Solution
0-12 years 13-19 years Total
Knee 14 9 23
Ankle 13 3 16
Quadricep 6 9 15
Shin 9 6 15
________________________________________
Total 42 27 69
Now we can calculate the chi square statistic given by:
[tex]\chi^2=\sum ^n_{i=1}\frac{(O-E)^2}{E}[/tex]The expected values for this case are:
0-12 years 13-19 years Total
Knee 14 9 23
Ankle 9.74 6.26 16
Quadricep 9.13 5.87 15
Shin 9.13 5.87 15
________________________________________
Total 42 27 69
Then the statistic is given by:
[tex]\chi^2=\frac{(14-14)^2}{14}+\frac{(9-9)^2}{9}+\frac{(13-9.74)^2}{9.74}+\frac{(3-6.26)^2}{6.26}+\frac{(6-9.13)^2}{9.13}+\frac{(9-5.87)^2}{5.87}+\frac{(9-9.13)^2}{9.13}+\frac{(6-5.87)^2}{5.87}[/tex]Solving we got:
[tex]\chi^2=1.09+1.70+1.07+1.67+0.00185+0.0028=5.53[/tex]The degrees of freedom are given by:
df = (2-1)*(4-1)= 3
And the p value is given by:
p = P(X > 5.53) = 0.137
Since the p value is higher than 0.05 we can conclude that we fail to reject the null hypothesis of independence
So then the correct answer is:
yes
