Chi square 2x3 table - Genotypes are related to disease?

Question

I want to conduct a Chi square test of independence.

I have three different genotypes (CC/CG/GG, independent variable) and "health conditions" (outcomes) which is basically a patient group with a certain disease and a control group without the disease.

Does it sound right if my null hypothesis states that "the distribution of these alleles in these "study groups" doesn't influence disease outcome"

Whereas the alternative hypothesis says that "the distribution of these alleles in these study groups does influence disease outcome".

Is i formulated in a right way? Because I can't state (with the Chi square) that there are a certain risk-allele that lies behind disease.

And lastly, I have heard that you can only use odds ratio if it is a 2x2 table, but if I use a "dominant" model of Pearsons Chi-square then the result would no longer be significant and there should no longer be any point to perform a OR.

Thankful for comments!

There is no such thing as independent variable in this test. You're not doing regression testing. You're testing whether there is any dependence between the levels of the variable. — SmallChess, May 26 '15 at 11:12
More precisely, the null hypothesis is that the levels of the variables are independent. The alternative is that they are dependent. — SmallChess, May 26 '15 at 11:13
Do you have absolute numbers or percentages? Are the number of patients and controls equal and what is the total sample size? — rnso, May 26 '15 at 11:14
To Student T: So basically it mean if I got a significant result, my alternative hypothesis would say something like "there is a significant association between the occurrence of these genotypes and occurrence of this certain disease"? — Zorua, May 26 '15 at 11:20
Of possible interest: http://stats.stackexchange.com/q/8774/930, http://stats.stackexchange.com/q/9062/930. — chl, May 26 '15 at 11:21
@Zorua. "There is dependence associated with your significance level between the two variables". — SmallChess, May 26 '15 at 11:23
@mso I have absolute numbers, so when I thought about which test a should use I assumed that my variables was qualitative (genotype and "health condition/study group". There are 36 patients and 61 controls, a total of 97. — Zorua, May 26 '15 at 11:34
@Student_T Hope I am doing it right now. So in my conclusion I can only state that there are a dependence/relationship between the two variables? Do I need to specify? — Zorua, May 26 '15 at 11:36
Test of independence only gives you the conclusion that there is dependence of the levels if you manage to reject the hypothesis. Wikipedia tells you more. — SmallChess, May 26 '15 at 11:37
@Student_T I may be lost know but I will try. My H0 states that allel distribution (CC/CG/GG) are independent of health condition (other word for study/control group?). H1 states these variables are dependent on each other. My result was (χ2 = 7,278>5,99, df = 2, p<0,05) and therefore I can reject my H0. My conclusion is that there is a dependence between alleles and health condition. — Zorua, May 26 '15 at 11:55

rnso · Answer 1 · 2015-05-26T13:05:07.697

Regarding calculating odds ratios:

> tt
        dss
genotype sick healthy
      CC   14      34
      CG   14      24
      GG    8       3
> 
> oddsratio(tt, rev='col')
$data
        dss
genotype healthy sick Total
   CC         34   14    48
   CG         24   14    38
   GG          3    8    11
   Total      61   36    97

$measure
        odds ratio with 95% C.I.
genotype estimate     lower    upper
      CC 1.000000        NA       NA
      CG 1.409998 0.5628447  3.55030
      GG 6.087096 1.4813184 32.87432

$p.value
        two-sided
genotype midp.exact fisher.exact  chi.square
      CC         NA           NA          NA
      CG 0.46208267   0.49321142 0.450640195
      GG 0.01142108   0.01320433 0.007043333

$correction
[1] FALSE

attr(,"method")
[1] "median-unbiased estimate & mid-p exact CI"
Warning message:
In chisq.test(xx, correct = correction) :
  Chi-squared approximation may be incorrect

The odds ratios are calculated for second and third genotypes with respect to first genotype.

Edit: I have edited the code above to have numbers as in comments. The odds ratio for GG (with respect to CC) being sick is 6.1 (95% CI 1.5,32.9; significant since it does not overlap 1). The odds of CG being sick (as compared with CC) is not significantly different 1. The P values are also shown in the output.

@mso Are you using R or any other kind of program? Because I have been using GraphPad and have not been able to calculate odds ratio (probably because I did not know about a reference genotype). But this is very interesting! Does it mean that people who have GG genotype has 2,76 higher risk to become sick? I have a huge problem to formulate correct null- and alternative hypothesis... — Zorua, May 26 '15 at 11:42
Yes, I am using R. The odds ratios above are for the fictitious numbers that I have used. What are your actual numbers? — rnso, May 26 '15 at 12:25
I am sorry, I have been spelling your user namn wrong this whole time. My numbers are for "sick": 14, 14, 8. And for the healthy: 34, 24, 3 (CC, CG, GG) @rnso — Zorua, May 26 '15 at 12:33

Chi square 2x3 table - Genotypes are related to disease?

1 Answers1