3 Marks Question

Question 13 Marks

While computing rank correlation coefficient between profits and investment for 10 years of a firm, the difference in rank for a year was taken as 7 instead of 5 by mistake and the value of rank correlation coefficient was computed as 0.80. What would be the correct value of rank correlation coefficient after rectifying the mistake ?

Answer

We are given that n = 10
$r_3=0.80$ and the wrong $d=7$ should be replaced by 5.
$r_{ s }=1-\frac{\Sigma 6 d^2}{n\left(n^2-1\right)}$
$ \begin{array}{ll} \Rightarrow & 0.80=1-\frac{6 \Sigma d^2}{10\left(10^2-1\right)} \\ \Rightarrow & 0.80=1-\frac{6 \Sigma d^2}{990} \\ \Rightarrow & \Sigma d^2=\frac{990(1-0.80)}{6}=33 \end{array} $
Corrected $\Sigma d^2=33-7^2+5^2=9$
Hence, rectified value of rank correlation coefficient
$=1-\frac{6 \times 9}{10 \times\left(10^2-1\right)}=0.95$

View full question & answer→

Question 23 Marks

For a number of towns, the coefficient of rank correlation between the people living below the poverty line and increase of population is 0.50. If the sum of squares of the differences in rank awarded to these factors is 82.50, find the number of towns.

Answer

As given
$r_s=0.50, d_t^2=82.50$
Thus, $\quad r_s=1-\frac{62 d^2}{n\left(n^2-1\right)}$
$\Rightarrow \quad 0.50=1-\frac{6 \times 82.50}{n\left(n^2-1\right)}$
$\Rightarrow \quad \frac{495}{n\left(n^2-1\right)}=1-0.50$
$\Rightarrow \quad \frac{495}{n\left(n^2-1\right)}=0.50$
$\begin{array}{l}\Rightarrow \quad n\left(n^2-1\right)=\frac{495}{0.50}=990 \\ \Rightarrow \quad n\left(n^2-1\right)=10\left(10^2-1\right)\end{array}$
Comparing both sides, we get
$n=10$ as $n$ must be a positive number.

View full question & answer→

Applied Maths 3 Marks Question

Test yourself on this topic