For Spearman's rank correlation, if the correlation coefficient is 0.7 and \(\rm \Sigma_{i=1}^n d_i^2=49.5\) then the value of sample size 'n' is:
10
Spearman's rank correlation coefficient, often denoted by \(\rho\) (rho) or \(r_s\), is a non-parametric measure used to assess the strength and direction of the monotonic relationship between two ranked variables. Unlike Pearson correlation, it does not assume the variables are measured on an interval or ratio scale, nor does it require a linear relationship. It works by ranking the data for each variable and then calculating the Pearson correlation coefficient on these ranks.
The formula for Spearman's rank correlation coefficient when there are no tied ranks (or a small number of ties) is:
\(\rho = 1 - \frac{6 \Sigma d_i^2}{n(n^2 - 1)}\)
Where:
In this problem, we are given the value of the Spearman's rank correlation coefficient and the sum of the squared differences in ranks, and we need to find the sample size 'n'.
Given values:
We can substitute these values into the formula:
\(0.7 = 1 - \frac{6 \times 49.5}{n(n^2 - 1)}\)
Now, let's solve for \(n(n^2 - 1)\).
\(0.7 - 1 = - \frac{297}{n(n^2 - 1)}\)
\(-0.3 = - \frac{297}{n(n^2 - 1)}\)
\(0.3 = \frac{297}{n(n^2 - 1)}\)
Rearrange the equation to isolate the term containing 'n':
\(n(n^2 - 1) = \frac{297}{0.3}\)
\(n(n^2 - 1) = 990\)
We need to find an integer value of 'n' that satisfies the equation \(n^3 - n = 990\). We can test the provided options to see which one fits.
Let's check the given options for 'n' in the equation \(n(n^2 - 1) = 990\).
\(99(99^2 - 1) = 99(9801 - 1) = 99(9800) = 970200\)
This is not equal to 990.
\(20(20^2 - 1) = 20(400 - 1) = 20(399) = 7980\)
This is not equal to 990.
\(10(10^2 - 1) = 10(100 - 1) = 10(99) = 990\)
This is equal to 990. This value satisfies the equation derived from the Spearman's rank correlation formula.
\(990(990^2 - 1)\)
This is a very large number, clearly not equal to 990.
Based on testing the options, the sample size 'n' that satisfies the equation \(n(n^2 - 1) = 990\) is 10.
Let's verify the calculation with n=10:
\(\rho = 1 - \frac{6 \Sigma d_i^2}{n(n^2 - 1)} = 1 - \frac{6 \times 49.5}{10(10^2 - 1)} = 1 - \frac{297}{10(100 - 1)} = 1 - \frac{297}{10(99)} = 1 - \frac{297}{990}\)
Now, calculate the fraction:
\(\frac{297}{990} = \frac{297 \div 99}{990 \div 99} = \frac{3}{10} = 0.3\)
So,
\(\rho = 1 - 0.3 = 0.7\)
This matches the given Spearman's rank correlation coefficient.
| Given Value | Symbol | Value |
|---|---|---|
| Spearman's Rho | \(\rho\) | 0.7 |
| Sum of Squared Differences | \(\Sigma d_i^2\) | 49.5 |
| Formula Applied | \(\rho = 1 - \frac{6 \Sigma d_i^2}{n(n^2 - 1)}\) | |
| Equation Derived | \(n(n^2 - 1) = 990\) | |
| Sample Size Found | \(n\) | 10 |
| Concept | Description | Formula (No ties) |
|---|---|---|
| Spearman's Rho (\(\rho\)) | Measures the strength and direction of monotonic association between two ranked variables. | \(\rho = 1 - \frac{6 \Sigma d_i^2}{n(n^2 - 1)}\) |
| \(d_i\) | Difference between the ranks of the \(i\)-th pair of observations. | |
| \(\Sigma d_i^2\) | Sum of the squares of the differences in ranks. | |
| \(n\) | Sample size (number of pairs of observations). |
It's helpful to understand how Spearman's rank correlation differs from Pearson correlation.
A monotonic relationship means that as one variable increases, the other variable either consistently increases (positive monotonic) or consistently decreases (negative monotonic), but not necessarily at a constant rate (which would be a linear relationship).
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