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 (0.63, 0.83).

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

narrower

wider

(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 (0.63, 0.83)?

We are 95% sure that the sample proportion falls in this interval.

We are 95% sure that the population proportion falls in this interval.

The probability is 0.95 that the sample proportion falls in this interval.

The probability is 0.95 that the population proportion falls in this interval.

Answer #1

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...

Use the sample information x¯ = 35, σ = 7, n = 16 to calculate
the following confidence intervals for μ assuming the sample is
from a normal population. (a) 90 percent confidence. (Round your
answers to 4 decimal places.) The 90% confidence interval is from
to (b) 95 percent confidence. (Round your answers to 4 decimal
places.) The 95% confidence interval is from to (c) 99 percent
confidence. (Round your answers to 4 decimal places.) The 99%
confidence interval...

A light bulb manufacturer wants to estimate the proportion of
light bulbs being produced that are defective. A random sample of
155 light bulbs is selected and 12 are defective. Write the
interval (Show all work and circle your final answer), and
a. Find the sample proportion ?̂.
b. Create a 95% confidence interval (round to 3 decimal places)
for the population proportion of defective lightbulbs being
produced.
c. Write a sentence interpreting this.
d. Using the sample proportion from...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals. A random
sample of 45 45 home theater systems has a mean price of $ 137.00
137.00. Assume the population standard deviation is $ 16.60 16.60.
Construct a 90% confidence interval for the population...

You are given the sample mean and the population standard
deviation. Use this information to construct the 90% and 95%
confidence intervals for the population mean. Interpret the results
and compare the widths of the confidence intervals. If convenient,
use technology to construct the confidence intervals.
A random sample of 35 home theater systems has a mean price of
$128.00 Assume the population standard deviation is $19.30
Construct a 90% confidence interval for the population
mean.
The 90% confidence interval...

A random sample of 14 observations is used to estimate the
population mean. The sample mean and the sample standard deviation
are calculated as 158.4 and 30.10, respectively. Assume that the
population is normally distributed. [You may find it useful
to reference the t table.]
a. Construct the 90% confidence interval for the
population mean. (Round intermediate calculations to at
least 4 decimal places. Round "t" value to 3 decimal
places and final answers to 2 decimal places.)
b. Construct...

he sample data below have been collected based on a simple
random sample from a normally distributed population. Complete
parts a and b.
7
5
0
7
6
5
9
8
9
3
a. Compute a 90% confidence interval estimate for the
population mean. The 90% confidence interval for the population
mean is from ______ to _________ (Round to two decimal places as
needed. Use ascending order.)
b. Show what the impact would be if the confidence level is
increased...

According to Runzheimer International, in a survey of relocation
administrators 63% of all workers who rejected relocation offers
did so for family considerations. Suppose this figure was obtained
by using a random sample of the files of 732 workers who had
rejected relocation offers. Use this information to construct a 95%
confidence interval to estimate the population proportion of
workers who reject relocation offers for family considerations.
round answer to 4 decimal places:
_____≤ p ≤ _______

A random sample of 23 items is drawn from a population whose
standard deviation is unknown. The sample mean is x¯x¯ = 770 and
the sample standard deviation is s = 25. Use Appendix D to
find the values of Student’s t.
(a) Construct an interval estimate of μ
with 95% confidence. (Round your answers to 3 decimal
places.)
The 95% confidence interval is from to
(b) Construct an interval estimate of μ
with 95% confidence, assuming that s =...

11. A random sample of 22 observations is used to estimate the
population mean. The sample mean and the sample standard deviation
are calculated as 135.5 and 30.50, respectively. Assume that the
population is normally distributed. [You may find it useful
to reference the t table.]
a. Construct the 90% confidence interval for the
population mean.
(Round intermediate calculations to at least 4 decimal
places. Round "t" value to 3 decimal places and final
answers to 2 decimal places.)
Confidence...

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