Question

Anystate Auto Insurance Company took a random sample of 384 insurance claims paid out during a 1-year period. The average claim paid was $1500. Assume σ = $258.

A) Find a 0.90 confidence interval for the mean claim payment. (Round your answers to two decimal places.)
lower limit $________

upper limit $_____

B) Find a 0.99 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

lower limit $_________

Upper limit $_________

Answer 1

A.
TRADITIONAL METHOD
given that,
standard deviation, σ =258
sample mean, x =1500
population size (n)=384
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 258/ sqrt ( 384) )
= 13.17
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 13.17
= 21.66
III.
CI = x ± margin of error
confidence interval = [ 1500 ± 21.66 ]
= [ 1478.34,1521.66 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =258
sample mean, x =1500
population size (n)=384
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1500 ± Z a/2 ( 258/ Sqrt ( 384) ) ]
= [ 1500 - 1.645 * (13.17) , 1500 + 1.645 * (13.17) ]
= [ 1478.34,1521.66 ]$
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [1478.34 , 1521.66 ] $ contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

B.
TRADITIONAL METHOD
given that,
standard deviation, σ =258
sample mean, x =1500
population size (n)=384
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 258/ sqrt ( 384) )
= 13.17
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 13.17
= 33.92
III.
CI = x ± margin of error
confidence interval = [ 1500 ± 33.92 ]
= [ 1466.08,1533.92 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =258
sample mean, x =1500
population size (n)=384
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1500 ± Z a/2 ( 258/ Sqrt ( 384) ) ]
= [ 1500 - 2.576 * (13.17) , 1500 + 2.576 * (13.17) ]
= [ 1466.08,1533.92 ]$
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [1466.08 , 1533.92 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean