- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 6. Hypothesis Testing
- Subject 8. Tests Concerning a Single Mean

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**CFA Practice Question**

A consumer group wants to prove that the 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 a-level of 5% and assumes a s of $50. The p-value is ______% (to the nearest 0.1%).

B. 0.2%

C. 99.8%

A. 5.0%

B. 0.2%

C. 99.8%

Correct Answer: B

The p-value is P(x-bar = 950 given a population mean of 931 or less). The z-score for 950 based on this assumption is 2.94. The table value for 2.94 is 0.9984. So, p = 1 - 0.9984 = 0.0016 = 0.2% (to the nearest 0.1%).

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**User Contributed Comments**
13

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

GeoffT |
Need z-table |

danlan |
(950-931)/(50/sqrt(60))=2.94, how to get 0.9984? |

lockedin |
From the z-table, half the area under the curve for 2.94 = 0.4984. The total area for 2.94 is 0.4984 x 2 = 0.9968. Therefore, p = 1 - 0.9968 = 0.32%. I think the 0.9984 from the answer is wrong. Anyone else? |

charlie |
The answer is correct. Notice this is a one-tailed test so 0.5 + 0.4984 = 0.9984. |

Janey |
YOu dont need to calculate this question. P is always less than the level of significance so answer has to be 0.2% |

xcye |
Janey, that's not true, p-value is only less than the level of significance if you reject the Ho! |

SuperKnight |
In this case you don't really need to calculate the p-value to know that it is lower than 5%, when you get the z-value of 2.94.. you know that its greater than 2.58 which makes up the 99% confidence interval (meaning alpha is 0.5% in each tail).. so you know that your value has to be even lower than 0.5%, in this case being 0.2%. |

bantoo |
great wisdom and common sense janey |

johntan1979 |
SuperKnight is THE MAN! |

gill15 |
Janey: P is not always less then significance. There can be large p values and that would indicate we would NOT reject the null hypothesis. Thats the point of calculating it. |

sgossett86 |
I get how we found the Z score, and I get that it gets plugged into the equation and that it will come out as a probability using (1-x), but I don't really understand anything else about it. It doesn't seem like we used Alpha at all in the problem.. and what is this answer really saying?? |

Shaan23 |
Sgoss - we are NOT using alpha in these questions. First all we're doing is finding the z critical value which is 2.94 here. That percentage when you look up in the Ztable is .2%. That .2% all it means is that value of significance is REQUIRED for the null hypothesis to be rejected. |

jgoff508 |
Wow, helpful question . |