An article in knee surgery, sports traumatology, arthroscopy (2005, vol. 13, pp. 273–279) "arthroscopic meniscal repair with an absorbable screw: results and surgical technique" showed that only 25 out of 37 tears (67.6%) located between 3 and 6 mm from the meniscus rim were healed. (a) calculate a 90% two-sided confidence interval on the proportion of such tears that will heal. round the answers to 3 decimal places.

66.600% to 68.600%

Explanation:

The formula for calculating confidence interval for a proportion is

p ± Z √(p(1-p)/n)

(p is p hat with ∧ over p)

p hat= number of events/ total number of attempts i.e. 25/37 = 0.676

Z=z i.e. alpha = 1-0.9 (from 90%) = 0.10

alpha/2= 0.10/2 = 0.05

z value for area of 0.0500 is o.13 (from z table)

putting these values in the above formula

0.676 ± 0.13 √(0.676(1-0.676)/37)

= 0.676 ±0.13(0.0769387)

=0.676± 0.10002034387

0.676+ 0.10002034387= 0.68600 or 68.600%

0.676-0.10002034387= 0.665998 or 66.600%

hence CI is 66.600% to 68.600%

i believe the answer would be negative peer pressure

im not sure looking for this 2 ik c is wrong