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**CFA Practice Question**

A sample of size 3,000 of apples has the total sum of weights equal to 1785 pounds, and total sum of squared weights equal to 3823470. The estimate of the standard error of sample mean is:

A. 0.004858

B. 0.004417

C. 0.005344

**Explanation:**We don't know what the population standard deviation is, but only have an estimate for it (which is probably a pretty good estimate given the very large sample size). std dev = ((sum of squared weights - (sum of weights)

^{2})

^{0.5})/Sample Size

standard error of sample mean = std dev / (size of sample

^{ 0.5})

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**User Contributed Comments**
5

User |
Comment |
---|---|

chantal |
Std deviation SB square root of variance variance sb the differences in squares divided by the sample size only once this is done do we take the square root of the variance so Variance = (3823470-3186225)/ 3000= 212.42 standard deviation = 14.56 standard error becomes 14.56 / 54.77 which is high |

lisab0131 |
if you follow their exact explanation though you get to the answer: ((3823470-3186225)^1/2)/3000 = 0.266 then divide by sqrt (3000) = 0.004858 |

kazec1 |
total sum of squared weights equal to 3823470: assuming uniform apples. Each apple weighs: sqrt(3823470/3000) = 35.7pound. hoho |

mary11 |
since its a sample isn't it n-1? |

ctschro |
i agree with chantal. this questions seems to be off. the answer choices are all far too small even for a sample of size 3000. also why divide first by n and then by sqrt(n)? my notes say divide by sqrt(n) twice. |