2. (6 pts) An article in Knee Surgery, Sports Technology Arthroscopy (2005, Vol
ID: 3356510 • Letter: 2
Question
2. (6 pts) An article in Knee Surgery, Sports Technology Arthroscopy (2005, Vol 13, pg 273-279) considered arthrescopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters 14 of 18 repairs were successful, but for shorter tears, 22 of 30 repairs were successful. Is there evidence that the success rate is greater for longer tears? a. What is your parameter of interest b. State your null and alternate hypothesis c. What test statistic will be used d. What is your rejection criteria? e What is your conclusion? Construct a 95% confidence interval on the difference in proportions.Explanation / Answer
Question 2(a) Here parameter of interest is difference of success rate between smaller and longer tears.
(b) H0 : Success rate for tears smaller than 25 mm is same as success rate for tears greater than 25 mm. p<25mm = p>25mm
Ha : Success rate for tears bigger than 25 mm is more than the success rate for tears smaller than 25 mm. p<25mm < p>25mm
(c) Here we will use Z statistic.
(d) Here we have rejection criteria for alpha = 0.05. so for Z > 1.96 , we shall reject the null hypothesis.
(e) Here p^<25mm = 22/30 = 0.7333 ; p^>25mm = 14/18 = 0.7778
Pooled estimate p0 = (22 + 14) / (30 + 18) = 0.75
Standard error of the proportion for difference sed= sqrt [p0 (1-p0) * (1/n1 + 1/n2 )] = sqrt [0.75 * 0.25 * (1/30 + 1/18) = 0.1291
Test statistic
Z = (p^>25mm - p^<25mm)/ sed = (0.7778 - 0.7333)/ 0.1291 = 0.345
So p - value = Pr(Z > 0.345) = 0.365
(f) 95% confidence interval = (p^>25mm - p^<25mm) +- Z95% se0
= (0.7778 - 0.7333) +- 1.96 * 0.1291
= 0.0445 +- 0.2530
= (-0.2085, 0.2975)
The 95% confidence interval does consists value of Zero.
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