CFA Practice Question
There are 434 practice questions for this study session.
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).
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.
User Contributed Comments 14
|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
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.
We're solving for E
|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?