Question

(Operations) The hospital has completed a medical operation for 48 patients and
37 of these have been successful. Suppose that each operation succeeds with an unknown probability p and that the successes of the different operations are independent of each other.
(a) Determine the conservative intermediate estimate (lecture note, formula (8.5)) for the probability of an operation using a 90% confidence level.
(b) How many operations should be performed to make the above-mentioned intermediate estimate width
would get five percentage points in their direction?

Answer 1

Let x be the number of successful operations and n is the total number of medical operations done by hospital.

We are given x = 37 and n = 48

Therefore = x/n = 37/48 = 0.7708

a) We are asked to find confidence interval estimate for population proportion P

Lower bound = - E

Upper bound = + E

E is margin of error =   

z is critical value follows standard normal distribution , we can find its value using z score table.

We are given confidence level = 0.90

Therefore α = 1 - 0.90 = 0.1 , 1 - (α/2) = 0.95

So we have to find z score corresponding to area 0.9500 on z score table

So z = 1.645

E =

E = 0.0998

Lower bound = - E = 0.7708 - 0.0998 = 0.6710

Upper bound = + E = 0.7708 + 0.0998 = 0.8706

Therefore 90% conservative intermediate estimate for the probability of an operation is (0.6710 , 0.8706 )

Part b) We are asked to find sample size n for intermediate estimate width = 0.05

intermediate estimate width = 2E

Therefore E = intermediate estimate width/2 = 0.05/2 =0.025

n = = = 764.91

n ~ 765

Therefor 765 operations should be performed to make the mentioned intermediate estimate width
would get five percentage points in their direction