- CFA Exams
- CFA Level I Exam
- Study Session 3. Quantitative Methods (2)
- Reading 10. Sampling and Estimation
- Subject 6. Confidence Intervals for the Population Mean

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**CFA Practice Question**

To estimate the average length of their employees' telephone calls, FoneJack, Inc. randomly sampled 15 employee phone calls. If the sample mean was 1.3 minutes and the sample standard deviation was 0.3 minutes (s is unknown) then a 90% confidence interval for the phone calls is ______.

A. 1.16 < m < 1.44

B. 1.2 < m < 1.4

C. 1.0 < m < 1.6

**Explanation:**For a 90% confidence interval, s unknown, we find t(0.05, 14), the cutoff for the top 5% of the t-distribution, df = 14. Looking in the t-table under column 0.05 and down to row 14, we get 1.761. Working with the formula for E, we get E = 0.14. So, the 90% confidence interval is 1.3 - 0.14 < m < 1.3 + 0.14 or 1.16 < m < 1.44.

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**User Contributed Comments**
5

User |
Comment |
---|---|

homersimpson |
n=15, t-table! |

praj24 |
hold on, so we don't get a t-table in the exam. How do you guys estimate it? I use the Z-test and use the close/logical answer? What's your process? also, am I on the right train of thought? Cheers |

Jpsmith835 |
agree with praj24, is this likely to come up since we have no t-table? |

AlVMc |
n = 15, but s is unknown, therefore df is n - 1 which implies that df = 14. Also, an excerpt of the table is given in the exam. |

dbalakos |
use the z critical value of 1.645 to find the interval and account for a small adjustment to "estimate" the interval of the t-distribution. Here the difference between 1.4 and 1.6 is too large so the most feasible answer would be A |