Answer to Question #310813 in Statistics and Probability for Kiddo

Question

A random sample of 200 travel insurance policies contain 29 of which the policy holders made claims in their most recent year of cover. Calculate a 99% confidence intervel for the proportion of policyholders who makes claims in a given year of cover.

Answer 1

$p=\frac{29}{200}=0.145.$

$99\%CI=(p-z_{0.005}\sqrt{\frac{p(1-p)}{n}}, p+z_{0.005}\sqrt{\frac{p(1-p)}{n}})=$

$=(0.145-2.576\sqrt{\frac{0.145(1-0.145)}{200}},0.145+2.576\sqrt{\frac{0.145(1-0.145)}{200}})=$

$=(0.0809, 0.2091).$