Hello I have 100 students who answered the question, time spent sitting and I divided it into three groups:
Group 1: (6-9) hours
Group 2: (10-13) hours
Group 3: (14-17) hours
Also the students answered the Boston questionnaire for symptom severity scale
Which is 11 questions, and each question has scores from 1 to 5.
Below shows Boston score divided into groups of time spent sitting.
(6-9 hours) group 1
11
14
12
23
16
and there is 78 more numbers in this group.
(10-13 hours) group 2
There is 15 numbers.
(14-17 hours) group 3
There is two numbers
17
17
So I want to compare the mean of these three groups, because I have unequal sample size which is very different between my groups I used welch ANOVA test. But balecause in group 3 the variance is 0 the welch test cannot be performed, how should I solve this problem?
I hope I explained it clearly, sorry for my English.
Thank you
Link to picture
https://drive.google.com/file/d/1wDjgXA1SnqGHhJq7PZrWpuzTWBBtDXIH/view?usp=drivesdk
Best Answer
This is probably the main source of your problem:
Breaking a continuous predictor into groups is not a good idea. If you keep the hours sitting as a continuous predictor you could use standard regression techniques to examine the relationship between your symptom severity score and hours sitting. You are otherwise making assumptions, for example, that there's a big difference between sitting for 9 hours versus 10 hours but no difference between 10 and 11 hours.
Added in response to comments:
If you have additional predictors to evaluate, perform multiple regression rather than separate regressions. Otherwise you run a risk of omitted variable bias that could affect all of your estimates of associations with outcome.
If for some reason you are forced to use some form of ANOVA for these data, despite its far-from-ideal applicability, you have a few choices:
You could just do a standard ANOVA, without the Welch modification, assuming equal underlying variances around group means. That's what Christian Hennig suggested in a comment on the question. You could then use standard techniques like normal qq plots to check that assumption with the residuals.
You could use the non-parametric Kruskal-Wallis test, which addresses similar analysis as ANOVA but makes no assumptions about error distributions. With only 2 cases in one of your groups, however, I don't suspect that will be helpful.
Instead of discarding the 2 cases in group 3, you could add them into group 2. The you could do a t-test between the groups (6-9 hours; 10+ hours). If you are concerned about equal variance between the 2 groups you can used the Welch modification.
I'm very reluctant to recommend discarding data unless you have strong evidence that there was something wrong in the way it was collected.