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An insurance company is trying to determine, using the 95% confidence level, what proportion of teenagers have an accid
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An insurance company is trying to determine, using the 95% confidence level, what proportion of teenagers have an accid
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Question 185745: 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?
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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