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 balance 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:

ACLinjurygroup 83.1 82.6 90.9 75.3 83.6 89.5 77.1 74.7 80.3 82.9 88.0 84.3

noinjurygroup 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.

(b) [1 mark] Calculate the observed value of the test statistic. State the distribution (and degrees of freedom if needed) it follows.

(c) [1 mark] Compute the p-value or provide a range of appropriate values for the p-value.

(d) [1 mark] Using the significance level α = 0.025, state your conclusions about performance on the “Y Balance Test” between the ACL injury group versus the group with no injuries.

Answer 1

a)

Answer:

$\text{parameter of interest}=\mu$

$\mu=\text{mean score on the "Y Balance Test"}$

Hypothesis

$H_0:\mu_1=\mu_2$

$H_1:\mu_1<\mu_2$

Explanation:

Parameter of Interest

A parameter is defined as the value that represents the whole population. In this scenario, we are comparing the average score on the "Y Balance Test" between the ACL injury group and the no injury group hence the mean value score on the "Y Balance Test" is the parameter of interest.

Hypothesis

The null hypothesis is defined as the mean score between the ACL injury group and the no injury group are equal and the alternative hypothesis test the claim that the mean score for ACL injury group less compared to the No injury group.

This is a left-tailed test.

b)

Answer:

Distribution: T distribution with 22 degrees of freedom.

The test statistic: t = -2.3942

Explanation:

Distribution

Since the population standard deviation is not known, the t distribution will be used to compare the mean scores hence the Two-sample t-test is used to compare the means assuming equal variance.

Test statistic

The t statistic is obtained using the formula,

$t=\frac{\overline{X}_1-\overline{X}_2}{\sqrt{\left ( \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} \right )\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}$

$t=\frac{82.7-88.2}{\sqrt{\left ( \frac{(12-1)5.2313^2+(12-1)6.0288^2}{12+12-2} \right )\left ( \frac{1}{12}+\frac{1}{12} \right )}}$

$t=-2.3942$

c)

Answer:

P-value = 0.0128

Explanation:

The p-value for t = -2.3942 is obtained from the t distribution table for the degree of freedom = n1+n2-2=12+12-2=22 and for left-tailed distribution. In excel use function =T.DIST(-2.3942,22,TRUE)

P-value = 0.0128

(or 0.025 < p-value < 0.01)

d)

Answer:

Reject the null hypothesis.

There is sufficient evidence to conclude that the mean score for the ACL injury group is significantly less than the mean score for the No injury group.

Explanation:

Since the p-value = 0.0128 is less than the significance level = 0.025, the null hypothesis is rejected at a 2.5% significance level. Hence there is sufficient evidence to conclude that the mean score for the ACL injury group is significantly less than the mean score for the No injury group.