CFA Practice Question
There are 434 practice questions for this study session.
CFA Practice Question
A portfolio with 10 stocks has a weighted average return of 5.5% and a variance of 25. What is the 85% confidence interval for the weighted average?
A. [-35.625%, 46.625%]
B. [-30.5%, 41.5%]
C. [3.22%, 7.78%]
Explanation: The variance is 25, which corresponds to a standard deviation of 5%. The 85% confidence interval is given by: X-bar +- 1.44s = 5.5 +- 1.44 x 5/10^0.5 = [3.22, 7.78]
User Contributed Comments 19
|cfahopeful||This seems to be wrong. Using Chebyshev's inequality, .85 = 1 - (1/k^2) k = 2.58 Therefore, 85% of observations will lie between 5.5 + 2.58 * 5 and 5.5 - 2.58 * 5 -7.4 to 12.4|
|tony1973||The answer's correct. The confidence interval calculated by Chebyshev's formula will always be wider.|
|mm04||So which one should we use for the exam? I found this from another question too.|
|garachen||Yes, Chebyshev is only when you have no idea of the underlying distribution. Since you can assume portfolio returns are normal you must use a normal distribution. I can't see remembering that 85% is 1.44 st deviations though.|
|synner||yeah, how do you know that 85% is 1.44 st.dev?|
|jack48084||we know 90% is 1.645, therefore 85% must be less. So answer C is the only one could be correct.|
|Pieter||One way to approach it would be to remember that +/-1 stdev covers 65%
+/-2 stdev covers 95%
Therefore, 85% is somewhere in between, approx. close to +/-1.5 stdev.
|eddeb||In the z table :
85% / 2 (two tailed) = 0.425
At 0.425 we have a 1.44 z score.
I just don't know why we use the z table.. N is under 30, so we should use the t table. Using a r numbers rule, I have obtained a t score of 1.605.
|danlan||Variance of 25 is the variance of the portfolio. It is not the variance of a sample.|
|Nightsurfer||Isn't 1.44 standard deviations 92.51%, not 85%?|
|EtnicPlaymaker||(1 - 0,9251) = 0,075
0,075 * 2 = 15. Everything is ok.
|uberstyle||stupid question here, but why is it given that if the portfolio variance is 25, standard error is 5?|
|jackwez||std deviaion is the sqrt of the variance|
|Yohan3109||This question is logique, no need to calcul. Just think logicaly. Mean is 5,5 and standard deviation 5. The mean of the interval is 5.5 add 2 standard deviation for each side -5.5% to 15.5%.
check A and B are totally wrong event 3 standard divation doesnt fit.
|Mariecfa||This is a two tail test. Alpa/2: (1-.85)=.15/2=.075
Probability of .075 on left tail from Z table gives -1.44 for the SD.
Probability of (1-.075)=0.925 (This is what the probability is that lies to the left of the Right-tail critical value. On the Z table 0.925 shows 1.44. This is what gives you the Standard deviations. Hope this helps.
|showmethemoney||gotta use t-test. the question doesn't assume normal dist. can't just use z test.|
|charlie||You already know the population variance, so it's a z test, not t test.|
|homersimpson||To sum up, we use Z test cos we know the population variance and we multiple it by 5 instead of standard error (5/10^2) because it is a population, not a sample.|
|Marinov||Even if we do know the population variance, we still need to divide it by the square root of n. This is the formula of the standard error.|