Question

Suppose a random group of 8 patients who had and didn't have knee injuries were asked to perform a “Y Balance Test”, where the patients performed the test once while standing on their injured leg, and then after a 5 minute rest period were asked to perform the test again while standing on their uninjured leg.

The results are recorded below:

1 2 3 5 6 7 Patient injured leg score uninjured leg score 86.3 89.7 85.1 86.0 79.2 80.4 4. 83.7 80.1 82.6 82.6 87.3 87.0 88.1 89.7 8 89.5 92.4

For patients with knee injuries are average scores on the “Y Balance Test” different while standing on the injured leg versus the uninjured leg? If needed, you may assume that performance on the “Y Balance Test” is normally distributed on both injured and uninjured legs.

(a) Define the parameter(s) of interest using the correct notation. Then, state the null and alternative hypotheses for this study.

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

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

(d) Using the significance level α = 0.10, state your conclusions about performance on the “Y Balance Test” for patients with knee injuries on their injured versus uninjured legs

Answer 1

Let us denote the difference

d = injured leg score - uninjured leg score

a) The parameter of interest : population mean of difference $\mu_d$

There is sufficient evidence to conclude that average scores on the “Y Balance Test” is significantly different while standing on the injured leg versus the uninjured leg.

Let ud denotes the population mean of difference. Hypotheses : To test null hypothesis Ho Jid =0 against alternative hypothesis H1 ud 70 This is a two tailed test as alternative hypothesis H contains only #sign. Let d denotes the sample mean of difference , sa denotes the sample standard deviation of difference and n denotes the sample size. Here, d = -0.7625, så= 4.740, $a = 2.1771, n=8 The test statistic can be written as: Vnd – 0.0) sd which under H, follows a t distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.1 level of significance if P-value < 0.1 or if tobs | > t0.05,n-1 = 1.894579 0.4- 0.3 0.2 0.1- 0.0 -3 + Now, Value of test statistic: The value of the test statistic tobs V8(-0.7625 - 0) 2.1771 -0.990613 -0.991 associated degrees of freedom = 7 The critical value tcritical = +1.894579 =+ 1.895 P-value = P(tn-11 > tobs) = 2* P(t7 < -0.990613) = 2 *0.177437 = 0.354874 0.3549 Conclusion : Since p-value > 0.1 and tobstcritical = 1.894579, so We fail to reject H, at 0.1 level of significance Hence, we can conclude that there is no significant difference between two population means.