Question

Anystate Auto Insurance Company took a random sample of 386 insurance claims paid out during a 1-year period. The average claim paid was $1585. Assume σ = $256.

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 $

please explain how to get the z score using the table. I can not figure it out.

please do not use excel. I am trying to learn how to do this.

Answer 1

Solution:
Given in the question
Population mean () = $1585
Population standard deviation () = $256
Number of sample (n) = 386
Confidence interval can be calculated as
+/- Zalpha/2 * /sqrt(n)
Solution(a)
at confidence level = 0.90
level of significance() = 1 - 0.90 = 0.10
alpha/2 = 0.1/2 = 0.05
Lower level alpha = 0.05, from Z table, Z-score = (-1.64-1.65)/2 = -1.645
upper level alpha = 0.90+0.05 = 0.95 from Z table Z-score = (1.64+1.65)/2 = 1.645
So from Z table we found Z-table = +/-1.645
and 90% confidence interval can be calculated as
+/- Zalpha/2 * /sqrt(n)
1585 +/- 1.645*256/sqrt(386)
1585 +/- 21.43
So 90% confidence interval is 1563.57 to 1606.43
Solution(b)
at confidence level = 0.99
level of significance() = 1 - 0.99 = 0.01
alpha/2 = 0.01/2 = 0.005
Lower level alpha = 0.005, from Z table, Z-score = (-2.57 - 2.58)/2 = -2.575
upper level alpha = 0.99+0.005 = 0.995 from Z table Z-score = (2.57 + 2.58)/2 = 2.575
So from Z table we found Z-table = +/-2.575
and 99% confidence interval can be calculated as
+/- Zalpha/2 * /sqrt(n)
1585 +/- 2.575*256/sqrt(386)
1585 +/- 33.55
So 99% confidence interval is 1551.45 to 1618.55