Question
Download Solution PDFFor 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:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF The Spearman's rank correlation is formulated as:
\(\rm ρ=1-\frac{6\Sigma d_i^2}{n(n^2-1)}\)
ρ = Spearman's rank correlation coefficient
di = different between the two ranks of each observation
n = number of observation
Given, ρ = 0.7 and \(\rm \Sigma_{i=1}^n d_i^2=49.5\)
Putting the given values into the formula, we get:
\(\rm \rho=1-\frac{6\Sigma_{i=1}^nd_i^2}{n(n^2-1)}\)
or, \(\rm 0.7=1-\frac{6\times 49.5}{n^3-n}\)
or, \(\rm 0.3=\frac{6\times 49.5}{n^3-n}\)
Simplifying the equation we have, \(n^3 - n - 990 = 0\)
Now, following the method of trial and error, we get:
- n = 99 ⇒ \(99^3 - 99 - 990 \neq 0\) ⇒ Violated
- n = 20 ⇒ \(20^3 - 20 - 990 \neq 0\) ⇒ Violated
- n = 10 ⇒ \(10^3 - 10 - 990 \neq 0\) ⇒ Satisfied
- n = 990 ⇒ \(990^3 - 990 - 990 \neq 0\) ⇒ Violated
Hence, the required sample size is 10.
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