What is the formula to calculate Spearman's rank correlation coefficient when ranks are not repeated.

R = \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R = 1 - \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R = - \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R = 1 + \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $R = 1 - \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}$
When ranks are not repeated Speraman's rank coefficient $= R = 1 - \frac{6 \sum d_{i}^{2}}{n (n^{2} - 1)}$
$^{'} n^{'}$ is the number of observations
$^{'} d_{i}^{'}$ is the difference in rank for $i^{t h}$ observation.
So option $B$ is corrrect.

Suggest Corrections

Similar questions

\sum d_{i}^{2} = 25

n = 6

d_{i}

i^{t h}

x^{14}

{(1 + 2 x + {3 x}^{2} \dots + {16 x}^{15})}^{2}

\sum_{r = 1}^{n} r (n - r) = \frac{n (n^{2} - 1)}{6}

n

n (n^{2} - 1)

24

n (n^{2} - 1) = n (n - 1) (n + 1)

(n - \frac{1}{n}) (n + \frac{1}{n}) (n^{2} + \frac{1}{n^{2}})

(n - \frac{1}{n}) (n + \frac{1}{n}) (n^{2} + \frac{1}{n^{2}})

Join BYJU'S Learning Program