A production process that fills 12-ounce cereal boxes is known to have a population standard deviation of 0.009 ounce. If a consumer protection agency would like to estimate the mean fill, in ounces, for 12-ounce cereal boxes with a confidence level of 92% and a margin of error of 0.001, what size sample must be used?

Given Information:  

Margin of error = 0.001

standard deviation = σ = 0.009

Confidence level = 92%

Required Information:

Population size = n = ?

Step-by-step explanation:

As we know

Margin of error = z*(σ/√n)

Re-arranging the equation to find n

√n = σz/margin of error

z-score corresponding to 92% confidence level = 1.72

√n = 0.009*1.72/0.001

√n = 15.48

square both sides

n = 15.48² = 239.63 = 240

n = 240 size sample should be used.

can i get the link to the assignment?

step-by-step explanation:

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.