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 83.182.690.9 75.3 83.6 89.5 77.1 74.7 80.3 82.9 no injury group 86.4 92.3 91.7 95.2 98.0 87.2 89.4 76.5 79.4 87.8 84.3 88.0 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 = 0.025
from standard normal table,left tailed t α/2 =2.201
since our test is left-tailed
reject Ho, if to < -2.201
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 2.201
we got |to| = 2.39413 & | t α | = 2.201
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 p0.025 > 0.0178,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
b.
test statistic: -2.394
critical value: -2.201
decision: reject Ho
c.
p-value: 0.0178
p value is greater than alpha value
we have enough evidence to support the claim that patients with ACL injuries have a lower average score on the Y Balance test than those without injuries.
d.
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 = 0.025
from standard normal table, two tailed t α/2 =2.593
since our test is two-tailed
reject Ho, if to < -2.593 OR if to > 2.593
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 2.593
we got |to| = 2.39413 & | t α | = 2.593
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.3941 ) = 0.036
hence value of p0.025 < 0.036,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.394
critical value: -2.593 , 2.593
decision: do not reject Ho
p-value: 0.036
we do not have enough evidence to support the claim that the Y Balance test between the ACL injuries group versus the group with no injuries.