ssrs 2012 barcode font Nonparametric Tests in Software

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CHAPTER 10 Nonparametric Tests
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Step 3. U mU Therefore, N1N2 2 N1N2 (9)(12) 2 Z N1(N1 2 54 U 1) R1 (9)(12) N2 12 80 54 14.07 1.85 (9)(10) 2 1) 73 80 198
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s2 U mU sU
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N1N2(N1
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(9)(12)(22) 12
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Since we wish to test the hypothesis H1 that the morning class performs worse than the afternoon class against the hypothesis H0 that there is no difference at the 0.05 level, a one-tailed test is needed. Referring to Fig. 10-2, which applies here, we have the decision rule: Accept H0 if z Reject H0 if z 1.645. 1.645.
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Since the actual value of z 1.85 1.645, we reject H0 and conclude that the morning class performed worse than the afternoon class at the 0.05 level. This conclusion cannot be reached, however, for the 0.01 level (see Problem 10.33).
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10.9. Find U for the data of Table 10-13 by using (a) formula (2) of this chapter, (b) the counting method (as described in Remark 4 of this chapter).
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(a) Arranging the data from both samples in an array in increasing order of magnitude and assigning ranks from 1 to 5 gives us Table 10-14. Replacing the data of Table 10-13 with the corresponding ranks gives us Table 10-15, from which the sums of the ranks are R1 5 and R2 10. Since N1 2 and N2 3, the value of U for sample 1 is U N1N2 N1(N1 2 1) R1 (2)(3) 2. (2)(3) 2 5 4
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The value U for sample 2 can be found similarly to be U
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Table 10-13
Sample 1 Sample 2 22 17 10 25 14 Data Rank
Table 10-14
10 1 14 2 17 3 22 4 25 5
Table 10-15
Sum of Ranks Sample 1 Sample 2 4 3 1 5 2 5 10
(b) Let us replace the sample values in Table 10-14 with I or II, depending on whether the value belongs to sample 1 or 2. Then the first line of Table 10-14 becomes Data From this we see that Number of sample 1 values preceding first sample 2 value Number of sample 1 values preceding second sample 2 value Number of sample 1 values preceding third sample 2 value Total 1 1 2 4 I II II I II
CHAPTER 10 Nonparametric Tests
Thus the value of U corresponding to the first sample is 4. Similarly, we have Number of sample 2 values preceding first sample 1 value Number of sample 2 values preceding second sample 1 value Total Thus the value of U corresponding to the second sample is 2. Note that since N1 2 and N2 3, these values satisfy U1 0 2 2
N1N2; that is, 4
(2)(3)
10.10. A population consists of the values 7, 12, and 15. Two samples are drawn without replacement from this population; sample 1, consisting of one value, and sample 2, consisting of two values. (Between them, the two samples exhaust the population.) (a) Find the sampling distribution of U. (b) Find the mean and variance of the distribution in part (a). (c) Verify the results found in part (b) by using formulas (3) of this chapter.
(a) We choose sampling without replacement to avoid ties which would occur if, for example, the value 12 were to appear in both samples. There are 3 2 6 possibilities for choosing the samples, as shown in Table 10-16. It should be noted that we could just as easily use ranks 1, 2, and 3 instead of 7, 12, and 15. The value U in Table 10-16 is that found for sample 1, but if U for sample 2 were used, the distribution would be the same. Table 10-16 Sample 1 7 7 12 12 15 15 Sample 2 12 15 7 15 7 12 15 12 15 7 12 7 U 2 2 1 1 0 0
(b) The mean and variance found from Table 10-15 are given by mU s2 U (2 1)2 (2 1)2 2 (1 2 1 6 1)2 6 (1 1)2 (0 1 0 0 1 1)2 (0 1)2 2 3
(c) By formulas (3), mU s2 U N1N2(N1 12 N1N2 2 N2 1) (1)(2) 2 (1)(2)(1 12 1 2 1) 2 3
showing agreement with part (a).
10.11. (a) Find the sampling distribution of U in Problem 10.9 and graph it. (b) Obtain the mean and variance of U directly from the results of part (a). (c) Verify part (b) by using formulas (3) of this chapter.
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