Question

Anystate Auto Insurance Company took a random sample of 360 insurance claims paid out during a 1-year period. The average claim paid was $1535. Assume σ = $252. Find a 0.90 confidence interval for the mean claim payment. (Round your answers to two decimal places.) lower limit $ upper limit $ Find a 0.99 confidence interval for the mean claim payment. (Round your answers to two decimal places.

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

lower limit	$
upper limit	$

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 that,

Point estimate = sample mean = = $1535

Population standard deviation =    = $ 252

Sample size = n =360

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 =1.645

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 252/  360 )

= 21.85

At 90% confidence interval estimate of the population mean is,

  ± E

1535  ± 21.85

( $1513.15, $ 1556.85 )

lower limit = $ 1513.15

upper limit = $ 1556.85

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 252/  360 )

= 34.21

At 99% confidence interval estimate of the population mean is,

  ± E

1535  ± 34.21   

( $1500.79, $ 1569.21 )

lower limit = $ 1500.79

upper limit = $ 1569.21