Question

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?

Answer 1

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.

Answer 2

Margin of error = 0.001

standard deviation = σ = 0.009

Confidence level = 92%  

• The calculation is as follows:

As we know that  

Margin of error = $z\times (\sigma\div \sqrt n)$

here

√n = σz/margin of error

z-score corresponding to 92% confidence level = 1.72

$\sqrt n = 0.009\times 1.72\div 0.001$

√n = 15.48

Now square both sides

n = 15.48² = 239.63 = 240

n = 240 size sample should be used.

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