Sample Size 8 9 4 3 4 3 2 5 3 Given the sample above, what is the mean, standard deviation, and 95% confidence interval.
All answers are rounded to the 3rd decimal. Mean = 4.125 s = 2.167 95% CI = 2.623, 5.627 This is sad. I haven’t reviewed quant in so long, that I am forgetting simple equations like these. Oh well, that’s what the mock exams are for.
You will never have to do this on the actual test. FYI
I had to look back to the material. Did not remember formula for calculating Standard Error. For benefit of those, who dont have material ready with them, I will solve it below. Wow, it was a great revision 
1. Sample Mean = (9+4+3+4+3+2+5+3)/8 = 4.125 2. Standard Deviation = SQRT( ((4.125-9)^2 + (4.125-4)^2 + (4.125-3)^2 + (4.125-4)^2 + (4.125-3)^2 + (4.125-2)^2 + (4.125-5)^2 + (4.125-3)^2) / ( 8 - 1) ) = 2.167 3. To get 95% Confidence Interval: a) First get Standard Error for the Sample Mean = ó/( (n)^1/2 ) where ó is Standard Deviation of the Population. But we dont know SD of Population, so we will use SD of the sample instead. = 2.167 / 8^1/2 = .766 b) Next, since Population variance is unknown and sample size is small (is less than 30) We will use t stats and not z stats. From t table: t value for 2 tailed 95% Confidence and degree of freedom 7 (which is 8-1) is = 2.365 NOTE: In the exam, if this t value is not provided, we will simply use z value of 1.96 c) So our 95% confidence interval is 4.125 - (2.365 * .766) and 4.125 + (2.365 * .766) So our 95% Interval is 2.313 and 5.937 with t stats and it is 2.623 and 5.627 using z stats. Also, t Stats are always more conservative than z stats and will give a wider interval for same Confidence Level.
Agree that you wont see such a question on the exam. However, the last line of rus1bus’ post must be emphasized: “t Stats are always more conservative than z stats and will give a wider interval for same Confidence Level”