- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 5. Sampling and Estimation
- Subject 5. Confidence Intervals for the Population Mean and Selection of Sample Size

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**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?

Correct Answer: 27,060.25

Since n = [z x σ/m]

^{2}, where z = 1.645, σ = 0.01 and m = 0.0001, n = 27,060.25.###
**User Contributed Comments**
12

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 |