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(In R) There were two materials (A and B )that were randomly assigned to the left...

(In R) There were two materials (A and B )that were randomly assigned to the left and right shoes of 10boys. the data are available in R ,is a list called shoes(MASS) a)Test the Hypothesis of no difference in mean shoe wear for materials A and B. Repeat using a non parametric test. Do the two tests agree? b) Assume now that there is in fact no difference between A and B. Test the hypothesis that the mean shoe wear from right feet is greater than that from left feet. Your answer should explain how you have dealt with the L-R information for boy 4 ,which is missing.

BOY A B
1 13.2(L) 14.0(R)
2 8.2(L) 8.8(R)
3 10.9(R) 11.2(L)
4 14.3 14.2
5 10.7(R) 11.8(L)
6 6.6(L) 6.4(R)
7 9.5(L) 9.8(R)
8 10.8(L) 11.3(R)
9 8.8(R) 9.3(L)
10 13.3(L) 13.6(R)
0 0
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Answer #1

let η be population median of difference between materials A and B The null hypothesis. Ho: η-0 us. Alternative hypothesis, H1 : η

C6 C1 BOY C2-T C3-T C4 C7 B Difference DiIDil C8 Rank 13.2 14.0 8.8 11.2 14.2 11.8 6.4 9.8 11.3 9.3 13.6 0.8 0.6 0.3 0.1 113.2(L) 14.0(R) 2 8.2(L) 8.8(R) 3 10.9(R) 11.2(L) 4 14.3 5 10.7(R)11.8(L) 6 6.6L) 6.4(R) 7 9.5(L) 9.8(R) 8 10.8(L)11.3(R) 9 8.8(R) 9.3(L) 10 13.3(L) 13.6(R) 9.0 8.0 4.0 1.0 10.0 2.0 4.0 6.5 0.6 0.3 0.1 10.9 14.3 10.7 6.6 9.5 10.8 8.8 13.3 14.2 0.2 0.3 -0.5 0.2 0.3 0.5 4.0

where, Di = difference between left and right feet Arrange the absolute values of Di in increasing order then Wilcoxon sign rank statistic = w = sum of the rank of positive D 1+2=3 p-value 0.014 〈 0.05 Hence we reject Ho at 5% level of significance and conclude that there is siqni ficant di f ference between materials A and B

We randomly assign (R, L) for boy y.

Case: 1 we assign 14.3 for R and 14.2 for L and then perform Wilcoxon sign rank test.

let θ be population median of difference between left and right feet The null hypothesis. Ho : θ = 0 us. Alternative hypothesis, H1 : θ < 0

сь CI Co C4 CT C2- Rank BOY Left Right Difference DiIDil 113.2(L) 14.0(R) 2 8.2(L) 8.8(R) 3 10.9(R) 11.2(L) 4 14.3 (L) 14.2(R) 5 10.7(R) 11.8(L) 6 6.6(L) 6.4(R) 7 9.5(L) 9.8(R) 8 10.8L) 11.3(R) 9 8.8(R)9.3(L) 10 13.3(L) 13.6(R) 13.2 8.2 11.2 14.3 11.8 6.6 9.5 10.8 14.0 8.8 10.9 14.2 10.7 6.4 9.8 11.3 8.8 13.6 0.8 0.6 0.3 0.1 9.0 8.0 4.0 1.0 10.0 2.0 4.0 0.6 0.3 0.1 0.2 0.3 0.5 0.5 0.3 0.2 0.3 0.5 13.3 0.3 4.0

where, Di- dif ference between left and right feet Arrange the absolute values of Di in increasing order then Wilcoxon sign rank statistic = w = sum of the rank of positive Di -4+110+2+6.5+6.530 p - value0.620 0.05 Hence we fail to reject Ho at 5% level of significance and conclude that no di f ference between means shoe wear from right feet and left feet

Case: 2 we assign 14.3 for L and 14.2 for R and then perform Wilcoxon sign rank test.

CT C2-1 C4 сь C/ BOY Left Right Difference DiIDil Rank 11.2 14.2 11.8 6.6 0.3 0.1 0.3 0.1 3 10.9(R) 11.2(L) 4 14.3(R) 14.2(L) 5 10.7(R) 11.8(L) 6 6.6(L) 6.4(R) 7 9.5(L) 9.8(R) 8 10.8) 11.3(R) 9 8.8(R) 9.3(L) 10 13.3(L) 13.6(R) 10.9 14.3 10.7 6.4 9.8 11.3 8.8 13.6 4.0 1.0 10.0 2.0 4.0 6.5 6.5 4.0 0.2 0.3 0.5 0.5 0.3 0.5 10.8 9.3 13.3 0.3

Wilcoxon sign rank statistic -w- sum of the rank of positive Di 4+10+2+6.56.5 26.5 p-value 0.480 > 0.05 Hence we fail to reject Ho at 5% level of significance and conclude that no dif ference between means shoe wear from right feet and left feet. Hence two tests have same conclusion

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