Question

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 99% two-sided confidence interval on the proportion of such tears that will heal. Round the answers to 3 decimal places.

--------------------- ≤p≤ --------------------------

(b) Calculate a 99% lower confidence bound on the proportion of such tears that will heal. Round the answer to 3 decimal places.

------------------- ≤p

Answer 1

Solution :

Given that,

n = 37

x = 25

Point estimate = sample proportion = = x / n = 25 / 37 = 0.676

1 - = 1 - 0.676 = 0.324

a) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576(((0.676 * 0.324) / 37 )

= 0.198

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.676 - 0.198 < p < 0.676 + 0.198

( 0.478 < p < 0.874 )

b) At 99% confidence level

= 1 - 99%

= 1 - 0.99

= 0.01

Z = Z0.01= 2.326

Margin of error = E = Z  * (( * (1 - )) / n)

= 2.326(((0.676 * 0.324) / 37 )

= 0.179

A 99% lower confidence interval for population proportion p is ,

- E < p

0.676 - 0.179 < p

( 0.497 < p )