Question

question number 3. determine the correlation coefficient for the data shown in this table:

y	x
29	4
28	10
40	15
34	13
43	16

o 0.8332
o 0.4564
o -0.9128
o 0.9128
o -0.8332
o none of the above

Explanation:

Step1: Recall correlation - coefficient formula

The Pearson correlation coefficient $r$ is given by the formula:
\[r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}\]
First, calculate the necessary sums:
Let $n = 6$.

$y$	$x$	$xy$	$x^{2}$	$y^{2}$
29	4	116	16	841
28	10	280	100	784
40	15	600	225	1600
34	13	442	169	1156
43	16	688	256	1849

$\sum x=2 + 4+10 + 15+13+16=60$
$\sum y=22 + 29+28+40+34+43=196$
$\sum xy=44 + 116+280+600+442+688=2170$
$\sum x^{2}=4 + 16+100+225+169+256=770$
$\sum y^{2}=484+841+784+1600+1156+1849=6714$

Step2: Substitute into the formula

\[n(\sum xy)-(\sum x)(\sum y)=6\times2170-60\times196\]
\[=13020 - 11760=1260\]

\[n\sum x^{2}-(\sum x)^{2}=6\times770-60^{2}\]
\[=4620 - 3600 = 1020\]

\[n\sum y^{2}-(\sum y)^{2}=6\times6714-196^{2}\]
\[=40284-38416 = 1868\]

\[r=\frac{1260}{\sqrt{1020\times1868}}\]
\[=\frac{1260}{\sqrt{1905360}}\]
\[=\frac{1260}{1380.3477}\approx0.9128\]

Answer:

0.9128