If the difference between the rank of the 4 observations are 2.5, 0.5, -1.5, -1.5, then Spearman's rank correlation coefficient equals to:
-0.1
Calculating Spearman's Rank Correlation Coefficient
Spearman's rank correlation coefficient, often denoted by \(\rho\) (rho) or \(r_s\), is a non-parametric measure of the strength and direction of association between two ranked variables. It assesses how well the relationship between two variables can be described using a monotonic function.
The formula to calculate Spearman's rank correlation coefficient when there are no tied ranks is:
\(\rho = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}\)
Where:
- \(\rho\) is the Spearman's rank correlation coefficient.
- \(d_i\) is the difference between the ranks of corresponding observations for the two variables.
- \(\sum d_i^2\) is the sum of the squares of these differences.
- \(n\) is the number of observations.
In this question, we are directly given the differences between the ranks (\(d_i\)) for 4 observations. The differences are 2.5, 0.5, -1.5, and -1.5. The number of observations, \(n\), is 4.
First, we need to calculate the square of each difference (\(d_i^2\)) and then find their sum (\(\sum d_i^2\)).
| Observation | Difference in Ranks (\(d_i\)) | Squared Difference (\(d_i^2\)) |
|---|---|---|
| 1 | 2.5 | \((2.5)^2 = 6.25\) |
| 2 | 0.5 | \((0.5)^2 = 0.25\) |
| 3 | -1.5 | \((-1.5)^2 = 2.25\) |
| 4 | -1.5 | \((-1.5)^2 = 2.25\) |
Now, let's calculate the sum of the squared differences:
\(\sum d_i^2 = 6.25 + 0.25 + 2.25 + 2.25 = 11\)
Next, we substitute the value of \(\sum d_i^2\) and \(n\) into the formula for Spearman's rank correlation coefficient:
\(\rho = 1 - \frac{6 \times 11}{4(4^2 - 1)}\)
Calculate the terms in the denominator:
- \(n^2 = 4^2 = 16\)
- \(n^2 - 1 = 16 - 1 = 15\)
- \(n(n^2 - 1) = 4 \times 15 = 60\)
Now, substitute these values back into the formula:
\(\rho = 1 - \frac{6 \times 11}{60}\)
\(\rho = 1 - \frac{66}{60}\)
Simplify the fraction:
\(\frac{66}{60} = \frac{11}{10} = 1.1\)
Finally, calculate \(\rho\):
\(\rho = 1 - 1.1\)
\(\rho = -0.1\)
Thus, Spearman's rank correlation coefficient is -0.1.
Revision Table: Spearman's Rank Correlation Calculation
| Step | Description | Calculation/Value |
|---|---|---|
| 1 | Number of observations (\(n\)) | 4 |
| 2 | Given differences in ranks (\(d_i\)) | 2.5, 0.5, -1.5, -1.5 |
| 3 | Square of differences (\(d_i^2\)) | 6.25, 0.25, 2.25, 2.25 |
| 4 | Sum of squared differences (\(\sum d_i^2\)) | \(6.25 + 0.25 + 2.25 + 2.25 = 11\) |
| 5 | Denominator term \(n(n^2-1)\) | \(4(4^2-1) = 4(16-1) = 4 \times 15 = 60\) |
| 6 | Calculate Spearman's \(\rho\) | \(1 - \frac{6 \sum d_i^2}{n(n^2-1)} = 1 - \frac{6 \times 11}{60} = 1 - \frac{66}{60} = 1 - 1.1 = -0.1\) |
Additional Information on Rank Correlation
Spearman's rank correlation coefficient ranges from -1 to +1. A value of +1 indicates a perfect positive monotonic relationship, -1 indicates a perfect negative monotonic relationship, and 0 indicates no monotonic relationship. It is based on the ranks of the data rather than the raw values, making it less sensitive to outliers than Pearson's correlation coefficient.
When ties exist in the ranks, a slightly adjusted formula for Spearman's rank correlation coefficient is sometimes used, which accounts for the ties. However, the basic formula \(1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}\) is often a good approximation even with ties, especially if the number of ties is small.
The rank difference method used here is a direct way to calculate Spearman's rho when the ranks or their differences are known. The steps involved are squaring the differences, summing them up, and applying the formula.
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Important Questions from Correlation and Regression
- For testing the significance of overall regression the test to be used is
- Including relevant lagged values of the dependent variable on the right hand side of a regression equation could lead to which one of the following ?
- If OLS is used in the presence of auto-correlation, which of the following will be likely consequences ?
i. Coefficients estimates may be misleading.
ii. Hypothesis tests could reach the wrong conclusions.
iii. Forecasts made from the model could be biased.
iv. Standard errors may be inappropriate. - Regression analysis is of particular interest
- Which of the following can be linked with those providing method for estimating black money in India ?
I. Kaldor’s approach
II. NIPFP method
III. O.P. Chopra’s method
IV. Arjun Sengupta
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