Question

Anystate Auto Insurance Company has shown that their paid claims have a standard deviation that is approximately​ $260. We would like to take a sample of their paid claims such that our estimate has a margin of error within ​$50 of the true mean. What sample size would we need to take if we wanted to use a​ 99% confidence​ level?

Confidence Level​ =

Standard Deviatio​n =

Width​ =

Sample Size​ =

Answer 1

Solution :

Given that,

standard deviation =   =$260

Margin of error = E = $50

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z/2* / E] 2

n = ( 2.58* 260 / 50 )2

n =178.98

n=180

Confidence Level​ =Z/2 = 2.58

Standard Deviatio​n =$260

Width​ =$100

Sample Size​ =180