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A TV poll A television news program conducts a call-in poll about a proposed city ban on handgun ownership. Of the 2372 calls, 1921 oppose the ban. The station, following recommended practice, makes a confidence statement: "81%of the Channel 13 Pulse Poll sample opposed the ban. We can be 95% confident that the true proportion of citizens opposing a handgun ban is within1.6% of the sample result."

(a) Is the station's quoted 1.6% margin of error correct? Explain.

(b) Is the station's conclusion justifed? Explain.

Short Answer

Expert verified

a) No

b) No

Step by step solution

01

Part (a) Step1 : Given Information

Need to explain the margin of error quoted is1.6%

02

Part (a) Step2: Explanation

No, because the sample is a voluntary response sample, the confidence interval method's randomness condition is not met, and so the claimed1.6 percent margin of error is wrong.

03

Part (b) Step 2: Given Information 

Need to explain the stations conclusion.

04

Explanation(Part b)

No, because the sample is a voluntary response sample, the confidence interval method's randomization condition is not met, and hence the conclusion based on the confidence interval is erroneous.

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Most popular questions from this chapter

57. Critical values What critical value t* from Table B would you use for a confidence interval for the population mean in each of the following situations?
(a) A 95%confidence interval based on  n=10 observations.
(b) A 99%confidence interval from an SRS of 20 observations.

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(b) Does this study show that living near power lines doesn't cause cancer? Explain.

5. NAEP scores Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source
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(b) Sketch the sampling distribution of x. Mark its mean and the values one, two, and three standard deviations on either side of the mean.
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(d) Whenever x¯falls in the region you shaded, the population mean μlies in the confidence intervalx±m. For what percent of all possible samples does the interval capture μ?

33. Going to the prom Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.
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(b) Check conditions for constructing a confidence interval for the parameter.

(c) Construct a 90% confidence interval for p. Show your method.
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A Census Bureau report on the income of Americans says that with 90% confidence the median income of all U.S. households in a recent year was \(57,005 with a margin of error of ±\)742. This means that

(a) 90% of all households had incomes in the range \(57,005 ±\)742.

(b) we can be sure that the median income for all households in the country lies in the range \(57,005 ±\)742.

(c) 90% of the households in the sample interviewed by the Census Bureau had incomes in the range \(57,005 ±\)742.

(d) the Census Bureau got the result \(57,005 ±\)742 using a method that will cover the true median income 90% of the time when used repeatedly.

(e) 90% of all possible samples of this same size would result in a sample median that falls within \(742 of \)57,005.

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