- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 8. Hypothesis Testing
- Subject 7. Test of a Single Mean

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**CFA Practice Question**

The supporters of Judge Hyde want to prove that she has the support of more than 60% of the voters in the district. To accomplish this, they take a random sample of 80 voters and find that 53 support Judge Hyde. For a hypothesis test at the 5% level of significance, the conclusion would be ______

B. Judge Hyde has 66.25% of the voters' support.

C. there is no evidence that Judge Hyde has more than 60% of the voters' support.

A. Judge Hyde has more than 60% of the voters' support.

B. Judge Hyde has 66.25% of the voters' support.

C. there is no evidence that Judge Hyde has more than 60% of the voters' support.

Correct Answer: C

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**User Contributed Comments**
18

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

kaliokale |
how did he get the test value? what formular is that? |

johans |
how did they derive the standard deviation? |

yliu |
bernoli distribution, variance = p * (1 - p) |

ehc0791 |
for sample proportion, the test value = z value (p - p-sub0)/s s = sqrt(p*(1-p)/n) |

tabulator |
yliu, don't confuse people. ehc0791 - well done. |

StanleyMo |
When the answer is too big, more than 4, we conclude that we do not have the evidence. |

chamad |
Why using bernouli trial in this case. |

labine |
bernoulli trial applies because you're dealing with the probability of sucess or failure(Hyde does or does not have voters' support). yliu is correct: variance for binomial distributions=p*(1-p), therefore the standard deviation is [p*(1-p)]^1/2 and the standard error of the mean is: [p*(1-p)]^1/2/(n^1/2)=[p*(1-p)/n]^1/2=[0.6(1-0.6)/80]^1/2 |

FayeMulvaney |
additionally, remember the rule that if the p-value is smaller than or equal to the level of significance (alpha) one generally rejects the null hypothesis. In this case, p is greater therefore we do not reject it. Could somebody please tell me how to calculate the critical value out of interest? |

KimActuary |
labine: to calculate the sample s.d., why not use .6625? Isn't that the p for the sample? I don't understand why you would use .6 when that is a test p. |

afficionado |
I thought the Bernomial distribution variance is: n*p(1-p)instead of p(1-p). Can someone comment on this please? |

johntan1979 |
There's been more than one similar question testing this formula before this in previous topics, so if you didn't get them back then, you will not get it now. Test value = (p'-p)/sqrt p(1-p)/n |

gill15 |
Did you guys skip all the other sections in this reading to not know where this comes from |

Thediceman |
I get correct answer with using wrong P value... same question as KimActuary. why use .60 instead of .6625 thank you |

sgossett86 |
THEDICEMAN & KIMACTUARY I keep having the same problem on these questions.. using sample as P ... |

Shaan23 |
I wouldnt use z test here. It specifically says in the notes if the population variance is UNKNOWN, use t score irrespective of sample size. So even though we have n=80 which is approaching the Z table values --- it isnt the same number. |

farrahkame |
Could we not use p = 1 - (table value of z-score) AKA table value of z-score = 1- 0.6625 = 0.3375 look it up in the table to find the closest z-score being -0.42 (table value 0.3372) and -0.42 would fall in the -1.645 < test value < 1.645 hence, fail to reject H0... does that make sense? |

SandraZake |
Guys isn't the critical value -1.645, confirmed by the previous question. Since the test value is 1.14 (positive) this is greater than the critical value and therefore the Ho is rejected. Let me know your thoughts. |