Spearman's Rank Correlation


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  1. State the null hypothesis i.e. "There is no relationship between the two sets of data."
  2. Rank both sets of data from the highest to the lowest. Make sure to check for tied ranks.
  3. Subtract the two sets of ranks to get the difference d.
  4. Square the values of d.
  5. Add the squared values of d to get Sigma d2.
  6. Use the formula Rs = 1-(6Sigma d2/n3-n) where n is the number of ranks you have.
  7. If the Rs value...
    ... is -1, there is a perfect negative correlation.
    ...falls between -1 and -0.5, there is a strong negative correlation.
    ...falls between -0.5 and 0, there is a weak negative correlation.
    ... is 0, there is no correlation
    ...falls between 0 and 0.5, there is a weak positive correlation.
    ...falls between 0.5 and 1, there is a strong positive correlation
    ...is 1, there is a perfect positive correlation
    between the 2 sets of data.
  8. If the Rs value is 0, state that null hypothesis is accepted. Otherwise, say it is rejected.

    Practical Example of Spearman's Rank Correlation

Question: Use the Spearman's Rank Correlation to establish whether there is any relationship between the distance away from school students live and the IB Geography grades they attain.
  1. Null Hypothesis: There is no relationship between the two sets of data.

  2. Distance From School (in miles)
    r
    IB Geography Grades Attained
    r
    d
    d^2
    3
    2
    4
    4
    2
    4
    7
    1
    4
    4
    3
    9
    2
    3
    7
    1
    2
    4
    2
    3
    6
    2
    1
    1
    1
    5
    5
    3
    2
    4
    Sigma d2 = 22
  3. Rs = 1-(6 Sigma d2 / n3-n).
  4. Rs = 1-(132/120)
  5. Rs = 1- 0.91
  6. Rs = 0.09
  7. There is a weak positive correlation between the two sets of data. The null hypothesis is rejected.
    Course Notes
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