# Quant Q

Can anyone explain this? I always thought that z test stat was calculated as : (X - mean)/(S.D) Why is the root(n) or sample size being included >??? A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than \$58,500 per year. What is the test statistic given a sample of 175 newly acquired CFA charterholders with a mean starting salary of \$67,000 and a standard deviation of \$5,200? A)1.63. B)-21.62. C)-1.63. D)21.62. With a large sample size (175) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (67,000 – 58,500) / (5,200 / 1751/2) = (8,500) / (5,200 / 13.22) = 21.62.

because you are doing distribution of sample means not the distribution of population

Yes, thunder is right: The variance of sample mean is VAR/n --> standard dev of sample mean is root(var/n)=s.d./n^0.5 So you standardize, but with the standard deviation of the sample mean and not of the population s.d.