the first line of Table 10-8. These values are numbered 1 to 18 in the second line, which gives us the ranks. Step 2. To find the sum of the ranks for each sample, rewrite Table 10-2 by using the associated ranks from Table 10-8; this gives us Table 10-9. The sum of the ranks is 106 for alloy I and 65 for alloy II.
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Table 10-8
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10.7 11.8 12.6 12.9 14.1 14.7 15.2 15.9 16.1 16.4 17.8 18.3 18.9 19.6 20.5 22.7 24.2 25.3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
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Step 3. Since the alloy I sample has the smaller size, N1 ranks are R1 106 and R2 65. Then U mU Thus sU N1N2 2 N1N2 (8)(10) 2 N1(N1 2 40 1) R1 (8)(10)
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8 and N2
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10. The corresponding sums of the
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N1N2(N1 12
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(8)(9) 106 10 2 1) (8)(10)(19) 12
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11.25 and Z U mU sU 10 40 11.25 2.67
Table 10-9 Alloy I Cable Strength 18.3 16.4 22.7 17.8 18.9 25.3 16.1 24.2 Rank 12 10 16 11 13 18 9 17 Sum 106 Alloy II Cable Strength 12.6 14.1 20.5 10.7 15.9 19.6 12.9 15.2 11.8 14.7 Rank 3 5 15 1 8 14 4 7 2 6 Sum 65
Since the hypothesis H0 that we are testing is whether there is no difference between the alloys, a two-tailed test is required. For the 0.05 significance level, we have the decision rule: Accept H0 if 1.96 Reject H0 otherwise. Because z z 1.96.
2.67, we reject H0 and conclude that there is a difference between the alloys at the 0.05 level.
10.6. Verify results (6) and (7) of this chapter for the data of Problem 10.5.
(a) Since samples 1 and 2 yield values for U given by U1 U2 N1N2 N1N2 N1(N1 1) 2 N2(N2 1) 2 R1 R2 (8)(10) (8)(10) (8)(9) 106 2 (10)(11) 65 2 10 70
CHAPTER 10 Nonparametric Tests
we have U1 U2 10 70 N(N 2 80, and N1N2 1) (N1 106 65 N2 1) 171 and (18)(19) 2 171
N2)(N1 2
10.7. Work Problem 10.5 by using the statistic U for the alloy II sample.
For the alloy II sample, U so that Z U mU sU 70 40 11.25 2.67 N1N2 N2(N2 2 1) R2 (8)(10) (10)(11) 2 65 70
This value of z is the negative of the z in Problem 10.5, and the right-hand tail of the normal distribution is used instead of the left-hand tail. Since this value of z also lies outside 1.96 z 1.96, the conclusion is the same as that for Problem 10.5.
10.8. A professor has two classes in psychology: a morning class of 9 students, and an afternoon class of 12 students. On a final examination scheduled at the same time for all students, the classes received the grades shown in Table 10-10. Can one conclude at the 0.05 significance level that the morning class performed worse than the afternoon class Table 10-10 Morning class Afternoon class 73 86 87 81 79 84 75 88 82 90 66 85 95 84 75 92 70 83 91 53 84
1 2 (5
Step I. Table 10-11 shows the array of grades and ranks. Note that the rank for the two grades of 75 is 1 6) 5.5, while the rank for the three grades of 84 is 3(11 12 13) 12. Step 2. Rewriting Table 10-10 in terms of ranks gives us Table 10-12. Check: R1 73, R2 158, and N N(N 2 1) N1 N2 (21)(22) 2 9 12 231 21; thus R1 R1 R2 R2 73 158 231 and
Table 10-11 53 66 70 73 75 75 79 81 82 83 84 84 84 85 86 87 88 90 91 92 95 1 2 3 4 5.5 7 8 9 10 12 14 15 16 17 18 19 20 21
Table 10-12
Sum of Ranks Morning class Afternoon class 4 15 16 8 7 12 5.5 17 9 18 2 14 21 12 5.5 20 3 10 19 1 12 73 158
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