CFA Practice Question

There are 434 practice questions for this study session.

CFA Practice Question

There are 2,000 eligible voters in a precinct. Despite protests from knowledgeable persons that a sample size of 500 was too large in relation to the total, the 500 selected at random were asked to indicate whether they planned to vote for the Democratic incumbent or the Republican challenger. Of the 500 surveyed, 350 said they were going to vote for the Democratic incumbent. Using the 0.99 confidence coefficient, what are the confidence limits for the proportion who plan to vote for the Democratic incumbent?
A. 0.397 and 0.797
B. 0.612 and 0.712
C. Neither of these answers are correct.
Explanation: The interval estimate can be found from np ± z[np(1-p)/n]0.5. Here we have n = 500, p = 350/500 = 0.7 and z = 2.58 (for 99%).

Therefore, 500* 0.7 ± 2.58*0.4583; we get 348.8177 and 351.1823.

User Contributed Comments 8

User Comment
danlan Can anyone help me on this?
orangejuice the number 0.4583 comes from variance = 0.7*0.3*500 = 105, sd = 10.2470 and standardize it by /500^0.5 = 0.4583.
AndyBear I thought the equation was [np(1-p)].
Where is the n in the denominator coming from?
[np(1-p)/n]
panvino AndyBear: Formula for standard error: standard dev divided by square root of n. np(1-p)=variance. So need to take square root of whole thing in this case.
mrpman when are you supposed to use the z table and when are you supposed to remember the z values (ie 1.65 for 95% confidence interval)
GBolt93 You don't need to solve this one really. It should be clear from looking at the answers that both are incorrect since they're not symmetric around the mean of .7
Patdotcom I dont understand this answer. They are asking for proportion and not for people, so the answer should be p+-z*((p*q)/n)^1/2
edrei7 The confidence interval for population proportions should be p +/- (z-value*sqrt(p(1-p)/n)).
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