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 $1520. Assume σ = $254.

(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

Solution :

Given that,

Point estimate = sample mean = = 1520

Population standard deviation =    = 254

Sample size = n =360

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 * ( 254 /   360)

= 22.02
At 90% confidence interval
is,

- E < < + E

1520 - 22.02 <   <1520 + 22.02

1497.98 <   < 1542.02

lower limit 1497.98

upper limit 1542.02

(b)

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* (254 / 360)

= 34.48

At 99% confidence interval is

- E < < + E

1520-34.48 < < 1520+34.48

1485.52< < 1554.48

lower limit 1485.52

upper limit 1554.48