10-84. + An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273–279) considered arthro- scopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 14 of 18 (78%) repairs were successful, but for shorter tears, 22 of 30 (73%) repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use a = 0.05. What is the P-value? %3D (b) Calculate a one-sided 95% confidence bound on the differ- ence in proportions that can be used to answer the question in part (a).

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 25EQ
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(In the other image, this is the correct answer, just to check)

1) The parameters of interest are the proportion of successes of surgical repairs for different tears, p. and p:
2) H; :
P1 =P2
3) H :
P > P2
4) Test statistic is
where
A1, 1A
=1.65 for a = 0.05
Z905
5) Reject the null hypothesis if z: >
where
6)
18
30
14
22
X2 =
0.73
X1 =
0.78
14 + 22
18 + 30
=0.75
0.78 - 0.73
=0.387
0.75(1- 0.75) 18* 301
7) Conclusion: Because 0.387 < 1.65, we fail to reject the null hypothesis at the 0.05 level of significance.
P-value = [1- P(Z < 0.387)] = 1 - 0.6517 = 0.35
b) 95% confidence interval on the difference:
SP:- P:
0.78(1- 0.78) 0.73(1- 0.73)
(0.78- 0.73) - 1.65,
Sp- P:
18
30
- 0.159 sp. - P:
Because this interval contains the value zero, there is not enough evidence to conclude that the success rate p, exceeds
Transcribed Image Text:1) The parameters of interest are the proportion of successes of surgical repairs for different tears, p. and p: 2) H; : P1 =P2 3) H : P > P2 4) Test statistic is where A1, 1A =1.65 for a = 0.05 Z905 5) Reject the null hypothesis if z: > where 6) 18 30 14 22 X2 = 0.73 X1 = 0.78 14 + 22 18 + 30 =0.75 0.78 - 0.73 =0.387 0.75(1- 0.75) 18* 301 7) Conclusion: Because 0.387 < 1.65, we fail to reject the null hypothesis at the 0.05 level of significance. P-value = [1- P(Z < 0.387)] = 1 - 0.6517 = 0.35 b) 95% confidence interval on the difference: SP:- P: 0.78(1- 0.78) 0.73(1- 0.73) (0.78- 0.73) - 1.65, Sp- P: 18 30 - 0.159 sp. - P: Because this interval contains the value zero, there is not enough evidence to conclude that the success rate p, exceeds
10-84. + An article in Knee Surgery, Sports Traumatology,
Arthroscopy (2005, Vol. 13, pp. 273–279) considered arthro-
scopic meniscal repair with an absorbable screw. Results
showed that for tears greater than 25 millimeters, 14 of 18
(78%) repairs were successful, but for shorter tears, 22 of 30
(73%) repairs were successful.
(a) Is there evidence that the success rate is greater for longer
tears? Use a = 0.05. What is the P-value?
(b) Calculate a one-sided 95% confidence bound on the differ-
ence in proportions that can be used to answer the question
in part (a).
Transcribed Image Text:10-84. + An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273–279) considered arthro- scopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 14 of 18 (78%) repairs were successful, but for shorter tears, 22 of 30 (73%) repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use a = 0.05. What is the P-value? (b) Calculate a one-sided 95% confidence bound on the differ- ence in proportions that can be used to answer the question in part (a).
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