Question

1. The anterior cruciate ligament (ACL) is one of the major ligaments in the knee. The ACL is especially important for many sports as it helps provide stability in the knee joint. A group of 24 participants (12 with ACL injuries and 12 without) were randomly selected to participate in a study to investigate balarice between those with ACL injuries and those without. Each participant was asked to perform a "Y Balance Test" in which they stand on a leg in the squat position in the case of the ACL injury group the participants stood on their injured leg) while reaching their other leg out as far as possible in each of three directions that are marked on the ground with tape in a "Y" shape. To control for variation in body sizes each participant had the measurements in each of the three directions compiled into a composite score that was then divided by the participants leg length to convert the measurement into a score out of 100. The results for the participants are recorded below: ACL injury group 83.182.6 90.9 75.3 83.6 89.5 77.1 74.7 80.3 82.9 88.0 84.3 no injury group 86.4 92.3 91.7 95.2 98.0 87.2 89.4 76.5 79.4 87.8 85.9 88.7 Do patients with ACL injuries have a lower average score on the "Y Balance Test" than those without injuries? If needed, you may assume that performance on the "Y Balance Test" is normally distributed, both for patients with ACL injuries and for those without injuries. (a) (2 marks] Define the parameter(s) of interest using the correct notation. Then, state the null and alternative hypotheses for this study.

Answer 1

1.

Given that,
mean(x)=82.6917
standard deviation , s.d1=5.2313
number(n1)=12
y(mean)=88.2083
standard deviation, s.d2 =6.0288
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.796
since our test is left-tailed
reject Ho, if to < -1.796
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =82.6917-88.2083/sqrt((27.3665/12)+(36.34643/12))
to =-2.394
| to | =2.394
critical value
the value of |t α| with min (n1-1, n2-1) i.e 11 d.f is 1.796
we got |to| = 2.39413 & | t α | = 1.796
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -2.3941 ) = 0.0178
hence value of p0.05 > 0.0178,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.394
critical value: -1.796
decision: reject Ho
p-value: 0.0178
we have enough evidence to support the claim that patients with ACL injury group have a lower average score on the Y balance test than those without injuries.