# z or t-value for confidence interval

The average salary for a sample of 61 CFA charterholders with 10 years experience is \$200,000, and the sample standard deviation is \$80,000. Assume the population is normally distributed. Which of the following is a 99 percent confidence interval for the population mean salary of CFA charterholders with 10 years of experience? A) \$160,000 to \$240,000. B) \$197,811 to \$202,189. C) \$172,514 to \$227,486. D) \$172,754 to \$227,246. This is a Schweser question. I went ahead and based it on a z-stat, but the answer, which is D, is based on the t-dist. Both are acceptable here, however, on the real test, will they be a little clearer about what to use? I don’t have time to waste on guessing what they want me to use?

BTW, this problem is coming up frequently in Schweser’s Q-bank, where they automatically use the t-dist when the pop. variance is not known, but the population is normally distributed and the sample size is >30.

how do you get D?

It’s completely clear that you use the t-stat instead of the z-stat and then you have an exact C.I… if "the pop. variance is not known, but the population is normally distributed "

You also use the t-stat if “pop. variance not known, distribution not known, n>30” correct, giving the same CI? or is it “distribution not known but assumed normal”?

Dimes27 Wrote: ------------------------------------------------------- > You also use the t-stat if “pop. variance not > known, distribution not known, n>30” correct, > giving the same CI? or is it “distribution not > known but assumed normal”? Assuming normal is the same thing as knowing normal since nobody ever really knows the dist’n is normal (well…). If you’re using the CLT, you should just use a z IMHO because your C.I. is approximate anyway. I’ve seen it done both ways though and it makes little practical difference.