Statistical Test on Anthropometric Data that varies with age, Z-test? I have what seems like it should be simple statistics problem. Given anthropometric data (height, weight, BMI, age, etc.), determine if a sample population has significantly different anthropomorphic characteristics than the larger population. The difficulty however lies in that this test must be done for a pediatric population, thus the anthropometric data is a function of age. 
For example, consider the following null-hypothesis: 
"The sample's height-per-age (growth-curve) is not significantly different than the World Health Org (WHO) database."
The WHO produced growth-curves (height as a function of age) for children, and offers a simple calculation to determine the z-score of an individual observation. But I want to make a claim about the sample as a whole, rather than individual measurements. This led me to believe I could simply take the mean of the z-scores from individual measurements, and look up the corresponding p-value on a z-table to test for significance, i.e. if Zmean < -1.96, then the sample is significantly different than the population. However, this calculation does not consider the sample size (n). I can't imagine making a statistical claim about the sample as a whole without considering the sample size. How can this null-hypothesis be evaluated?
Also, the WHO software produces a the below graph from the z-scores, which leads me to wonder if I could simply perform a t-test from the z-scores to determine if the sample is statistically different than the WHO data population.
 A: You should simply calculate differences between the height of people in a sample and average height of a person  given age (WHO growth curve). So if a height of an 2 months old person $i$ in a sample is 55 cm and average height of a 2 months old person based on WHO growth curve is 58 cm then $diff_i=-3$. Then you should simply run a T-test where:
$T_0: \quad diff=0$
$T_1: \quad diff \neq0$
But you can also just calculate t-test based on Z scores. Z score is simply a difference between an observation $x_i$ (height of person $i$ in a sample) and population mean $\bar{x}$ (given the age) normalized for standard deviation $s$:
$$Z=\frac{x_i-\bar{x}}{s}$$
But the meaning is slightly different. Z-value measures deviation from the mean expressed in SD. Consequently $T_0$ assumes deviations of the sample heights are equal to 0 and $T_1$ proposes that deviations are not 0. You will also notice that if SD is constant and independent of age (this is not true in this particular case), both T-tests give identical results. 
