ssrs 2012 barcode font Sum of Ranks 24 12 14 9 19 1 6.5 23 14 17.5 6.5 2.5 21 4 6.5 70 48.5 93 40.5 73 in Software

Creation QR in Software Sum of Ranks 24 12 14 9 19 1 6.5 23 14 17.5 6.5 2.5 21 4 6.5 70 48.5 93 40.5 73

Table 10-19 Sum of Ranks 24 12 14 9 19 1 6.5 23 14 17.5 6.5 2.5 21 4 6.5 70 48.5 93 40.5 73
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Since there are five samples (A, B, C, D, and E ), k 5. And since each sample consists of five values, we have N1 N2 N3 N4 N5 5, and N N1 N2 N3 N4 N5 25. By arranging all the values in increasing order of magnitude and assigning appropriate ranks to the ties, we replace Table 10-18 with Table 10-19, the right-hand column of which shows the sum of the ranks. We see from Table 10-19 that R1 70, R2 48.5, R3 93, R4 40.5, and R5 73. Thus H
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k R2 j 12 aN N(N 1) j 1 j
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3(N (48.5)2 5
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1) (93)2 5 (40.5)2 5 (73)2 R 5 3(26) 6.44
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For k 1 4 degrees of freedom at the 0.05 significance level, from Appendix E we have x2 9.49. 0.95 Since 6.44 9.49, we cannot reject the hypothesis of no difference between the machines at the 0.05 level and therefore certainly cannot reject it at the 0.01 level. In other words, we can accept the hypothesis (or reserve judgment) that there is no difference between the machines at both levels. Note that we have already worked this problem by using analysis of variance (see Problem 9.8) and have arrived at the same conclusion.
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CHAPTER 10 Nonparametric Tests
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10.15. Work Problem 10.14 if a correction for ties is made.
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Table 10-20 shows the number of ties corresponding to each of the tied observations. For example, 48 occurs two times, whereby T 2, and 53 occurs four times, whereby T 4. By calculating T3 T for each of these values of T and adding, we find that g(T 3 T ) 6 60 24 6 24 120, as shown in Table 10-20. Then, since N 25, the correction factor is 1 a (T 3 T ) N3 N 1 120 (25)3 25 0.9923
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Table 10-20 Observation Number of ties (T ) T3 T 48 2 6 53 4 60 64 3 24 68 2 6 72 3 24 g(T 3 T) 120
and the corrected value of H is Hc 6.44 0.9923 6.49
This correction is not sufficient to change the decision made in Problem 10.14.
10.16. Three samples are chosen at random from a population. Arranging the data according to rank gives us Table 10-21. Determine whether there is any difference between the samples at the (a) 0.05, (b) 0.01 significance levels. Table 10-21 Sample 1 Sample 2 Sample 3 7 11 5 4 9 1 6 12 3 8 2 10
Here k 3, N1 4, N2 3, N3 R2 11 9 12 32, and R3 5 H
k R2 j 12 aN N(N 1) j 1 j
5, 1 1)
N1 N2 N3 12, 8 2 19. Thus (32)2 3
(27)2 12 c (12)(13) 4
(19)2 d 5
3(13)
(a) For k 1 3 1 2 degrees of freedom, x2 5.99. Thus, since 6.83 5.99, we can conclude 0.95 that there is a significant difference between the samples at the 0.05 level. (b) For 2 degrees of freedom, x2 9.21. Thus, since 6.83 9.21, we cannot conclude that there is a 0.95 difference between the samples at the 0.01 level.
The runs test for randomness 10.17. In 30 tosses of a coin, the following sequence of heads (H) and tails (T) is obtained:
H T T H T H H H T H H T T H T H T H H T H T T H T H H T H T (a) Determine the number of runs, V. (b) Test at the 0.05 significance level whether the sequence is random.
CHAPTER 10 Nonparametric Tests
(a) Using a vertical bar to indicate a run, we see from H u T T u H u T u H H H u T u H H u T T u H u T u H u T u H H u T u H u T T u H u T u H H u T u H u T u that the number of runs is V 22.
(b) There are N1 16 heads and N2 14 tails in the given sample of tosses, and from part (a), the number of runs is V 22. Thus from formulas (13) of this chapter we have mV or sV 2(16)(14) 16 14 1 15.93 s2 V 2(16)(14)[2(16)(14) 16 14] (16 14)2(16 14 1) 22 runs is therefore 15.93 2.679 2.27 7.175
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