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Spearman's Rank Correlation Coefficient Simplified Revision Notes

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17.1.2 Spearman's Rank Correlation Coefficient

Spearman's Rank Correlation Coefficient:

Sometimes, instead of using raw data, it's more appropriate to analyze the ranking of the data, especially when comparing subjective judgments or when raw scores are inconsistent.

lightbulbExample

Example Data: The table shows the scores that two judges gave five contestants in a dance competition.

Contestant12345
Judge A7.64.23.89.68.7
Judge B21141169

Clearly, the judges have different views on how to score system works, the raw data seem less meaningful. However, what is important is low well the judges think constants did compared to each other.

Calculate Spearman's Rank Correlation Coefficient:

Ranking the Data:

Using a consistent ranking system:

Contestant12345
Judge A (Rank)34512
Judge B (Rank)12354

Differences and Squared Differences:

Calculate the difference in ranks, dd, and then the square of the differences, d2d^2:

Contestant12345
dd222-4-2
d2d^2444164

Total d2=:highlight[32]\sum d^2 = :highlight[32].

Apply the Formula:

The formula for Spearman's Rank Correlation Coefficient rsr_s:

rs=16d2n(n21)r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}

Substituting the values:

d2=32rs=16(32)5(521)=:highlight[0.6]\sum d^2 = 32 \Rightarrow r_s = 1 - \frac{6(32)}{5(5^2 - 1)} = :highlight[-0.6]

Conclusion:

  • Conclude (using the word "agreement" rather than "correlation".
  • There seems to be fairly strong disagreement between the two judges about the relative performance of the dancers. image

Note that "no agreement" is different than "disagreement".

Calculating rsr_s on a calculator:

Follow the method for 'rr' but simply input rankings.

Least Squares Regression Line for Coded Data

Coding is the transformation of data to make it easier to work with, read, and analyze.

lightbulbExample

Example:

  • The following data is to be analyzed and the least squares regression line calculated:
h0.020.030.040.05h1002100710091006\begin{array}{c:c:c:c:c:c} h & 0.02 & 0.03 &0.04 &0.05\\ \hline h & 1002 & 1007 & 1009 & 1006\\ \end{array}

To make the data easier to input, it makes sense to transform it using coding:

Let H=:highlight[100h]H = :highlight[100h] and let L=:highlight[l1000]L = :highlight[l - 1000].

H2345L2796\begin{array}{c:c:c:c:c:c} H & 2 & 3 &4 &5\\ \hline L & 2 & 7 & 9 & 6\\ \end{array}

Step 1: This variable appears to have been controlled, so we should calculate LL on HH.

Using the regression formula L=a+bHL = a + bH:

Thus, the equation becomes:

L=:highlight[1.1+1.4H]L = :highlight[1.1 + 1.4H]

Step 2: To get the original equation back, simply reverse the substitution:

L=l1000L = l - 1000

l1000=1.1+1.4(100h)l - 1000 = 1.1 + 1.4(100h)

l=:highlight[1001.1+140h]l = :highlight[1001.1 + 140h]

Verifying:

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