Question

1. Bottled drinking water distributors want to estimate the contents of the water contained in a 1-gallon bottle purchased from a well-known national company. The bottling company's specifications state that the standard deviation of the water content is 0.04 gallons. A random sample of 50 bottles was chosen, and the average sample content of one gallon of water content was 0.927 gallons.

a. Arrange 95% confidence intervals for the average population of water contents in a one-gallon bottle.
b. On the basis of these results, do you think that the distributor has the right to complain to a water bottling company? Why?
c. Arrange 90% confidence intervals. Does this change the answer to (b)?

Answer 1

a)

sample mean, xbar = 0.927
sample standard deviation, σ = 0.04
sample size, n = 50

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 0.04/sqrt(50)
ME = 0.01

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (0.927 - 1.96 * 0.04/sqrt(50) , 0.927 + 1.96 * 0.04/sqrt(50))
CI = (0.916 , 0.938)

b)

yes, because confidence interval does not contain in the interval

c)
sample mean, xbar = 0.927
sample standard deviation, σ = 0.04
sample size, n = 50

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 0.04/sqrt(50)
ME = 0.01

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (0.927 - 1.64 * 0.04/sqrt(50) , 0.927 + 1.64 * 0.04/sqrt(50))
CI = (0.918 , 0.936)

no, answer is same as b)