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# Calculating the t-statistic for Hypothesis Testing on Correlation

The Spearman’s rank correlation coefficient is a non-parametric statistical test used to examine whether there is a significant relationship between two sets of data. Like the correlation coefficient, Spearman’s rank correlation can have any value between –1 and +1. A value of 0 indicates no relationship, and values of +1 indicate a perfectly positive correlation, while -1 indicates a perfectly negative correlation.

### Steps of Calculating Spearman’s Rank Correlation Coefficient

1. Rank the values of variables X and Y in descending order.
2. Find the difference (denoted as $$d_i$$) between the rank of the pair values of X and Y.
3. Square the difference and calculate the sum of the difference, that is $$\sum d_i$$.
4.  Calculate the Spearman’s rank correlation coefficient as follows:

$$r_{s}=1-\frac{6 \sum d_{i}^{2}}{n\left(n^{2}-1\right)}$$

Example: Calculatig Spearman’s Rank Correlation Coefficient

An analyst is studying the relationship between returns for two sectors, steel and cement over past 5 years by using spearman’s rank correlation coefficient. The returns of both sectors are provided as under.

$$\begin{array}{c|c|c} \textbf { Year } & \textbf { Steel sector returns } & \textbf { Cement sector returns } \\ \hline 1 & 2.5 \% & 3.2 \% \\ \hline 2 & 5 \% & 4.5 \% \\ \hline 3 & 5.6 \% & 4.2 \% \\ \hline 4 & -3 \% & -1.7 \% \\ \hline 5 & 0.5 \% & 1.1 \% \\ \end{array}$$

The Spearman’s rank correlation coefficient is closest to:

Solution

$$\begin{array}{c|c|c|c|c|c|c} \textbf { Year } & \begin{array}{l} \textbf { Steel } \\ \textbf { sector } \\ \textbf { returns } \\ \textbf { (X) } \end{array} & \begin{array}{l} \textbf { Cement } \\ \textbf { sector } \\ \textbf { returns } \\ \textbf{ (Y) } \end{array} & \begin{array}{l} \textbf { Rank of } \\ \textbf { X } \end{array} & \begin{array}{l} \textbf { Rank of } \\ \mathrm{Y} \end{array} & \bf{d} & \bf{d^2} \\ \hline 1 & 2.50 \% & 1.60 \% & 3 & 4 & -1 & 1 \\ \hline 2 & 5 \% & 4.50 \% & 2 & 1 & 1 & 1 \\ \hline 3 & 5.60 \% & 4.20 \% & 1 & 2 & -1 & 1 \\ \hline 4 & -3 \% & -1.70 \% & 5 & 5 & 0 & 0 \\ \hline 5 & 0.50 \% & 2.20 \% & 4 & 3 & 1 & 1 \\ \hline & & & & & \textbf { Sum } & 4 \\ \end{array}$$

We can now use the formula:

\begin{align} r_{s}&=1-\frac{6 \sum d_{i}^{2}}{n\left(n^{2}-1\right)} \\ &=1- \left[\frac{(6 \times 4)}{ 5 \times (5^2 – 1)}\right] \\ &= 0.8 \end{align}

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