written 4.9 years ago by | • modified 4.9 years ago |
Solution:
Given
$\begin{array}{|c|c|c|c|} \hline X& 10 & 12 & 18&18&15&40 \\ \hline y& 12& 18&25&25&50&25 \\ \hline \end{array}$
$\begin{array}{|c|c|} \hline sr.no& X & Rank(R1) & y&Rank(R2)&D^2 =(R1-R2)^2 \\ \hline 1& 10& 1&12&1&0& \\ \hline 2& 12 & 2 & 18 & 2 & 0 \\ \hline 3& 18& 4.5&25&4&0.25 \\ \hline 4& 18& 4.5&25&4&0.25 \\ \hline 5& 15& 3&50&6&9 \\ \hline 6& 40& 6&25&4&4 \\ \hline & N=6& &&& \sum D^2=13.50 \\ \hline\end{array}$
There are two items in X series having equal values at the rank 4 .
$\therefore $They given the rank = $\frac{4+5}{2}=4.5$
Similarly there are 3 items in y series at the rank 3.each of the given rank =$\frac {3+4+5}{3}=4$
Since there are equal ranks we are using a formula ,
$R= 1-\frac{ 6[\sum di^2 +\frac{1}{12} (m_1^3-m_1)+\frac{1}{12} (m_2^3-m_2)+....]}{N^3-N}$
$\sum D^2 =13.50,m_1=2,m_2=3,N=6$
$R= 1-\frac{ 6[13.50+\frac{1}{12} (2^3-2)+\frac{1}{12} (3^3-3)+....]}{6^3-6}$
=$1-0.4571$
=$0.5429$