Rank Correlation Coefficient Assignment Help | Rank Correlation Coefficient Homework Help


The Karl Pearson’s method is based on the assumption that the population being studied is normally distributed. When it is known that the population is not normal, or when the shape of the distribution is not known, there is need for a measure of correlation that involves no assumption about the parameters of the population.

It is possible to avoid making any assumption about the populations being studied by ranking the observations according to size and basing the calculations on the ranks rather than upon the original observations. It does not matter which way the items are ranked, item number one may be the largest or it may be the smallest. Using ranks rather than actual observations gives the coefficient of rank correlation.

This method of finding out covariability or the lack of it between two variables was developed by the British psychologist, Charles Edward Spearman in 1904. This measure is especially useful when quantitative measures for certain factors (such as in the evaluation of leadership ability or the judgement of female beauty) cannot be fixed, but the individuals in the group can be arranged in order thereby obtaining for each individual a number indicating his (her) rank in the group. Spearman’s rank correlation coefficient if defined as:

  R = 1 _    6 D2
                 N3 - N

where R denotes rank coefficient of correlation and D refers to the difference of ranks between paired items in two series.

The value of this coefficient, interpreted in the same way as Karl Pearson’s correlation coefficient,
ranges between +1 and -1. When rs   is + 1 there is complete agreement in the order of the ranks and the ranks are in the same direction. When rs is – 1 there is complete disagreement in the order of the ranks and  they are in opposite direction. This shall be clear from the following:

Rank Correlation Coefficient

R = 1 _  6 D2                                                                                       R = 1 _  6 D2
             N3 - N                                                                                                     N3 - N
  1 _  6 X 0    = 1 _ 0 = 1                                                                         = 1 _   6 X 8      = 1 _ 2 =  _ 1
       33 _ 3                                                                                                        33 _ 3

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