Question 185747: can someone please help me solve this? thank you so much.
An insurance company is trying to determine, using the 95% confidence level, what proportion of teenagers have an accident. Using a sample of 582 accidents, the company discovers that teenagers were involved in 91 accidents. What is confidence interval for this proportion?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An insurance company is trying to determine, using the 95% confidence level, what proportion of teenagers have an accident. Using a sample of 582 accidents, the company discovers that teenagers were involved in 91 accidents. What is confidence interval for this proportion?
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sample proportion = 91/582 = 0.1564
Standard error = E = z*sqrt[pq/n] = 1.96sqrt[0.1564*0.8436/582] = 0.0295
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95% CI: 0.1564 - 0.0295 < p < 0.1564 + 0.0295
95% CI: 0.1269 < p < 0.1859
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Do you understand this procedure for constructing confidence intervals?
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Cheers,
Stan H.
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