Question

A manager in a production line of cereal wants to estimate the mean weight of the boxes of cereal produced under his supervision. He randomly selects a sample of 28 boxes and finds that the sample mean xbar = 12.07 ounces and sample standard deviation s = 0.23ounces .

(a) What is the population? What is the sample?

(b) What is the population mean with 90% confidence?

Answer 1

a)
Population : The boxes of cereals produced under manager's supervision

Sample: 28 boxes which is taken from the population

b)
sample mean, xbar = 12.07
sample standard deviation, s = 0.23
sample size, n = 28
degrees of freedom, df = n - 1 = 27

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.703

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12.07 - 1.703 * 0.23/sqrt(28) , 12.07 + 1.703 * 0.23/sqrt(28))
CI = (12.00 , 12.14)