Answer to Question #156711 in Statistics and Probability for Eugine Haweza

Question

Consider the two results of Mr Nyirenda courses: Medical miclo-biology I (X) and miclo-biology II (Y):

The coefficient of correlation between X and Y for 20 students is 0.3, Mean of X and Y are 15 and 20 respectively and also the standard deviation of X and Y are 4 and 5 respectively.

At the time of calculation one item 27 has wrongly been taken as 17 for X and 35 instead of 30 for Y. As a biostatistics student find the correct coefficient of correlation between X and Y for 20 students.

Answer 1

We are given that n = 20 and the original r=0.3,

$\overline{X}$ = 15,

$\overline{Y}$ = 20,

$S x = 4 \\and\\S y = 5$

$r= \frac{Cov(x,y)}{S_xS_y}$ =$\frac{cov(x,y)}{4*5}$

Hence cov(x,y)= 0.3*20= 6

$\frac{∑xy}{n}-\overline{X}*\overline{Y}=6\\\frac{∑xy}{20}-{15}*20=6\\$

$\frac{∑xy}{20}-300=6$

$\sum{xy}=6120$

Hence, corrected = 6120 – 17 × 35 + 27 × 30 = 6335

$Also, S_x^2= 16\\= (∑x^2/ 20) – 15^2= 16\\ ∑x^2= 4820\\Similarly, S_y^2= 25$

$= (∑y^2/ 20) – 20^2= 25\\ ∑y^2= 8500$

Thus corrected

∑x = n x – wrong x value + correct x value

=20*15-17+27=310

similarly corrected

$∑y = 20 × 20 – 35 + 30 = 395\\Corrected ∑x^2= 4820 – 17^2+ 27^2= 5260\\Corrected ∑y^2= 8500 – 35^2+ 30^2= 8175$

Thus corrected value of the correlation coefficient by applying formula

$r =\frac{n\sum xy-\sum x\sum y}{\sqrt{(n\sum x^2)-(\sum x)^2(n\sum y^2)-(\sum y)^2)}}$

=$\frac{20(6335)-(310*395)}{\sqrt{((20*5260)-(310)^2)((20*8175)-(395)^2))}}$ =$=\frac{4250}{\sqrt{9100*7475}}=0.515$