If you only sample the people with the disease, then there is no variation in disease status within the sample, so you cannot correlate variation in disease status with variation in any other characteristic.

Or have I missed something?

Dear Ronan Michael conroy thank you so much for your answer.

Actually I want to correlate pes planus with knee Osteoarthritis these both are independent conditions. I am taking knee Osteoarthritis patients as my sample of study , I will check the presence of pes planus in them by applying several physical assessment test.

I want to know will my study be correlation analytical cross sectional study or not?

If you are doing a study to see if pes planus is a risk factor for knee OA, then you are in the territory of a case-control study. What you have to show is that pes planus is more common in patients with knee OA than it is in people (probably of a similar age) from the same population who do not have knee OA.

Just studying the patient population will give you the prevalence of pes planus but it won't tell you if this prevalence is unexpectedly high or low. To know that, you need to establish the prevalence of pes planus in a group from the same population who do not have knee OA.

Does that make sense?

Yes sir I am getting your point since we don't have any control group and we cannot justify that the prevalence of pes planus in knee Osteoarthritis is higher than prevalence of pes planus in normal population so this would not be an analytical cross sectional survey instead it will comes in prevalence based descriptive cross sectional study design

If I am correct ?

If you know the prevalence of pes planus in the general population, you could use a 1-sample z-test. Here is an online calculator:

• https://epitools.ausvet.com.au/ztestone

But if your sample size is small, you'll want an exact (binomial) test instead. (What is your sample size?)

Alternatively (or additionally), you could estimate the 95% CI for your prevalence estimate in the knee OA group. The same set of online tools has a nice calculator for that too:

• https://epitools.ausvet.com.au/ciproportion

I like that the default setting gives the Wilson (score) CI for a proportion.

HTH.

Thank you @Ronan Michael conrey and @ Bruce Weaver for your assistance

Regards

If you don't have a comparison group, in this case you have to know the rate of pes planus in the general population. If you multiply this rate by your sample size, you will get the expected frequency of pes planus of your sample. If the observed (real frequency) is significantly higher than the expected value, in this case you can say that there is a significant association. Chi square test can be used.

Thank you Namir G Al Tawil for your answer .

This will develop weak association as compare to if we add control group but in my situation ( already collecting data ) this method seems eaisy to do however technically right.

Can you please provide me any reference or website to study this method in more detail.

I shall be thank full.

Regards

You can use the link of Bruce (above) for the Z test.