- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 5. Sampling and Estimation
- Subject 5. Confidence Intervals for the Population Mean and Selection of Sample Size
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 Comments 8
| 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. |