ssrs 2012 barcode font Nonparametric Tests in Software

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CHAPTER 10 Nonparametric Tests
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Another method As in 5, the variance is given by s2 V # V2 # V2 (2)(2)2 (3)(3)2 10 (4)(4)2 (1)(5)2
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10.21. Work Problem 10.20 by using formulas (13) of this chapter.
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Since there are three a s and two b s, we have N1 (a) (b) s2 V mV 2N1N2 N1 N2 1 3 and N2 2(3)(2) 3 2 2. Thus 1 17 5
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2N1N2(2N1N2 N1 N2) (N1 N2)2 (N1 N2 1)
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2(3)(2)[2(3)(2) 3 2] (3 2)2(3 2 1)
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Further applications of the runs test 10.22. Referring to Problem 10.3, and assuming a significance level of 0.05, determine whether the sample lifetimes of the batteries produced by the PQR Company are random.
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Table 10-24 shows the batteries lifetimes in increasing order of magnitude. Since there are 24 entries in the table, the median is obtained from the middle two entries, 253 and 262, as 1(253 262) 257.5. Rewriting 2 the data of Table 10-23 by using an a if the entry is above the median and a b if it is below the median, we obtain Table 10-25, in which we have 12 a s, 12 b s, and 15 runs. Thus N1 12, N2 12, N 24, V 15, and we have mV so that 2N1N2 N1 N2 1 2(12)(12) 12 12 Z V mV sV 1 13 15 13 2.396 s2 V 2(12)(12)(264) (24)2(23) 5.739
Using a two-tailed test at the 0.05 significance level, we would accept the hypothesis of randomness if 1.96 z 1.96. Since 0.835 falls within this range, we conclude that the sample is random.
Table 10-24 198 243 272 211 252 275 216 253 282 219 253 284 224 262 288 225 264 291 230 268 294 236 271 295 a b a b b a b a b
Table 10-25 a a b a b a b a b a b a b b a
10.23. Work Problem 10.5 by using the runs test for randomness.
The arrangement of all values from both samples already appears in line 1 of Table 10-8. Using the symbols a and b for the data from samples I and II, respectively, the arrangement becomes b b b b b b b Since there are four runs, we have V mV s2 V so that 4, N1 b a a a a a b b a a a
8, and N2 1 N1 N2 4
10. Then 1 9.889 4.125
2N1N2 N1 N2
2(8)(10) 18 N2) 1)
2N1N2(2N1N2 (N1 N2)2(N1 Z V
2(8)(10)(142) (18)2(17) 2.90
mV sV
9.889 2.031
CHAPTER 10 Nonparametric Tests
If H0 is the hypothesis that there is no difference between the alloys, it is also the hypothesis that the above sequence is random. We would accept this hypothesis if 1.96 z 1.96 and would reject it otherwise. Since 2.90 lies outside this interval, we reject H0 and reach the same conclusion as for Problem 10.5. Note that if a correction is made for continuity, Z V mV sV (4 0.5) 9.889 2.031 2.65
and we reach the same conclusion.
Rank correlation 10.24. Table 10-26 shows how 10 students, arranged in alphabetical order, were ranked according to their achievements in both the laboratory and lecture sections of a biology course. Find the coefficient of rank correlation.
Table 10-26 Laboratory Lecture 8 9 3 5 9 10 2 1 7 8 10 7 4 3 6 4 1 2 5 6
The difference in ranks, D, in the laboratory and lecture sections for each student is given in Table 10-27, which also gives D2 and gD2. Thus rs 1 6 a D2 N(N 2 1) 1 6(24) 10(102 1) 0.8545
indicating that there is a marked relationship between the achievements in the course s laboratory and lecture sections. Table 10-27 Difference of ranks (D) D2 1 1 2 4 1 1 1 1 1 1 3 9 1 1 2 4 1 1 1 1 gD2 24
10.25. Table 10-28 shows the heights of a sample of 12 fathers and their oldest adult sons. Find the coefficient of rank correlation. Table 10-28 Height of father (inches) Height of son (inches) 65 68 63 66 67 68 64 65 68 69 62 66 70 68 66 65 68 71 67 67 69 68 71 70
Arranged in ascending order of magnitude, the fathers heights are 62 63 64 65 66 67 67 68 68 69 71 (18)
Since the sixth and seventh places in this array represent the same height (67 inches), we assign a mean rank 1 7) 6.5 to these places. Similarly, the eighth and ninth places are assigned the rank 1(8 9) 8.5. 2 (6 2 Thus the fathers heights are assigned the ranks 1 2 3 4 5 6.5 6.5 8.5 8.5 10 11 12 (19)
Similarly, arranged in ascending order of magnitude, the sons heights are 65 65 66 66 67 68 68 68 68 69 70 71 (20)
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