- 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
To estimate the average SAT scores for entering freshmen at universities, a random sample of 14 SAT scores is collected. If the sample mean produced a score of 1150 and the sample standard deviation is 150 points (σ is unknown) then a 95% confidence interval is ______.
A. 1079.4 < μ < 1220.6
B. 1063.4 < μ < 1236.6
C. 1079.0 < μ < 1221.0
Explanation: For a 95% confidence interval, σ unknown, we find t(0.025, 13), the cutoff for the top 2.5% of the t-distribution, df = 13. Looking in the t-table under column 0.025 and down to row 13, we get 2.16. Working with the formula for E, we get E = 86.6. So, the 90% confidence interval is 1150 - 86.6 < μ < 1150 + 86.6 or 1063.4 < μ < 1236.6.
User Contributed Comments 8
| User | Comment |
|---|---|
| danlan | The t value should be greater than 1.96 so that's the only possible choice |
| miso | where does 2.16 come from it should be 2.106 |
| labsbamb | How to find the "t" if we don`t have the "t" table? |
| dimanyc | Danlan is right. it should be more than z value for 95%. and since the sample is so small (n=14), quite significantly more. |
| Adkins | I have the same issue as Miso. Thought the t-value was 2.106 but then when I check here: http://www.socr.ucla.edu/Applets.dir/T-table.html the t-value is 2.16. The answer is then correct. |
| uberstyle | remember that t-test here is taking sqrt of n in calculating E, not n-1 |
| thekobe | since t is wider than z, B is the only possible outcome |
| dbalakos | find the interval with 1.96, we have a wider interval with the t distribution, so the only option is B |