question number 3. determine the correlation coefficient for the data shown in this table: \n| y | x…

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 $n = 6$.

$y$	$x$	$xy$	$x^{2}$	$y^{2}$
22	2	44	4	484
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