Question

A report states that "many hospital administrators are leery of the push toward mandatory reporting of medical errors". Specifically, in a survey conducted between 2002 and 2003, a certain proportion of chief executives and operating officers in various states said that a state-run, mandatory, non-confidential reporting system would encourage lawsuits, despite evidence that patients are less likely to sue if doctors admit their mistakes and apologize. Based on the observed sample proportion, a 95% confidence interval for the proportion of all hospital administrators who believe such a system would encourage lawsuits is (0.60, 0.88). For this question I have gotten the correct answers for A and B but cannot get C and D please help!

(a) Would a 90% confidence interval be wider or narrower than a 95% confidence interval?

answer: Narrower

(b) The point estimate for population proportion must be at the center of the reported interval. What is it? (Round your answer to two decimal places.)

answer: .74

(c) What is the margin of error in this estimate? (Round your answer to two decimal places.)

(d) What is the approximate standard deviation of the sample proportion? (Round your answer to two decimal places.)

Answer 1

Solution:

Given a 95% confidence interval for the proportion is (0.60, 0.88)

a)  Narrower

b) The point estimate = (Upper limit + Lower Limit)/2

= (0.88+0.60)/2

= 0.74

c) Margin of error = (Upper limit - Lower Limit)/2

= (0.88-0.60)/2

= 0.14

Margin of error = 0.14

d) For 95% confidence , c = 95% = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.025

$\therefore$ = 1.96 is the critical value.

Now , we know,

Margin of error = * Standard error

$\therefore$ 0.14 = 1.96 * Standard error

$\therefore$ Standard error = 0.14/1.96 = 0.07142857142 = 0.07

Standard error is nothing but the standard deviation of sampling distribution,.

Answer : 0.07