1.

The coefficient of rank correlation of the marks obtained by 10 students in two particular subjects was found to be 0.5. Later on, it was found that one of the differences of the ranks of a student was 7 but it was taken as 3. Find the corrected value of the correlation coefficient.

Answer»

Here, n = 8;

r = 0.5
True difference = 7.

False difference = 3

Now.,  \(r = 1 – \cfrac{6 (\Sigma d^{2}+CF)}{n\left(n^{2}-1\right)}\)

Putting n = 10 and r = 0.5 in the formula.

\(0.5 = 1 – \frac{6Σd^2}{10(10^2−1)}\)

\(∴ \frac{6Σd^2}{10(100−1) }= 1 – 0.5\)

\(∴ \frac{6Σd^2}{990 }= 0.5\)

\(∴ Σd^2 = \frac{0.5×990}{6} = 82.5\)

Corrected Σd2 = 82.5 – (False d)2 + (True d)2
= 82.5 – (3)2 + (7)2
= 82.5 – 9 + 49
= 122.5

Corrected rank correlation coefficient:

\(r = 1 – \cfrac{6( Corrected \,Σd^2)}{n(n^2−1)}\)

Putting n = 10 and corrected Σd2 = 122.5 in the formula.

\(r = 1 – \cfrac{6(122.5)}{10(10^2−1)}\)

= \(1 – \frac{735}{990}\)

= 1 – 0.74

= 0.26

Hence, the corrected value of the coefficient of rank correlation obtained is 0.26.



Discussion

No Comment Found