# Risk factor question

#### Xaaan

##### New Member
Hi!

I have some problem with calculating relative risk for particular acquired kidney disease. My data are as follows:

4 subjects have disease of both kidneys
7 subjects have disease of left kindey only
5 subjects have disease of right kindey only
184 subjects have no kindey disease.

I try to find out if having disease of one kidney is a risk factor for disease in 2nd kindey. I could calculate right kidney disease as a risk factor to left kindey disease and then left kindey disease as a risk factor to the right kindey disease and then add the values, but is it even a correct approach? Is risk ratio even an appropiate thing to use here? I mostly just want to find out whether there is a relation between disease of left and right kidney or the bilateral fiding is accidental.

#### j58

##### Active Member
Asking if having disease in one kidney affects the probability of having disease in the other kidney implies the null hypothesis that disease in the right kidney (R) and the left kidney (L) are independent. If this null hypothesis is true, then P(R and L) = P(R)*P(L). From your data,

P(R) = (5 + 4) / 200 = .045;
P((L) = (7 + 4) / 200 = .055;
P(R and L) = 4 / 200 = .02; and
P(R)*P(L) = .045 * .055 = .0025.

P(R and L) is quite a bit larger than P(R)*P(L), suggesting that the null hypothesis should be rejected. The classic test for this is the chi-squared test of independence, but the expected cell counts are a bit too small for this test to be accurate, suggesting the need for an "exact" test. Significance tests are usually named after famous statisticians; in this case, think of an "exact" test named after a famous statistician whose last name begins with "F." That should be enough food for thought. If you need more help, convince me that this is not a homework problem.

If you want an estimate of the "relative risk," it can be calculated using the same contingency table required for the above test. I suggest calculating the odds ratio, because Odds(R)/Odds(L)=Odds(L)/Odds(R), whereas the estimated relative risk will depend on whether you use right- or left-kidney disease as the reference category.

Last edited:

#### hlsmith

##### Not a robit
As @j58 mention, kidneys inside the same person are likely not independent. I will add a little more information, relative risks are traditionally reserved for prospective data collection since it is an incidence ratio. Odds ratios are suppose to be used with cross-sectional, retrospective, or data that lacks knowledge of a temporal relationship.

Secondarily, you need confidence intervals on estimates since you could easily mislead readers about statistical significance using such small samples. Lastly, if you are conducting multiple comparisons you should correct your alpha levels to prevent false discovery. So likely use a smaller alpha, which would mean broadening you confidence intervals closer toward 100% instead of say 95% CIs.

#### Xaaan

##### New Member
Thank you very much for your help!

I assure you this is not for homework but actual registry analysis - although I simplified the values a bit to make them easier to discuss. I am familiar with chi-square test limitations and Fisher's exact test. The problem is, I am medical person rather than professional biostatistician and was only taught how to use Statistica/online calcs and for these you usually need 2x2 table with actual values, and here is my struggle:
I clearly have a group that had a risk factor (left/right kidney disease) and got sick (4)
I also have a group that had a risk factor (left/right kidney disease) and didn't get sick (7+5=12)
And what next? By definition there won't be any subjects who had bilateral kindey disease and no left/right kidney disease.

That contingency table makes perfect sense but I still have no idea how to get p value for significance test or calculate odds ratio with 95% CI. I can clearly see that the null hypothesis is most probably false with these probabilities and sample size but I need to prove it somehow.

#### j58

##### Active Member
Thank you very much for your help!

I assure you this is not for homework but actual registry analysis - although I simplified the values a bit to make them easier to discuss. I am familiar with chi-square test limitations and Fisher's exact test. The problem is, I am medical person rather than professional biostatistician and was only taught how to use Statistica/online calcs and for these you usually need 2x2 table with actual values, and here is my struggle:
I clearly have a group that had a risk factor (left/right kidney disease) and got sick (4)
I also have a group that had a risk factor (left/right kidney disease) and didn't get sick (7+5=12)
And what next? By definition there won't be any subjects who had bilateral kindey disease and no left/right kidney disease.
As I said in the original post, if disease in one kidney not is a risk factor for bilateral kidney disease, then your null hypothesis is that disease in the left kidney is independent of disease in the right kidney. That implies a 2x2 contingency table where (say) the rows are presence (+) or absence (–) of disease in the right kidney and columns are presence (+) or absence (–) of disease in the left kindey. That results in 4 cells: +/–, –/–, +/+, –/+, –/– (where, eg, +/– means disease in the left kidney only).

That contingency table makes perfect sense but I still have no idea how to get p value for significance test or calculate odds ratio with 95% CI. I can clearly see that the null hypothesis is most probably false with these probabilities and sample size but I need to prove it somehow.
If you intend to publish your restults, you should have a biostatistican as a co-investigator.

Last edited:

#### Xaaan

##### New Member
Thank you very much! Now that I know the answer it's much easier to understand. I had the wrong approach and couldn't stop thinking in terms of 3 groups (bilateral, unilateral, none) instead of 4.

Secondarily, you need confidence intervals on estimates since you could easily mislead readers about statistical significance using such small samples. Lastly, if you are conducting multiple comparisons you should correct your alpha levels to prevent false discovery. So likely use a smaller alpha, which would mean broadening you confidence intervals closer toward 100% instead of say 95% CIs.
Thank you for this advice! I am comparing multiple factors bilaterally and fortunately most of them have higher sample sizes. Broadening confidence interval seems like a good idea!

If you intend to publish your restults, you should have a biostatistican as a co-investigator.
I agree that including biostatisticans as co-investigators should be more common. In this registry most calculations were really simple so I was trying handling them myself, this was the only one that gave me some trouble so far.