Refer to the Spearman rank correlation coefficient of
Chapter 15, Problem 78SE(choose chapter or problem)
Refer to the Spearman rank correlation coefficient of Section 15.10. Show that, when there are no ties in either the x observations or the y observations, then
\(r_{S}=\frac{n \sum_{i=1}^{n} R\left(x_{1}\right) R\left(y_{1}\right)-\left[\sum_{i=1}^{n} R\left(x_{1}\right)\right]\left[\sum_{i=1}^{n} R\left(y_{i}\right)\right]}{\sqrt{\left\{n \sum_{i=1}^{n}\left[R\left(x_{i}\right)\right]^{2}-\left[\sum_{i=1}^{n} R\left(x_{1}\right)\right]^{2}\right\}\left\{n \sum_{i=1}^{n}\left[R\left(y_{1}\right)\right]^{2}-\left[\sum_{i=1}^{n} R\left(y_{1}\right)\right]^{2}\right\}}}\)
\(=1-\frac{6 \Sigma_{i=1}^{n} d_{i}^{2}}{n\left(n^{2}-1\right)}\),
where di = R (xi) - R (yi).
Equation Transcription:
Text Transcription:
r{S}=frac{n sum{i=1}^{n} R(x{1}) R(y{1})-[\sum{i=1}^{n} R(x{1})][sum{i=1}^{n} R(y{i})]}{sqrt{\{n sum{i=1}^{n}[R(x{i})]^{2}-[sum{i=1}^{n} R(x{1})]^{2}}{n sum{i=1}^{n}[R(y{1})]^{2}-[sum{i=1}^{n} R(y{1}]^{2}}}}
= 1 - 6 the sum from i=1 to n d_i^2/n(n^2-1)
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer