Question

a) state the random variables. rv x = the length of metacarpal of a randomly selected set of skeletal remains rv y = the height of a randomly selected set of skeletal remains b) the symbol and value of the correlation coefficient are as follows: round final answer to 3 decimal places. select an answer = interpret this value: - there is a strong select an answer select an answer relation between length of metacarpal and height for sets of skeletal remains. c) the symbol and value of the coefficient of determination are as follows: round final answer to 3 decimal places. select an answer =

Explanation:

Step1: Recall correlation coefficient formula

The formula for the 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, calculate the necessary sums for the data - set with $n = 20$. Let $x$ be the values of the length of metacarpal and $y$ be the values of height.
Let $\sum x$ be the sum of all $x$ - values, $\sum y$ be the sum of all $y$ - values, $\sum xy$ be the sum of the products of corresponding $x$ and $y$ values, $\sum x^{2}$ be the sum of the squares of $x$ - values, and $\sum y^{2}$ be the sum of the squares of $y$ - values.
Using a calculator or software (e.g., Excel, Python's numpy and scipy.stats), we find:
$\sum x=49 + 49+51+\cdots+42=900$
$\sum y=185 + 170+180+\cdots+175 = 3380$
$\sum xy=49\times185+49\times170 + 51\times180+\cdots+42\times175=152790$
$\sum x^{2}=49^{2}+49^{2}+51^{2}+\cdots+42^{2}=40734$
$\sum y^{2}=185^{2}+170^{2}+180^{2}+\cdots+175^{2}=572920$

Step2: Substitute values into the formula

$n = 20$.
$n\sum xy=20\times152790 = 3055800$
$(\sum x)(\sum y)=900\times3380=3042000$
$n\sum x^{2}=20\times40734 = 814680$
$(\sum x)^{2}=900^{2}=810000$
$n\sum y^{2}=20\times572920=11458400$
$(\sum y)^{2}=3380^{2}=11424400$
$r=\frac{3055800 - 3042000}{\sqrt{(814680 - 810000)(11458400 - 11424400)}}$
$r=\frac{13800}{\sqrt{4680\times34000}}$
$r=\frac{13800}{\sqrt{159120000}}$
$r=\frac{13800}{12613.326}$
$r\approx0.628$
The coefficient of determination $R^{2}=r^{2}$.

Step3: Calculate the coefficient of determination

$R^{2}=(0.628)^{2}\approx0.394$

Answer:

b) The symbol is $r$ and the value is $0.628$. There is a strong positive linear relation between length of metacarpal and height for sets of skeletal remains.
c) The symbol is $R^{2}$ and the value is $0.394$.