- CFA Exams
- CFA Level I Exam
- Study Session 3. Quantitative Methods (2)
- Reading 10. Sampling and Estimation
- Subject 6. Confidence Intervals for the Population Mean

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**CFA Practice Question**

A researcher randomly samples 100 residents of Florida and finds that 95 of the citizens are literate. The researcher then finds a 90% confidence interval. Which of the following is false?

A. The researcher is 90% sure that the percentage of literate citizens is between 91.4% and 98.6%.

B. The sample proportion p' is 0.95 and the margin of error is 3.6%.

C. The confidence interval procedure is invalid because Np' and N(1 - p') are not both greater than 5.

**Explanation:**The procedure used for confidence intervals assumes that Np' > 5 and that N(1 -p') > 5. Because this is not the case and 100(1 - 0.95) = 5, the researcher cannot conclude with 90% confidence that the percentage of Florida citizens that are literate is between 91.4% and 98.6%.

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**User Contributed Comments**
7

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

haarlemmer |
How to caculate the margin of error though? |

tengo |
variance = np(1-p) 100*.95*.05 =4.75 |

MUSK |
What is this "NP, N(1-P) both not being higher than 5?" Is there a rule like that for determining the intervals? Please reply. |

aggabad |
margin of error= 1.645*(.05*.95/100)^1/2 |

Epix |
Where in the book does it talk about the rule for C. I cannot find it. |

Mariecfa |
Margin of Error E = Test Statistic*(P*(1-P)/(SD)/n^1/2)) aggabad forgot the standard deviation 2.179. E = 1.645*(.95*.05)/(2.179/(100^1/2)) or you use the variance from tengo's suggestion. E = 1.645*(.95*.05)/(4.75/100)^1/2 |

GBolt93 |
Mariecfa, those give you the wrong answer. you end up with a error of .358 when it should be .0358 which you get from aggabad's calculation. .95*.05 is the variance, so not sure why you're dividing that by itself. |