### CFA Practice Question

There are 434 practice questions for this study session.

### CFA Practice Question

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

Since n = [z x σ/m]2, where z = 1.645, σ = 0.01 and m = 0.0001, n = 27,060.25.

User Comment
tinku good formula to remember
quincy don't understand this formula, can anyone explain it a little bit more?
Gina confidence interval for u = estimate +/- margin of error (m) if you just look at second part of the formula, the margin of error: m = z (stddev)/n^1/2 solve for n: n = [z(stddev)/m]^2
jb1969 1.645 x 0.01/n^0.5 = 0.0001

solving for n : n= 27060.25
StanleyMo To explain further,

Confidence interval = Point estimate (mean) + realibility factor x standard factor

= X + Z * (stddev)/ n ^ 1/2

margin of error = Z * stddev / n ^ 1/2

so, n ^ 1/2 = (Z * stddev) / margin or error

n = ((Z * stddev)/ margin of error ) ^ 2
apiccion Look, it's not necessary to remember another formula; the solution is not always memorization.

Remember that the confidence interval is:
u - zσ' <= u <= u + zσ'; where σ' = σ/sqrt(n)

The question is asking you for what value of n will the margin of error will be 0.0001. In other words, for what value of n will
u + zσ' - u = 0.0001
or
zσ' = 0.0001
zσ/sqrt(n) = 0.0001
zσ/0.0001 = sqrt(n)
(zσ/0.0001)^2 = n
Allen88 Neat, thanks apiccion!
ybavly Neat, thanks apiccion!
leftcoast Neat, thanks apiccion!
johntan1979 I know this is probably insignificant, but the more precise answer is 27,061 times, since you can't have 0.25 times of repetition, and why round up is to make sure you have 90% included.
sgossett86 Guys it's just algebra.
sgossett86 Solve for N on the standard error formula

Om=O/N^.5
N=(O/Om)^2