Question

Suppose that X = (Xi, X2, . . . , Xn) and Y = (y,Y2, . . . ,Yn) are random samples from continuous distributions F and G, respectively. Wilcoxons two-sample test statistic W = W(X,Y) is defined to be Σ-ngi Ri where Ri is the rank of in the combined sample.5. The paper Measuring the exposure of infants to tobacco smoke, (New England Journal of Medicine, 1984, pp. 1075-1078) compared infants who had been exposed to household smoke with those that hadnt. The data are observations of urinary concentra- tion of cotamine, a major metabolite of nicotine. Adapt the Wilcoxon rank sum statistic, W, to test whether the mean cotanine level in the population of exposed children is more than 25 units higher than that of unexposed children. Use W to form a confidence interval for the difference. Д. between the means of the two populations Unexposed: 8,11,12, 14, 20, 43, 111 Exposed: 35, 56, 83,92, 128, 150, 176, 208 You will find exact critical values of W on the web, in standard statistical textbooks and in statistical software, such as R. You can find an approximation to the correct p-value by doing appropriate random permutations of the samples in R. Alternatively you can use the normal approrimation to the null distribution of W.

Unexposed: 8,11,12,14,20,43,111

Exposed: 35,56,83,92,128,150,176,208

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Answer #1

Steps:

1. Let the sample X(sample size n) be the exposed sample and sample Y(sample size m) will the unexposed sample.

2. Now, combine both the samples and rank them in increasing order, keeping track of which observation belongs to which sample.

3. Compute W(N)= Sum of ranks associated with X sample values in the combined ranking scheme.

here, n=8 and m=7 , N=m+n=15

Ho: Both the sample means are equal

H1: mean of exposed sample is 25 units higher than mean of unexposed sample.

After combining and ranking, the data will look like: 8,11,12,14,20,35,43,56,83,92,111,128,150,176,208

and the sample from which each obs belongs is: Y,Y,Y,Y,Y,X,Y,X,X,X,Y,X,X,X,X

following above steps we get W(N)= 87

From table at alpha = 5% , critical value= 16

Since, W(n)> W(c) we reject the null hypothesis.

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