- 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**

For a 90% confidence interval for the population proportion, p, if the sample proportion, p', is 0.4 and the sample size is n = 100 then the error term, E, is ______ (to the nearest 0.001).

B. 0.004

C. 0.081

A. 0.049

B. 0.004

C. 0.081

Correct Answer: C

The error term is some number of standard deviations, in this case, 1.645 (the cutoff for the top 5% of the normal distribution). The standard deviation of the sampling distribution is sqrt[(0.4)(1 - 0.4)/100] = 0.049. Now, E = 1.64(0.049) = 0.081. Thus, the confidence interval for p is 0.4 - 0.081 < p < 0.4 + 0.081.

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**User Contributed Comments**
14

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

danlan |
How to get sqrt(0.4*0.6/100)? |

akanimo |
am lost here ... can anyone explain how variance = s^2 = p'(1 - p') i havent seen this formula anywhere yet??? |

sunilcfa |
its done with binomial dis |

surob |
Didn't understand. In binomial, variance is equal to np(1-p), not p(1-P). Can someone clarify? |

weiwei |
see the http://mathworld.wolfram.com/SampleProportion.html |

Tomcat82 |
I get it. The formula for var of binomial is var=np(1-p) The formula for see=sd/(n^(1/2)), so var is this squared. since you have n in the numerator, and n^2 in the demoninator, they cancel out, so you're just left with n in the demoninator, and no n in the numerator. Now remember, this is the formula for the variance. to get the SEE, you have to take the sqare root, thus you're left with the equation from above |

StanleyMo |
Normal Approximations for Counts and Proportions For large values of n, the distributions of the count X and the sample proportion are approximately normal. This result follows from the Central Limit Theorem. The mean and variance for the approximately normal distribution of X are np and np(1-p), identical to the mean and variance of the binomial(n,p) distribution. Similarly, the mean and variance for the approximately normal distribution of the sample proportion are p and (p(1-p)/n). |

cfabuzz |
simply, variance of distribution mean = s^2/n the variance(s^2)in binomial is np(1-p) so we get np(1-p)/n --> n cancel out and we get var = p(1-p), so standard deviation(s)is square root of p(1-p) E = 1.645* s/ square root of n E = 1.645* [p(1-p)/n]^1/2 |

panvino |
Good question! |

johntan1979 |
http://www.stat.yale.edu/Courses/1997-98/101/binom.htm pop mean, var = np, np(1-p) sample mean, var = p, [p(1-p)]/n Too easy to assume sample var is divided by only one n to become p(1-p) like the mean. |

sgossett86 |
I had to kinda backtrack to learn this one.. We're talking about a proportion, so assume that the Single Bernoulli Variable rules apply. They were one of the important formulas we went over regarding binomials. Review it if necessary. Variance=p(1-p)=.4(1-.4) sd=(.4(1-.4))^.5=.48990 We're solving for E E=Z(a/2)*(sd/N^.5)=(1.645)*(.48990/100^.5)=.08059 |

sgossett86 |
I couldn't got this without initial guidance from all your suggestions above, but that's the formula breakdown. |

lndcobra |
What makes you realise that this is talking about a Binomial Distribution, and hence arriving at var =s^2 = np(1-p) |

lawlee |
I still don't get how the n in the numerator is offset, squarting n^1/2 on the denominator get you to n, not n^2. Can someone clarify? |