CFA Practice Question

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CFA Practice Question

Calculate a 90% confidence interval for a population mean. You have a sample of 21, a sample mean of -25%, and a sample standard deviation of 10%. The sample appears to be approximately normally distributed.
A. [-28.9%, -21.1%]
B. [-27.9%, -22.1%]
C. [-28.1%, -21.9%]
Explanation: Based on the data given, we should use the t-distribution. The critical value will be based at t_(0.05, 20) and is 1.725. Our confidence interval will then be [-25% - 1.725*(10%)/(210.5), -25% + 1.725*(10%)/(210.5)] = [-28.9%, -21.1%].

User Contributed Comments 15

User Comment
danlan Since we use sample standard deviation, we suppose the population variance is unknown, so we use t-distribution.
samsonov should they have 20 or 21 in devisor?
Criticull can we get critical values with the BAII Plus? If not I assume they provide tables.
markhuang 21. 20 is the degree of freedom when getting the critical value, but 21 is used to calculate standard error.
mpapwa22 Thanks Markhung, this is very clear now, was getting really confused with the 20 or 21.
sargutin I am using the equation in the explanation and get a [-28.76 ,-21.236] interval. Am I doing something wrong, or the multiple choice is just really not precise?
jpducros I got the same as you Sargutin. I seems they used 20 instead of 21 in their formula to reach their result.
maria15 Can someone please explain how to get 1.725 from the t-distribution? I completely forgot. Thanks!
maria15 Never mind. I checked my book and got it. Thanks!
Shaan23 N - 1 DF guys.
chris297 So the fact that it is approximately normally distributed would not lead to the usage of normal distribution table?
SalimBouch no because you should estimate the standard deviation of the population using the standard deviation of the sample
thanhb91 I heard they wont give you the table, so we should be remember some critical number!!
dbalakos Such questions with critical values of t-distribution wont be in the exam since we wont be provided with tables. Just memorize important levels of significance from z-table and you will be fine.
Patdotcom Are we sure that t is 1,725? I used it first and the figures were as sargutin said(-28,8;21,2). I tried then with t(0,1,20) t=1,325 and then option b was exactly...
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