1
$\begingroup$

if I have a table like this

enter image description here

In which,

site = different sites of measurement (tissue or organ)

type = different types of virus

And I want to ask something about the data

1.What test can I use to test if the distribution of type is similar within site 1, site 2, site 3, etc. It looks similar, but I don't know to proof it statistically. To compare between two sites, I think we can use chi square goodness of fit test, but what if I want to do it simultaneously across all sites ?

2.Is there any ways to tell us that the distribution of type is dominated by only one type ? For example in this data set, we can observe that type 2 is the most dominant type across all sites. How can we do it statistically ?

Thanks

$\endgroup$
2
$\begingroup$

In answering this, I am assuming that the contents of the data table are counts of how many times each combination occurred. If they are measurements, or percents, then my answer won't be right.

For your first question, a chi-square test of independence (not goodness of fit) is the standard technique. It tests whether the types are independent of sites. However, the data for types 1, 5, and 3 are too sparse to use that test (and note you have absolutely no information about type 1). So I would suggest applying the chi-square test only to the 5 x 3 subtable obtained by ignoring columns 1, 3, and 5. You can also compare one site with another using a chi-square test for independence of the corresponding 2 x 3 table (again omitting columns 1, 3, and 5.

For the second question, it is a chi-square goodness-of-fit test for the uniform distribution, based on the 6 column totals. In this test, the observed totals are tested against the expected totals under uniformity - namely, each expected total is the grand total divided by 6.

$\endgroup$
  • $\begingroup$ Actually, I got more than 6 types and more than 5 sites. The problem is I can't remove or pool the small values. So, I guess is that appropriate to use Fisher test with monte carlo as an alternative to chi square test of independece ? $\endgroup$ – csb_cvr Aug 21 '14 at 3:11
  • $\begingroup$ I don't know. I'll defer to other suggestions. $\endgroup$ – Russ Lenth Aug 21 '14 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.