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**CFA Practice Question**

A random sample of 85 group leaders, supervisors, and similar personnel revealed that on the average a person spent 6.5 years on the job before being promoted. The standard deviation of the sample was 1.7 years. Using the 0.95 degree of confidence, what is the confidence interval within which the population mean lies?

A. 6.49 and 7.49

B. 4.15 and 7.15

C. 6.14 and 6.86

**Explanation:**Interval estimate can be found from x_bar +/-z*s/(n

^{0.5}). Here we have n = 85, x_bar = 6.5 and z = 1.96 (for 95%) and s = 1.7. Therefore 6.5+/-1.96*1.7/9.22 and we get 6.14 and 6.86.

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**User Contributed Comments**
12

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

Alastair |
of just pick the answer where the mean is 6.5.... |

julamo |
ahah Alastair, didn't think about that but that's really the smart way to do it! |

dimanyc |
Alastair, thank you. that's a great shortcut! |

wink44 |
Except on the test you can bet they'll all have the same mean :/ |

Kuki |
thanks Alastair... |

AkashKB |
Thats what I did. Kudos Alastair |

hannovanwyk |
in the test they'l have the same means, and if your calculation fails then...ye.... |

takor |
Alastair, thx for the trick. Saves me good time! |

shiva5555 |
Lol, I just assumed that 6.5 would be the mean of all the answers, so I did the calculations. When i saw the answers I was annoyed. Surely they won't make it this easy on the test. |

andrewmorgan |
I saw a question very similar to this in schweser, you worked it out based on 1.96, which was wrong because you cant assume this distribution is normal! fundamental error of this question |

Shcote |
You have to use normal distribution for a sample of 30 or more |

tzanchan |
In the real test they will almost definitely have the same mean for all three answers. It doesn't hurt to try if you have no better method, but not a good idea to rely on it. |