Question

any state auto insurance company took a random sample of 360 insurance claims paid out during a one year. The average claim paid was $1,575 assume the name equals $238 find a 0.90 confidence interval for the mean claim round your answers to two decimal places lower limit of the find a 0.99 confidence interval for the mean claim payment round your answers to two decimal places lower limit upper limit

Answer 1

Solution:

Given:

Sample size = n = 360

Sample mean=

Population standard deviation =

Part a) Find a 0.90 confidence interval for the mean claim:

Formula:

where

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

Thus

Thus

Lower Limit = $ 1554.37

Upper Limit = $ 1595.63

Part b) find a 0.99 confidence interval for the mean claim

Formula:

where

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Thus

Thus

Lower Limit = $ 1542.70

Upper Limit = $ 1607.30