0
$\begingroup$

I am need some help. I have two outcomes and 4 different variables and I want to see which outcome is much more likely with a particular variable. Please see this table. I am unsure what statistical test to use on SPSS

                  No alterations 1 alteration 2 alterations  >3 alterations

tumour Present        14           18           16             12

Not present           66           2            0               0
$\endgroup$
  • $\begingroup$ What is your research question? $\endgroup$ – user2974951 Aug 6 '19 at 12:44
  • $\begingroup$ I would like to know whether if you have tumour present are you more likely to have more alterations $\endgroup$ – Anu Jayaram Aug 6 '19 at 13:50
  • $\begingroup$ Please clarify this minor point. You said you have two outcomes. I see one outcome in the table above: tumor status. That is one binary outcome, meaning that 0 is no tumor and 1 is tumor present. Do you actually have a second, different outcome? Most people would not say that tumor no tumor are two outcomes. $\endgroup$ – Weiwen Ng Aug 6 '19 at 14:50
  • $\begingroup$ Yes I mean Tumour present versus tumour not present $\endgroup$ – Anu Jayaram Aug 6 '19 at 14:54
0
$\begingroup$

You said in comments that you'd like to know: if you have a tumor present, do you have more alterations? So, number of alterations is actually the outcome, or $y$ variable, and tumor is the independent variable, or $x$ variable. You may have meant the reverse.

Regardless of which you are treating as the outcome, for basic descriptive statistics, many of us would use the chi-squared test or Fisher's Exact test (difference explained later). This would show if tumor status has any association with number of alterations. You could take the p-value and compare the number of cases in each cell. If you ran a chi-squared test on your data above and you got a $p$-value < 0.05, you could say that there is an association, and from inspection of the table, it appears that a higher number of alterations is associated with tumor status. Note that the test itself does not tell you the direction of association. I'm pretty sure these data would also return a chi-square p-value < 0.05:

                      No alterations 1 alteration 2 alterations  >3 alterations

tumour Present        48               0           0               12

Not present           48               6           6               0

I'd say that was a bit of an odd association, and I'm not sure how to describe it.

In your case, you have some cells with low numbers, so you would use Fisher's Exact test, not the chi-squared. You can read the Wikipedia link for the rationale, and for a general decision rule about when to prefer Fisher's over chi-squared.

In a regression context, you could use logistic regression. Regression enables you to account for the effect of confounding variables that you've measured. Furthermore, if you meant that category of alterations was the outcome, (ordered) logistic regression lets you account for the directionality (i.e. a higher category means more alterations). If you actually meant that the category of alterations was your outcome, you would be using ordered logistic regression. It would tell you if tumor status was associated with a higher category of alterations.

If you meant that tumor status was the outcome, then you would use logistic regression. Here, alteration category would be your dependent variable, and you would use it as a categorical covariate. Likely you'd select no alterations as the base category. One problem here is that there's no way to indicate that a categorical covariate has any sort of direction.

In both these cases, you have some zero cells. This is a major problem for logistic regression. The reasons are more technical, and you should read up on logistic regression in general to understand them.

If you prefer more basic analysis but you need to account for a categorical confounder, you could run chi-square or Fishers tests by stratum.

$\endgroup$
  • $\begingroup$ Thank you so much for your clear answer. You have made it very clear. Thank you $\endgroup$ – Anu Jayaram Aug 6 '19 at 16:23

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.