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**Basic Question 1 of 10**

A consumer group wants to prove that average hospital costs are more than $931 per day. The group randomly samples 60 accounts and finds a sample mean of $950. The hypothesis test is at an α-level of 5% and assumes a σ of $50. The critical value is ______ standard deviations.

B. -1.645

C. 1.28

A. 1.645

B. -1.645

C. 1.28

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**User Contributed Comments**
7

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

Bibhu |
One can refer to www.mathsnet.net/asa2/modules/s33binom.html for further reading on this subject. The easier way is sample mean 950 above 931. So +Z and then 5% means 1.645. |

whiteknight |
we can approach the problem like this...for 5% level of significance in a one tail test the z value is 1.645...i.e the right hand portion of the normal distribution....thus for any Z greater than 1.645 this statement will hold true ... |

raymondg |
could somebodyplease explain how to arrive to 1.645 ? |

meghanchloe |
I THOUGHT 5% LEVEL OF SIGNIFICANT IS 1.96 |

ebayer |
5% one tail = 10% both tails -> use 1.645 instead of 1.96. |

johntan1979 |
This test is clearly one-tailed (>931). So answer should be 1.645 if alpha is .05. two-tailed would be 1.96. |

Sp1993 |
Hi johntan1979, could you please clarify why a two-tailed test would give an answer of 1.96! Thanks! |

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**Learning Outcome Statements**

explain hypothesis testing and its components, including statistical significance, Type I and Type II errors, and the power of a test

*CFA® 2024 Level I Curriculum, Volume 1, Module 8.*