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**CFA Practice Question**

A statistician has framed his hypothesis testing problem as:

H

_{0}: mean >= 1; H_{1}: mean < 1For the given sample, he calculates the z-statistic. Then, the region of rejection at the 95% level is given by:

A. z-statistic < -1.96

B. z-statistic < -1.645

C. z-statistic > +1.96

**Explanation:**Since the alternative hypothesis is directional and to the left, the rejection region for the null is also to the left. Hence, you should use a one-tailed test. 5% of the probability mass in a normal distribution lies in the left tail with values which are less than -1.645. Therefore, the rejection region comprises of z-values less than -1.645.

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**User Contributed Comments**
5

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

brujita94 |
I supose it should be answer A value of 1.96 negative. It could be B is the region was 90%! |

Will1868 |
1.645 is for the 90% level right? |

snider |
1.96 is for 95% but two tails. 1.64 is for 90% two tails. So 1.64 is for 90% one tail which is applicable to this question. |

volkovv |
snider i think you meant to say 1.64 is for 95% one tail you can think of it this way: Z value for 1.96 is .975, so its 97.5% for one tail or 95% for two tail, since you chop .025 from the other side as well Z value for 1.64 is .95, so its 95% for one tail or 90% for two tail since you chop .05 from the other side the question here is for 95% and you have one tail test, so the correct Z value is 1.645 |

birdperson |
1.645 is 90% with 5% in each tail... so if you are one sided test and just want 5% in a tail... that's the magic number |