A statistical quality control process for cereal production measures the weight of a cereal box.The population standard deviation is known to be .06 ounces.In order to achieve a 97% confidence with a margin of error of .02 ounces,how large a sample should be used?
A)32
B)43
C)7
D)664

Question

Quizplus · Accepted Answer

We need to use the formula:
n = (z*σ / E)^2
where
z = z-score for 97% confidence level = 1.8808 (from the Z-table or calculator)
σ = population standard deviation = 0.06 ounces
E = margin of error = 0.02 ounces

Plugging in the values, we get:
n = (1.8808*0.06 / 0.02)^2
n = 42.73

Rounding up to the nearest whole number, we get:
n = 43

Therefore, we need a sample size of 43 to achieve a 97% confidence with a margin of error of 0.02 ounces.

A Statistical Quality Control Process for Cereal Production Measures the Weight