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2.679. The z score corresponding to V Z V mV sV 22
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Now for a two-tailed test at the 0.05 significance level, we would accept the hypothesis H0 of randomness if 1.96 z 1.96 and would reject it otherwise (see Fig. 10-4). Since the calculated value of z is 2.27 1.96, we conclude that the tosses are not random at the 0.05 level. The test shows that there are too many runs, indicating a cyclic pattern in the tosses.
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Fig. 10-4
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If a correction for continuity is used, the above z score is replaced by z (22 0.5) 15.93 2.679 2.08
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and the same conclusion is reached.
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10.18. A sample of 48 tools produced by a machine shows the following sequence of good (G) and defective (D) tools: G G G G G G D D G G G G G G G G G G D D D D G G G G G G D G G G G G G G G G D D G G G G G D G G Test the randomness of the sequence at the 0.05 significance level.
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The numbers of D s and G s are N1 the mean and variance are given by mV 2(10)(38) 10 38 1 10 and N2 38, respectively, and the number of runs is V 2(10)(38)[2(10)(38) 10 38] (10 38)2(10 38 1) 11. Thus
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s2 V
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so that sV 2.235. For a two-tailed test at the 0.05 level, we would accept the hypothesis H0 of randomness if 1.96 (see Fig. 10-4) and would reject if otherwise. Since the z score corresponding to V 11 is Z and 2.61 V mV sV 11 16.83 2.235 2.61
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1.96, we can reject H0 at the 0.05 level.
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CHAPTER 10 Nonparametric Tests
The test shows that there are too few runs, indicating a clustering (or bunching) of defective tools. In other words, there seems to be a trend pattern in the production of defective tools. Further examination of the production process is warranted.
10.19. (a) Form all possible sequences consisting of three a s and two b s, and give the numbers of runs, V, corresponding to each sequence. (b) Obtain the sampling distribution of V. (c) Obtain the probability distribution of V.
(a) The number of possible sequences consisting of three a s and two b s is 5 a b 2 5! 2!3! 10
These sequences are shown in Table 10-22, along with the number of runs corresponding to each sequence. Table 10-22 Sequence a a a a a a b b b b a a a b b b b a a a a b b a b a a b a a b a b b a a a a a b b b a a a b a a b a Runs (V ) 2 4 3 5 3 4 2 4 3 4
Table 10-23
V 2 3 4 5 f 2 3 4 1
(b) The sampling distribution of V is given in Table 10-23 (obtained from Table 10-21), where V denotes the number of runs and f denotes the frequency. For example, Table 10-23 shows that there is one 5, four 4s, etc. (c) The probability distribution of V is obtained from Table 10-23 by dividing each frequency by the total 1 frequency 2 3 4 1 10. For example, Pr5V 56 0.1. 10
10.20. Find (a) the mean, (b) the variance of the number of runs in Problem 10.19 directly from the results obtained there.
(a) From Table 10-22 we have mV 2 4 3 5 3 10 4 2 4 3 4 17 5
Another method From Table 10-22 the grouped-data method gives mV afV af (2)(2) (3)(3) 2 3 (4)(4) 4 1 (1)(5) 17 5
(b) Using the grouped-data method for computing the variance, from Table 10-23 we have s2 V a f (V af # V )2 1 c (2)a2 10 17 2 b 5 (3)a3 17 2 b 5 (4)a4 17 2 b 5 (1)a5 17 2 b d 2 21 25
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