Question #256369

 Marks of 12 students in Arithmetic and Algebra are given below:

Arithmetic 60 34 40 50 45 40 22 43 42 66 64 46

Algebra 75 32 33 40 45 33 12 30 34 72 41 57

Calculate the rank correlation coefficient


1
Expert's answer
2021-10-26T08:12:14-0400
xRank(x)yRank(y)60107512342323403.5334.5509407457459403.5334.52211214363024253466612721164114184685710Sum=7878\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} & x & Rank(x)& y & Rank(y) \\ \hline & 60 & 10 & 75 & 12 \\ \hdashline & 34 & 2 & 32 & 3 \\ \hdashline & 40 & 3.5 & 33 & 4.5 \\ \hdashline & 50 & 9 & 40 & 7 \\ \hdashline & 45 & 7 & 45 & 9 \\ \hdashline & 40 & 3.5 & 33 & 4.5 \\ \hdashline & 22 & 1 & 12 & 1 \\ \hdashline & 43 & 6 & 30 & 2 \\ \hdashline & 42 & 5 & 34 & 6 \\ \hdashline & 66 & 12 & 72 & 11 \\ \hdashline & 64 & 11 & 41& 8 \\ \hdashline & 46 & 8 & 57 & 10 \\ \hdashline Sum= & & 78 & & 78 \end{array}



Rank(x)Rank(x)(Rank(x))2(Rank(y))212010014464915.7512.2520.2563814963498115.7512.2520.251111236430253613214412188121648064100Sum=626.5649.5649.5\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Rank(x) Rank(x) & ( Rank(x))^2 & ( Rank(y))^2 \\ \hline & 120 & 100 & 144 \\ \hdashline & 6 & 4 & 9 \\ \hdashline & 15.75 & 12.25 & 20.25 \\ \hdashline & 63 & 81 & 49 \\ \hdashline & 63 & 49 & 81 \\ \hdashline & 15.75 & 12.25 & 20.25 \\ \hdashline & 1 & 1 & 1 \\ \hdashline & 12 & 36 & 4 \\ \hdashline & 30 & 25 & 36 \\ \hdashline & 132 & 144 & 121 \\ \hdashline & 88 & 121& 64 \\ \hdashline & 80 & 64 & 100 \\ \hdashline Sum= & 626.5& 649.5 & 649.5 \\ \end{array}


SSxx=i=1n(Rank(xi))21n(i=1nRank(xi))2SS_{xx}=\displaystyle\sum_{i=1}^n(Rank(x_i))^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(x_i))^2

=649.5112(78)2=142.5=649.5-\dfrac{1}{12}(78)^2=142.5


SSyy=i=1n(Rank(yi))21n(i=1nRank(yi))2SS_{yy}=\displaystyle\sum_{i=1}^n(Rank(y_i))^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(y_i))^2

=649.5112(78)2=142.5=649.5-\dfrac{1}{12}(78)^2=142.5



SSxy=i=1nRank(xi)Rank(yi)SS_{xy}=\displaystyle\sum_{i=1}^nRank(x_i)Rank(y_i)

1n(i=1nRank(xi))(i=1nRank(yi))-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nRank(x_i))(\displaystyle\sum_{i=1}^nRank(y_i))

=626.5112(78)(78)=119.5=626.5-\dfrac{1}{12}(78)(78)=119.5



Therefore, based on this information, the sample Spearman correlation coefficient is computed as follows


rS=SSxySSxxSSyy=119.5142.5142.50.8386r_S=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}=\dfrac{119.5}{\sqrt{142.5}\sqrt{142.5}}\approx0.8386


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS