An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp.

Question

An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273-279), considered arthroscopic meniscal able screw. Results showed that for tears greater than 25 mill meters, 14 of 1 repairs were successful repair with an absorb while for shorter tears, 22 of 30 repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use 0,05 what is the P -value? (b) Calculate a one-sided 95% confidence bound on the difference in proportions that can be used to answer the question in part (a) (a) There is not that the success rate is tears. p is evidence greater for longer The Value Round your answer to four decimal places (e.g. 98.7654). S p 1 D2 Round your answer to three decimal (b) The one-sided 95% confidence bound is places (e.g. 98.765)

Explanation / Answer

For tears greater than 25 mm, successful tears ^p>25mm = 14/19 = 0.7368

For tears shorter than 25 mm, successful tears ^p<25mm = 22/30 = 0.7333

H0 : p>25mm = p<25mm

Ha : p>25mm > p<25mm

(a) Test statistic.

pooled estimate p = (14 + 22)/ (19 + 30) = 0.7347

Standard error of the proportion se0 = sqrt [p(1-p) (1/n1 + 1/n2 )] = sqrt [0.7347 * 0.2653 * (1/19 + 1/30)] = 0.1294

Z = ( ^p>25mm - ^p<25mm)/ se0 = (0.7368 - 0.7333)/ 0.1294 = 0.027

p - value = 0.489

(b) One sided 95% confidence bound on the difference in proportions = (p>25mm -  p<25mm) - Z95% se0

= 0.0035 - 1.96 * sqrt [0.7368 * 0.2632/ 19 + 0.7333 *0.2667/30] = - 0.250

so 95% confidence lower bound is [-0.250, infinity] which contains the value of 0. So, we cannot reject the null hypothesis.

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