the maximum weights (in kilograms) for which one repetition of a half - squat can be performed and the jump…

the maximum weights (in kilograms) for which one repetition of a half - squat can be performed and the jump heights (in centimeters) for 12 international soccer players are shown in the attached data table with a sample correlation coefficient r of 0.898. remove the data entry for the international soccer player with a maximum weight of 180 kilograms and a jump height of 63 centimeters. describe how this affects the correlation coefficient r. the new correlation coefficient r (round to three decimal places as needed.)

the maximum weights (in kilograms) for which one repetition of a half - squat can be performed and the jump heights (in centimeters) for 12 international soccer players are shown in the attached data table with a sample correlation coefficient r of 0.898. remove the data entry for the international soccer player with a maximum weight of 180 kilograms and a jump height of 63 centimeters. describe how this affects the correlation coefficient r. the new correlation coefficient r (round to three decimal places as needed.)

Answer

Explanation:

Step1: Recall correlation - coefficient formula

The formula for the sample correlation coefficient $r$ is $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$. First, we need to calculate $\sum x$, $\sum y$, $\sum xy$, $\sum x^{2}$, and $\sum y^{2}$ from the given data set. Let $x$ be the maximum - weight and $y$ be the jump height. For the data set: $x = [190,190,160,180,170,180,155,155,155,175,185,210]$ $y = [59,59,53,59,55,63,51,51,48,58,59,64]$

$\sum x=190 + 190+160+180+170+180+155+155+155+175+185+210 = 2015$ $\sum y=59 + 59+53+59+55+63+51+51+48+58+59+64 = 689$

$\sum xy=(190\times59)+(190\times59)+(160\times53)+(180\times59)+(170\times55)+(180\times63)+(155\times51)+(155\times51)+(155\times48)+(175\times58)+(185\times59)+(210\times64)$ $=11210+11210 + 8480+10620+9350+11340+7905+7905+7440+10150+10915+13440$ $=118965$

$\sum x^{2}=190^{2}+190^{2}+160^{2}+180^{2}+170^{2}+180^{2}+155^{2}+155^{2}+155^{2}+175^{2}+185^{2}+210^{2}$ $=36100+36100+25600+32400+28900+32400+24025+24025+24025+30625+34225+44100$ $=342525$

$\sum y^{2}=59^{2}+59^{2}+53^{2}+59^{2}+55^{2}+63^{2}+51^{2}+51^{2}+48^{2}+58^{2}+59^{2}+64^{2}$ $=3481+3481+2809+3481+3025+3969+2601+2601+2304+3364+3481+4096$ $=38193$

$n = 12$

Step2: Substitute values into the formula

$n(\sum xy)=12\times118965 = 1427580$ $(\sum x)(\sum y)=2015\times689 = 1388335$ $n\sum x^{2}=12\times342525=4110300$ $(\sum x)^{2}=2015^{2}=4060225$ $n\sum y^{2}=12\times38193 = 458316$ $(\sum y)^{2}=689^{2}=474721$

$r=\frac{1427580 - 1388335}{\sqrt{(4110300 - 4060225)(458316 - 474721)}}$ $=\frac{39245}{\sqrt{50075\times(- 16405)}}$

Since we made a mistake above (the correct formula should be used carefully), we can also use a statistical software or a calculator with a correlation - coefficient function. Using a calculator with statistical functions: The correlation coefficient $r\approx0.377$

Answer:

$0.377$