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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.enter image description here

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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.

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  • $\begingroup$ thanks. Unfortunately I cannot find the actual function determining the growth curves, so am not able to evaluate the curve numerically. But along the same line of what you are suggesting, would it be correct to run a T-test with the individual z-scores? $\endgroup$ – user157637 Apr 17 '17 at 18:54
  • $\begingroup$ You can use charts like this who.int/childgrowth/standards/… $\endgroup$ – Vivaldi Apr 17 '17 at 19:05
  • $\begingroup$ You don't need the actual function. You just need to match based on age the values from a growth chart on 50 percentile with those in the sample and then just take a difference. $\endgroup$ – Vivaldi Apr 17 '17 at 19:08
  • $\begingroup$ Thanks! I see how this could work. Unfortunately it is not feasible for large samples because of the manual table look-up. Perhaps I could build a function to be evaluated from the 50th percentiles for every age. Alternatively, Is there any apparent problem that you see with doing a t-test on the individual z-scores? $\endgroup$ – user157637 Apr 17 '17 at 19:12
  • $\begingroup$ Okay, I edited the answer based on your input. Kind regards! $\endgroup$ – Vivaldi Apr 17 '17 at 19:30

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