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Suppose that observations X₁, X₂,鈥�, X_n are made on a process at times 1, 2,鈥�, n. On the basis of this data, we wish to test H₀: the X_i鈥檚 constitute an independent and identically distributed sequence versus H_a: X_i11 tends to be larger than X_i for i =1,鈥�, n (an increasing trend) Suppose the X_i鈥檚 are ranked from 1 to n. Then when Ha is true, larger ranks tend to occur later in the sequence, whereas if H₀ is true, large and small ranks tend to be mixed together. Let Ri be the rank of Xi and consider the test statistic D = on i= 1(R_i = i)₂.

Then small values of D give support to Ha (e.g., the smallest value is 0 for R₁ = 10, R₂ = 2,鈥�, R_n= n). When H₀ is true, any sequence of ranks has probability 1yn!. Use this to determine the P-value in the case n = 4, d= 2. (Hint: List the 4! rank sequences, compute d for each one, and then obtain the null distribution of D. See the Lehmann book (in the chapter bibliography), p. 290, for more information.).

Step by step solution

01

Wilcoxon rank sum test

The Wilcoxon test is based upon ranking the nA + nB observations of the combined sample. Each observation has a rank: the smallest has rank 1, the 2nd smallest rank 2, and so on. The Wilcoxon rank-sum test statistic is the sum of the ranks for observations from one of the samples.

02

Finding the value of c

Use n=4 and d=2. The following table represents all 4! rank sequences as well as

\(D = \sum\limits_{i = 1}^4 {({R_i}} - i{)^2}\)

\({r_1}\)	\({r_2}\)	\({r_3}\)	\({r_4}\)	d
1	2	3	4	0
1	2	4	3	2
1	3	2	4	2
1	3	4	2	6
1	4	2	3	6
1	4	3	2	8
2	1	3	4	2
2	1	4	3	4
2	3	1	4	6
2	3	4	1	12
2	4	1	3	10
2	4	3	1	14
3	1	2	4	6
3	1	4	2	10
3	2	1	4	8
3	2	4	1	14
3	4	1	2	16
3	4	2	1	18
4	1	2	3	12
4	1	3	2	14
4	2	1	3	14
4	2	3	1	18
4	3	1	2	18
4	3	2	1	20

Assume that H₀ is true.The rank sequence is likely to happen. For example:

P(D=2)=P{1243}U{{1324}U{2134}}=1/24

d	0	2	4	6	8	10	12	14	16	18	20
p(d)	1/24	3/24	1/24	4/24	2/24	2/24	2/24	4/24	1/24	3/24	1/24

Value of c such that P(D鈮�c)鈮�0.1

So the closest to the 0.1 significance level is c=2.

Hence, the final answer is c=2.

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