### CFA Practice Question

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### CFA Practice Question

The mean of a sample with a known population variance of 25 was estimated as 12.5. The sample size was 22 and the sample variance was 28. Which of the following represents the appropriate 95% confidence interval for the normally distributed population?
A. 10.41; 14.59
B. 10.29; 14.71
C. 10.74; 14.26
Explanation: Since the population variance is known and it is normally distributed, we will ignore the sample variance. A 95% confidence interval corresponds to a z-value of 1.96. Thus, CI is given by 12.5 ± 1.96 x (25/22)1/2 or [10.41, 14.59].

### User Contributed Comments8

User Comment
eddeb n is less than 30, so i thought it was appropriate to use t table.

Is is beacause pop variance is given?
tabulator normally distributed population, and population standard deviation is known: use z-statistics irrespective of sample size.
jerrick to use the z value you should have,
- the population variance "OR"
- sample size greater than 30
far1080 Alright you bright sparks: why is analyst note's standard error formula ^0.5 rather than square root of sample size (20)?
fanDango x^0.5 = sqrt(x)
MoviusCFA I've been looking at this too long. Why are they multiplying by ((25/22)^0.5), shouldn't the formula be (25/(22^0.5))?
sharkie @MoviusCFA: because 25 is the variance and you need to get standard deviation!
GBolt93 they're simplifying. you could also do 5/(22^.5), which is the same.
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