In: Math
Anystate Auto Insurance Company took a random sample of 358
insurance claims paid out during a 1-year period. The average claim
paid was $1530. Assume σ = $230.
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 | $ |
Solution :
Given that,
= $1530
= $230
n = 358
a) At 0.90 confidence level the z is,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1/ 2 = 0.05
Z/2 = Z0.05 =1.645
Margin of error = E = Z/2* ( /n)
= 1.645 * (230 / 358)
= 20
At 0.90 confidence interval estimate of the population mean is,
- E < < + E
1530 - 20 < < 1530 + 20
1510 < < 1550
(1510, 1550)
Lower limit = $1510
Upper limit = $1550
b) At 0.99 confidence level the z is,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01/ 2 = 0.005
Z/2 = Z0.005 =2.576
Margin of error = E = Z/2* ( /n)
= 2.576 * (230 / 358)
= 31.31
At 0.99 confidence interval estimate of the population mean is,
- E < < + E
1530 - 31.31 < < 1530 + 31.31
1498.69 < < 1561.31
(1498.69, 1561.31)
Lower limit = $1498.69
Upper limit = $1561.31
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