Question

shifty trends
find the correlation coefficient for the given data. round to the nearest hundredth.
% remaining minutes
41 90
63 65
88 22
82 30
79 45
38 99
93 15
50 75
35 111
47 80

Explanation:

Step1: Label the variables

Let $x$ be the % remaining and $y$ be the minutes.

Step2: Calculate the means

$\bar{x}=\frac{41 + 63+88+82+79+38+93+50+35+47}{10}=\frac{616}{10}=61.6$
$\bar{y}=\frac{90 + 65+22+30+45+99+15+75+111+80}{10}=\frac{632}{10}=63.2$

Step3: Calculate the numerator of the correlation - coefficient formula

$S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})$
$(41 - 61.6)(90 - 63.2)+(63 - 61.6)(65 - 63.2)+(88 - 61.6)(22 - 63.2)+(82 - 61.6)(30 - 63.2)+(79 - 61.6)(45 - 63.2)+(38 - 61.6)(99 - 63.2)+(93 - 61.6)(15 - 63.2)+(50 - 61.6)(75 - 63.2)+(35 - 61.6)(111 - 63.2)+(47 - 61.6)(80 - 63.2)$
$=(-20.6)\times26.8 + 1.4\times1.8+26.4\times(-41.2)+20.4\times(-33.2)+17.4\times(-18.2)+(-23.6)\times35.8+31.4\times(-48.2)+(-11.6)\times11.8+(-26.6)\times47.8+(-14.6)\times16.8$
$=- 552.08+2.52 - 1087.68-677.28 - 316.68-844.88-1513.48-136.88-1271.48-245.28$
$=-6542.12$

Step4: Calculate the denominator of the correlation - coefficient formula

$S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$
$(41 - 61.6)^{2}+(63 - 61.6)^{2}+(88 - 61.6)^{2}+(82 - 61.6)^{2}+(79 - 61.6)^{2}+(38 - 61.6)^{2}+(93 - 61.6)^{2}+(50 - 61.6)^{2}+(35 - 61.6)^{2}+(47 - 61.6)^{2}$
$=(-20.6)^{2}+1.4^{2}+26.4^{2}+20.4^{2}+17.4^{2}+(-23.6)^{2}+31.4^{2}+(-11.6)^{2}+(-26.6)^{2}+(-14.6)^{2}$
$=424.36 + 1.96+696.96+416.16+302.76+556.96+985.96+134.56+707.56+213.16$
$=4439.4$
$S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}$
$(90 - 63.2)^{2}+(65 - 63.2)^{2}+(22 - 63.2)^{2}+(30 - 63.2)^{2}+(45 - 63.2)^{2}+(99 - 63.2)^{2}+(15 - 63.2)^{2}+(75 - 63.2)^{2}+(111 - 63.2)^{2}+(80 - 63.2)^{2}$
$=26.8^{2}+1.8^{2}+(-41.2)^{2}+(-33.2)^{2}+(-18.2)^{2}+35.8^{2}+(-48.2)^{2}+11.8^{2}+47.8^{2}+16.8^{2}$
$=718.24+3.24 + 1697.44+1102.24+331.24+1281.64+2323.24+139.24+2284.84+282.24$
$=10163.6$
$\sqrt{S_{xx}S_{yy}}=\sqrt{4439.4\times10163.6}\approx\sqrt{45140336.84}\approx6718.66$

Step5: Calculate the correlation coefficient $r$

$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-6542.12}{6718.66}\approx - 0.97$

Answer:

$-0.97$