### CFA Practice Question

Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for president. Using the 0.95 degree of confidence, what is the interval estimate for the population proportion (to the nearest tenth of a percent)?
A. 76.5% to 83.5%
B. 78.2% to 81.8%
C. 69.2% to 86.4%
Explanation: Interval estimate can be found from p +/- z[p(1-p)/n]0.5. Here we have n = 2000, p = 1600/2000 = 0.8 and z = 1.96 (for 95%). Therefore, 0.8 +/- 1.96*0.008944 and we get 0.7825 and 0.8175.

User Comment
humphrey do we have Z tables or T tables to consult during exam?
EK65 if this is a binomonal variance it should be np(1-p), not p(1-p)/n. Please explain.
murli I think it is from binomial linked normal distribution!
vivi2005 for a single Bernoulli random variable Y which takes on the value 1 with probabiity p and the value 0 with probability 1-p, its mean is p and its variance is p(1-p);
but then shouldn't its stanadard devidation be [p(1-p)]^0.5 ?
bawejate ok..i don't know abt which probability theorem it is but here is a self-made fluked easy logic:

z(0.95) = 1.96

(80-1.96) = 78.04
(80+1.96) = 81.96

now just choose the closest match!
sf2va For a normal distribution remember the three main confidence intervals:
90% confidence interval is 1.65 Std. Dev. z=1.65
95% confidence interval is 1.96 Std. Dev. z=1.96
99% confidence interval is 2.58 Std. Dev. z=2.58

Remembering these three Z values will get you pretty far on a multiple choice exam.
swift should it not be *n*p(1-p)/root n?
armanaziz p(1-p)=s^2....bernoulli random
So, s = [p(1-p)]^0.5
Now for hypothesis testing st. error = s/n^0.5
So here st. error =
[p(1-p)]^0.5/n^0.5
=[p(1-p)/n]^0.5

makes sense!!!
bhawa Thnx armanaziz, you are absolutely correct!!!!
Murashi Your standard error of 0.008944 is quite bizarre because no where is the sample standard deviation and sample size mentioned in the question.
dimanyc armanaziz, thx for your post - very helpful. one thing to add is that in a bermoulli trial n=1, that's why your bernoulli s={p*q}^0.5, and not s=p*q*n, where q=1-p
Quan makes sense to use n=1 in this case 'cause the questions asks interval in percentage not in # of people.
achu Provided solution is right; the one confusing part is that you have n=2000 for the sample n BUT for the variance of this Bernouli (special case of Binomial Var), the "N=1" (n * p * (1-p) = var for a binomial variable. Other than the 2 n letters running around this is fairly straight forward solution given.

thekapila all variance for bernouli = np(1-p)
now since its sample so s dev = var /sqrt n

sd =( np(1-p) / sqrt n) 1/2 =( p(1-p) / n ) 1/2

mean +_ z std = interval
6YASHWIN BAWEJATE where did u get the 80 from
shiva5555 How did they get .000894?
mindi for binomial distributions, mean = np and variance = p(1-p)/N
drb2234 I'm still not understanding how they got 0.000894...can someone please present the figures from the problem in the actual equation?
maurlamf It looks like they used plain STDEV instead of SE here.

.8(1-.8)2000 = 320 (variance)
sqrt(320) = 17.88 (STDEV)

The answer is asked for in % terms though, so you need to do 17.88/2000 = .008944

or, .8944%

so... 80% +/- 1.96(.8944%) = answer
emergis ((0.80)*(0.20)/2000)1/2 = .008944
acemaj well said maurlamf
moneyguy @6Washwin: 80% probability voters will vote Republican
will080912 I don´t understand why n=1 and not n=2000