I'm confused. An auto insurance company checks police records on 561 accidents slected at random and notes that teenagers were at the wheel in 81 of them.
Construct the 95% confidence interval for the percentage of all auto accidents that involve teenage drivers. 95% CI=(_ %,_ %) How do I figure this out?
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Ref: http://en.wikipedia.org/wiki/Binomial_pr…Positives Q=81 out of sample N=561, so estimate p=14.4%95% confidence interval spans the 2.5% to the 97.5% range.From a normal probability table, these would correspond toZ=+/-1.960 times sigma points.Therefore from the Wiki pageCI = p +/- Z*sqrt(p*(1-p)/N)= 0.144 +/- 1.960*sqrt((0.144)(1-0.144)/561)= 14.4% +/- 2.9% Source(s): http://en.wikipedia.org/wiki/Binomial_pr…
Ref: http://en.wikipedia.org/wiki/Binomial_pr…Positives Q=81 out of sample N=561, so estimate p=14.4%95% confidence interval spans the 2.5% to the 97.5% range.From a normal probability table, these would correspond toZ=+/-1.960 times sigma points.Therefore from the Wiki pageCI = p +/- Z*sqrt(p*(1-p)/N)= 0.144 +/- 1.960*sqrt((0.144)(1-0.144)/561)= 14.4% +/- 2.9% Source(s): http://en.wikipedia.org/wiki/Binomial_pr…