# Write a two-way table of observed counts for sex and whether a participant in soccer had a lower-extremity injury or not. Determine a two-way table of expected counts for these data. Show calculations verifying that the value of the chi-square statistic is 2.51.

Write a two-way table of observed counts for sex and whether a participant in soccer had a lower-extremity injury or not. Determine a two-way table of expected counts for these data. Show calculations verifying that the value of the chi-square statistic is 2.51.
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joshyoung05M

a) Here is the table of observed counts:
$\begin{array}{cccc}\text{Gender / Had injury?}& \text{Yes}& \text{No}& \text{Total}\\ \text{Male}& 153& 1514& 1667\\ \text{Female}& 74& 925& 999\\ \text{Total}& 227& 2439& 2666\end{array}$
The numbers in the ”No” column are simply differences between last column and first column , and the numbers in the last row are sums of columns
b)The expected count for each cell is calculated as
$\text{Expected}=\frac{\text{Total row count}\cdot \text{Total column count}}{\text{Sample size}}$
Here is the table of the expected counts:
$\begin{array}{cccc}\text{Gender / Had injury?}& \text{Yes}& \text{No}& \text{Total}\\ \text{Male}& \frac{1667\cdot 227}{2666}=141.94& \frac{1667\cdot 2439}{2666}=1525.06& 1667\\ \text{Female}& \frac{999\cdot 227}{2666}=85.06& \frac{999\cdot 2439}{2666}=913.94& 999\\ \text{Total}& 227& 2439& 2666\end{array}$
c) The value of chi-square statistics is
${x}^{2}=\sum _{\text{all cells}}\frac{\left(\text{Observed count}-\text{Expected count}{\right)}^{2}}{\text{Expected count}}$
$=\frac{\left(153-141.94{\right)}^{2}}{141.94}+\frac{\left(1514-1525.06{\right)}^{2}}{1525.06}+\frac{\left(74-85.06{\right)}^{2}}{85.06}+\frac{\left(925-913.94{\right)}^{2}}{913.94}$
$=0.8618+0.0802+1.4381+0.1338$
$=2.5139$