Question

Perform rank correlation analysis on the following data set: -5 у -253 -182 -43 -101 -3.55 -2.8 -1.8 -0.95 -0.2 0.8 1.5 2.25 3.15 3.95 4.9 5.75 -56 -3 -5 -5 10 -8 -13 15 30 42 78 The scatterplot looks like this:

Answer 1

The following scatter plot is obtained based on the data provided: Scatter Plot of X and Y 97.35 47.35 -6.61 -4.61 -2.61 1 3.39 5.39 х -52.65 - 102.65 - 152.65 -202.65 -252.65 -302.65

X Rank(X) Y Rank(y) -5 -253 -4.3 2 -182 2 -3.55 3 -101 3 -2.8 4 -56 4 -1.8 5 -3 9 -0.95 6 -5 7.5 -0.2 7 -5 7.5 0.8 8 10 10 1.5 9 -8 6 2.25 10 -13 5 3.15 11 15 11 3.95 12 30 12 4.9 13 42 13 5.75 14 78 14 Also, the following calculations involving the ranks are needed to compute the Spearman correlation coefficient

Rank(X) Rank(y) Rank(X)*Ra nk(Y) Rank(x)2 Rank(y) 1 1 1 1 1 2 2 4 4 4 3 3 9 9 9 4 4 16 16 16 5 9 45 25 81 6 7.5 45 36 56.25 7 7.5 52.5 49 56.25 8 10 80 100 64 81 9 6 54 36 10 5 50 100 25 11 11 121 121 121 12 12 144 144 144 13 13 169 169 169 14 14 196 196 196 Sum = 105 105 986.5 1015 1014.5

72 72 SSxy= Rank(X;)Rank(Y;) Rank(X;) Rank(Y) n i=1 i=1 2 n SS i=1 Rankx )° - + (Rank(x3) 588 = Renkx» - (Renky) 2 In this case, based on the data provided, we get that SSxy = 986.5 1 (105 x 105) = 986.5 14 1 SS i = 1015 ia (105)2 = 1015 SSýý = 1014.5 1 (105)2 = 1014.5 14 Therefore, based on this information, the sample Spearman correlation coefficient is computed as follows SS XY T's = 986.5 1015 x 1014.5 0.876 SS SSyÝ

The sample size is n = 14, so then the number of degrees of freedom is df = n – 2 = 14 – 2 = 12 The corresponding critical correlation value re for a significance level of a = 0.05, for a two-tailed test is: rc = 0.532

Observe that in this case, the null hypothesis Ho : p=0 is rejected if r>rc = 0.532. Based on the sample correlation provided, we have that rl = 0.876 > rc = 0.532, from which is concluded that the null hypothesis is rejected.

Yes, the rank correlation coefficient is larger (in absolute value) than the critical rho value.