Question 933936: A poll asked for people's opinion on whether closing local newspapers would hurt civic life; 430 of 1001 respondents said it would hurt civic life a lot.
A. Find the proportion of the respondents who said closing local papers would hurt civic life a lot.
(round to three decimal places as needed)
B. Find a 95% confidence interval for the population proportion who believed closing newspaper would hurt civic life a lot. Assume the poll used a simple random sample (SRS) (In fact, it used random sampling, but a more complex method than SRS
(, ) round to three decimal places as needed
C. Find an 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot.
(, ) round to three decimal places as needed
D. Which interval is wider?
Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A poll asked for people's opinion on whether closing local newspapers would hurt civic life; 430 of 1001 respondents said it would hurt civic life a lot.
A. Find the proportion of the respondents who said closing local papers would hurt civic life a lot. (round to three decimal places as needed)
Ans: 430/1001 = 0.430 or 430 out of 1000
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B. Find a 95% confidence interval for the population proportion who believed closing newspaper would hurt civic life a lot. Assume the poll used a simple random sample (SRS) (In fact, it used random sampling, but a more complex method than SRS (, ) round to three decimal places as needed
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p-hat = 0.43
ME = 1.96*sqrt[0.43*0.57/1001] = 0.031
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95% CI:: 0.43-0.031 < p < 0.43+0.031
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C. Find an 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot.
(, ) round to three decimal places as needed
ME:: 1.2820*sqrt[0.43*0.57/1001] = 0.02
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80% CI: 0.43-0.02 < p < 0.43+0.02
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D. Which interval is wider?
95% CI is wider than 80% CI.
The more confidence you want, the wider the CI must be.
Cheers,
Stan H.
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