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Let's say I have growth parameters for a population. People who exceed 95th percentile are marked with 1, and people who are otherwise below it are marked as zero. I will then use crosstabs to compare the number of >95th across different regions. My question is, I did not take an equal amount of data from each region. Region 1 could have more >95th because I took more cases from it. Does Chi-Square account for difference in number of cases, I mean does it use the percentage of "1" among ones and zeros in a region or does it just use the 1 and not account for its percentage in that specific region, when outputting the 2-tailed sig.

I also did analysis with BMI, height, weight Z scores that are >1.96 and <-1.96 which is also equivalent to 95th and 5th, and then compared mean Z values of different regions using anova for >2 regions, and 2 sample independent t-tests for dichotomous variables like gender.

Which of the methods is the best way to go in your opinion? I feel that converting my continuous percentile height, bmi and weight data to nominal variables(count the >95th) for the sake of chisquare analysis among regions or gender would yield inferior quality results.

which one of these:

1) Prevalence of tall stature in region 1 vs 2 was 1% vs 2% respectively with a mean z-score/percentile of xx vs xx (P_value from t-test of 2 means of z/percentile scores with NO CHISQUARE p-values)

2) Among those who are >95th, mean z/percentile score was 96% in group 1 vs 98% in group 2(p_value from t-test of 2 means of z-score NO overall prevalence so NO CHISQUARE TOO)

3) 1% tall stature in region 1 vs 2% in region 2(P-value from crosstabs chisquare)

Which do you think is the best method to put it in terms of significance and study strength. If you think of other ways, PLEASE do tell me.

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It appears you want to (1) compare relative heights of people from different regions and (2) also to determine overall upper limit of "normal height" or cut-off for "tall stature". For first aim, you should perform t-test or ANOVA to determine if the heights are different in different regions. For overall average, you have to decide if you want all regions to be represented equally or in proportion to their population. Then either combine means from different regions or combine means*population from different regions. Comparison of proportions with tall stature will also depend on which method you choose for determining overall definition of 'tall stature'. On the whole it will be simpler and also make communication easier if you talk in terms of absolute height.

For comparison of proportion of persons above a certain cutoff you can perform test of proportions using all numbers. For example, if two regions have population of 1000 and 3000 respectively, and number of tall persons in these 2 regions is 50 and 85, one can test as follows in R:

> prop.test(c(50,85), c(1000, 3000))

        2-sample test for equality of proportions with continuity correction

data:  c(50, 85) out of c(1000, 3000)
X-squared = 10.142, df = 1, p-value = 0.001449
alternative hypothesis: two.sided
95 percent confidence interval:
 0.006244601 0.037088732
sample estimates:
    prop 1     prop 2 
0.05000000 0.02833333 

First region has significantly more proportion of tall persons.

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    $\begingroup$ The method I used for determining cutoffs is the CDC SAS app cdc.gov/nccdphp/dnpao/growthcharts/resources/sas.htm. The formula outputs Z scores for the values I put in so I could compare my sample to CDC growth charts. They define tall stature as >95th or Z>1.96 and the other way around for short stature. I have performed an ANOVA on all Z values to determine regional differences. As you said, I am now attempting to perform analysis on cut-off Z-values across regions. I have to do counts on cutoffs to determine numbers of cases in region. What about comparing the Zs beyond cutoffs? $\endgroup$ – Ned321 Jul 11 '15 at 9:24
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    $\begingroup$ "tall stature as >95th or Z>1.96" is well accepted definition. The issues is which values you put in initially. The regions will be represented in the ratio of the numbers of samples from that region. Hence, number of samples from each region should be in proportion to population of those regions. $\endgroup$ – rnso Jul 11 '15 at 9:54

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