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The ranking procedure described in Exercise 35 is somewhat asymmetric, because the smallest observation receives rank 1, whereas the largest receives rank 2, and so on. Suppose both the smallest and the largest receive rank 1, the second smallest and second largest receive rank 2, and so on, and let W鈥欌�� be the sum of the X ranks. The null distribution of W鈥欌�� is not identical to the null distribution of W, so different tables are needed. Consider the case m = 3, n = 4. List all 35 possible orderings of the three X values among the seven observations (e.g., 1, 3, 7 or 4, 5, 6), assign ranks in the manner described, compute the value of W鈥欌�� for each possibility, and the tabulate the null distribution of W0. What is the P-value if 飞鈥欌赌� = 9? This is the Ansari-Bradley test; for additional information, see the book by Hollander and Wolfe in the chapter bibliography.

Step by step solution

01

analyzing using table:

W鈥欌�� is the sum of ranks. List all 35 ordering of X, assign ranks, sum them. The following table contain everything mentioned.

i	X ordering	3 ranks	飞鈥欌赌�
1	123	123	6
2	124	124	7
3	125	123	6
4	126	122	5
5	127	121	4
6	134	134	8
7	135	133	7
8	136	132	6
9	137	131	5
10	145	143	8
11	146	142	7
12	147	141	6
13	156	132	6
14	157	131	5
15	167	121	4
16	234	234	9
17	235	233	8
18	236	232	7
19	237	231	6
20	245	243	9
21	246	242	8
22	247	241	7
23	256	232	7
24	257	231	6
25	267	221	5
26	345	343	10
27	346	342	9
28	347	341	8
29	356	332	8
30	357	331	7
31	367	321	6
32	456	432	9
33	457	431	8
34	467	421	7
35	567	321	6

02

finding probability:

The probability of any of ordering is equal and it is 1/35, obtain the null distribution od W鈥欌��

飞鈥欌赌�	4	5	6	7	8	9	10
p(飞鈥欌赌�)	\(\frac{2}{{35}}\)	\(\frac{4}{{35}}\)	\(\frac{9}{{35}}\)	\(\frac{8}{{35}}\)	\(\frac{7}{{35}}\)	\(\frac{4}{{35}}\)	\(\frac{1}{{35}}\)

If 飞鈥欌赌� = 9, P- value is

The cumulative density function of random variable W鈥欌�� is

飞鈥欌赌�	4	5	6	7	8	9	10
P\(W'' < w''\)	\(\frac{2}{{35}}\)	\(\frac{6}{{35}}\)	\(\frac{{15}}{{35}}\)	\(\frac{{23}}{{35}}\)	\(\frac{{30}}{{35}}\)	\(\frac{{34}}{{35}}\)	\(\frac{{35}}{{35}}\)

\(P\left( {W'' \ge w''} \right) = P\left( {W \ge 9} \right) = \frac{5}{{35}} = 0.143\)

Hence, P=0.143

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