Question

The USDA is interested in knowing if the Tastes Good Peanut Company is really keeping to their claim of actually providing 50 lbs. of Peanuts in their “50 Pound Bag” that they sell to the general public. To test this, the auditor takes a sample of 100 bags of peanuts and calculates their mean weight. He finds from his sample of 100 bags that the Sample Mean weight is 49.8 lbs. and that the Sample Standard Deviation is 1.5 lbs.

1. Calculate a 90% Confidence Interval around the mean.
2. As the auditor, do you find this to be a plausible value to support that the company is putting at least 50 lbs of peanuts in their “50 Pound Bag” given the result of your confidence interval? State your conclusion in the language of the problem
3. If you were to use a 80% Confidence Level instead, would that change your decision? Why or why not?

Answer 1

a)
sample mean, xbar = 49.8
sample standard deviation, s = 1.5
sample size, n = 100
degrees of freedom, df = n - 1 = 99

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.66

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (49.8 - 1.66 * 1.5/sqrt(100) , 49.8 + 1.66 * 1.5/sqrt(100))
CI = (49.55 , 50.05)

b)
As 50 is included in the calculated CI, we can say that the company is putting at least 50 lbs of peanuts in their “50 Pound Bag".

c)
Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, tc = t(α/2, df) = 1.29

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (49.8 - 1.29 * 1.5/sqrt(100) , 49.8 + 1.29 * 1.5/sqrt(100))
CI = (49.61 , 49.99)

As CI does not include 50, we can not conclude that the company is putting at least 50 lbs of peanuts in their "50 pound bag"