### CFA Practice Question

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### 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 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