Question

Part 1. Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from $2.99 to $3.99. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.)

x = $
E = $

Part 2. Anystate Auto Insurance Company took a random sample of 380 insurance claims paid out during a 1-year period. The average claim paid was $1510. Assume σ = $254.

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

1)

x = (upper + lower)value/2
= ( 3.99 + 2.99)/2
= 3.49

E = Upper value - x
= 3.99 - 3.49
= 0.50

2)

sample mean, xbar = 1510
sample standard deviation, s = 254
sample size, n = 380
degrees of freedom, df = n - 1 = 379

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.65

ME = tc * s/sqrt(n)
ME = 1.65 * 254/sqrt(380)
ME = 21.499

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1510 - 1.65 * 254/sqrt(380) , 1510 + 1.65 * 254/sqrt(380))
CI = (1488.50 , 1531.50)

Lower limit = 1488.50
Upper limit = 1531.50

sample mean, xbar = 1510
sample standard deviation, s = 254
sample size, n = 380
degrees of freedom, df = n - 1 = 379

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.59

ME = tc * s/sqrt(n)
ME = 2.59 * 254/sqrt(380)
ME = 33.748

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (1510 - 2.59 * 254/sqrt(380) , 1510 + 2.59 * 254/sqrt(380))
CI = (1476.25 , 1543.75)

Lower limit = 1476.25
Upper limit = 1543.75