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 (0.69, 0.85).
(a) Would a 90% confidence interval be wider or narrower than a 95% confidence interval?
wider
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.)
(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.)
(e) Which one of these is the correct interpretation of the
interval (0.69, 0.85)?
The probability is 0.95 that the population proportion falls in this interval.
We are 95% sure that the sample proportion falls in this interval.
The probability is 0.95 that the sample proportion falls in this interval.
We are 95% sure that the population proportion falls in this interval.
a) 95% confidence interval will be wider that 90% confidence interval because 95% CI has larger margin of error.
b) Point estimate = (0.69 + 0.85)/2 = 0.77
c) Margin of error = 0.77 - 0.69 = 0.08
d) For 95% z = 1.96
Margin of error = z*(standard deviation of sample proportion)
0.08 = 1.96*(standard deviation of sample proportion)
Standard deviation of sample proportion = 0.08/1.96 = 0.04 (rounded to 2 decimal places)
e) Option D) We are 95% sure that the population proportion falls in this interval. is correct choice.
Please comment if any doubt. Thank you.
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