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 $1595. Assume σ = $270.

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 = = 1595

Population standard deviation = = 270

Sample size = n =384

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645  
Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 270 /  384 )

= 22.66

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

- E < < + E

1595 - 22.66 <   < 1595 + 22.66

1572.34<   <1617.66

( 1572.34,1617.66 )

Lower limit = 1572.34

Upper limit = 1617.66

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z_{0.005 = 2.576}

Z/2 = 2.576

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

= 2.576 * ( 270 /  384 )

= 35.49

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

- E < < + E

1595 -35.49 <   < 1595 + 35.49

1559.91<   <1630.49

(  1559.91,1630.49)

Lower limit =  1559.91

Upper limit = 1630.49