Question

9. Two arthroscopic surgeries were compared among patients suffering from osteoarthritis who had at least moderate knee pain. The lavage type has the joint flushed with fluid and no instrument is used to remove tissue; the debridement type has the joint flushed with fluid and an instrument is used to remove tissue. Knee pain scores, ranging from 0 to 100 with higher scores indicating severe pain, were obtained for patients randomly assigned to the two groups after the surgery and the results are given below: Sample size sample mean sample standard deviation Lavage 11 59 24 Debridement 13 49 Determine a 95% confidence interval for the difference between the mean knee pain score for the lavage group and that for the debridement group. (5 points)

Answer 1

9)

$df = \frac{(s1^2/n1 + s2^2/n2)^2}{\frac{(s1^2/n1)^2}{n1 - 1} + \frac{(s2^2/n2)^2}{n2 - 1}}$

    $= \frac{((24)^2/11 + (23)^2/13)^2}{\frac{((24)^2/11)^2}{10} + \frac{((23)^2/13)^2}{12}}$

  

At 95% confidence level, the critical value is t_0.025,21 = 2.080

The 95% confidence interval is

$(\bar x1 - \bar x2) \pm t_{0.025,21} * \sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}$

$= (59 - 49) \pm 2.08 * \sqrt{\frac{(24)^2}{11} + \frac{(23)^2}{13}}$

$=10 \pm 20.065$

7)

HO : μα = 0

  

0 + Pr: 11

$\bar d$ = (-4 + (-9) + 2 + (-3) + (-12))/5 = -5.2

s_d = sqrt(((-4 + 5.2)^2 + (-9 + 5.2)^2 + (2 + 5.2)^2 + (-3 + 5.2)^2 + (-12 + 5.2)^2)/4) = 5.45

The test statistic is

t= d - D sd/n

-5.2 -0 5.45/5

= -2.133

At alpha = 0.01, the critical values are +/- t_0.005,4 = +/- 4.610

Since the test statistic value is not less than the lower critical value(-2.133 > -4.610), so we should not reject the null hypothesis.

At 0.01 significance level, there is not sufficient evidence to conclude that the mean score before tutoring differs from the mean score after tutoring.