CFA Practice Question
There are 434 practice questions for this study session.
CFA Practice Question
If the number of days in a 92-day summer period in which a thunderstorm occurred is 28, then a 90% confidence interval for the percentage of days in summer that have thunderstorms, p, is ______ (to nearest 0.1%).
A. 21.0% < p < 39.8%
B. 25.6% < p < 35.2%
C. 22.5% < p < 38.3%
Explanation: The confidence interval is p' - E < p < p' + E. Let x count the number of days with thunderstorms; p' is x/n = 28/92 = 0.304. The computation of E is shown at the top right. So, E = 0.079 and the confidence interval is 22.5% < p < 38.5%.
User Contributed Comments 12
|schandri||how do we get sqrt(n) in the denominator??
isn't variance n.p.(1-p)??
|dealsoutlook||i dont get this answer..can someone please explain. Shouldn't variance be np(1-p) like the schandri posted and then we calculate sd for it? And then CI would be mean +- 1.96 (std error)? Please expain|
|uberstyle||This question is killing me. Why is n not included in the numerator? Is it because .304 is in terms of percentage of success, but an np(1-p) would be in terms of days, thus a .304 +_z(SE) would be combining of non-like units?|
|shavkat||They are apparently treating this as a Bernoulli trial, for which variance would be p(1-p). My question is then, how to decide, when to use Bernoulli trials and when Binominal distribution to calculate mean and variance?|
|Mariecfa||This question deals with proportions instead of means. The proportion is 28/92=.304. You would use n the denominator in the equation to find the mean but not for a proportion.
This can be very tricky. That is why it is important to know if they are asking to find the proportion, the mean on the variance to set up the correct equation.
|prajacti||i think you have to use bernoulli trial to find the std dev [sd = sqrt p(1-p) in a bernoulli trial] and then used the regular confidence interval formula. trick is to understand how to calculate std dev|
|atemple315||Hmm - I'm going to have to guess this - way tooo late to try and remember this formula. It just isnt sinking in!!|
|teje||Can someone please explain???|
|NickNT||I guess that variance of n*p*(p-1) is for binomial random variable B(n,p) where n is expected number of successes (which 92 in this case is not). So one shall use variance for a Bernoulli random variable p(1-p) to solve this task.|
|Shaan23||Very simple. Understand one thing and the TWO are not related
POPULATION Variance of binomial = npq
SAMPLE variance = p(1 - p) /n
just know that and then its easy.
|Boydbri1||Is a population variance as they are talking about the summer and not sampling from the summer period.
[.304(1-.304)*92]^.5=4.412 (if this were sample you would divide by 92^.5)