To assess the accuracy of a kitchen scale, a standard weight, known to weigh 1 gram, is weighed a total of n times, and the mean of the n weight measurements is computed. Suppose the scale readings are normally distributed with unknown mean μ and standard deviation σ = 0.01 gram. How large should n be so that a 90% confidence interval for μ has a margin of error of E = ± 0.0001 gram?

Question

To assess the accuracy of a kitchen scale, a standard weight, known to weigh 1 gram, is weighed a total of n times, and the mean  of the n weight measurements is computed. Suppose the scale readings are normally distributed with unknown mean μ and standard deviation σ = 0.01 gram. How large should n be so that a 90% confidence interval for μ has a margin of error of E = ± 0.0001 gram?

anonymous · Answer

Choices are:
A. 165
B. 27,061
C. 38,416

anonymous · Answer

Margin of error = z-score for 90% confidence*standard deviation/sqrt n
0.0001 = 1.645*0.01/sqrt n
sqrt n = 1.645*0.01/0.0001 = 164.5
on squaring both sides
n = 27061
Therefore answer to the question is CHOICE (B)...27,061