A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $62,500 per year. What is the test statistic given a sample of 125 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $2,600? A) 10.75. B) 0.96. C) -10.75. D) -0.96. I get so confused about which formula to use when.

sample has known variance (s.d), sample is greater than 30 participants, --> use z-statistics. 65000-62500=2500 -2500/2600 = z score = -.96 I am not sure about the sign, there is a great table on page 34 of the secret sauce to help with this… msg me if you want me to send it to you…

B? Z=(65-62.5)/2.6=0.96 As n>30, we use Z score

I have the secret sauce. I just get so confused with so much quant stuff to remember. This one we need to use the standard error of the sample statistic. The test statistic = (sample mean – hypothesized mean) / (sample standard deviation / (sample size1/2)) = (X − µ) / (s / n^1/2) = (65,000 – 62,500) / (2,600 / 125^1/2) = (2,500) / (2,600 / 11.18) = 10.75.

C

t-stat = (sample - claim) / standard error

ah crap I mixed it up - sign is off. It is A , isn’t it.

I get C z = (62 500 - 65 000)/(2600/125^0.5)= - 10.75

I get A test=(65,000-62,500)/sqrt(2,600^2/125)=10,75