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 percent 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

To find the sample size (n), we use the formula:
n = [(z*σ)/E]^2
Where z is the z-value for the desired confidence level (97% = 1.88), σ is the population standard deviation (.06 ounces), and E is the margin of error (.02 ounces).
Plugging in the values, we get:
n = [(1.88*0.06)/0.02]^2
n = 42.336
Since we can't have a fraction of a sample size, we round up to the nearest whole number, which gives us a sample size of 43. Therefore, the best choice is B.

A Statistical Quality Control Process for Cereal Production Measures the Weight