How to check whether an ordinal variable is dependent on a nominal variable (dichotomous)?

[Nico Dekker]

Hello,

I am quite new to Stata and statistics in general, so I am not the smartest out there.
I have a data set with two variables, one being gender (male or female) and one being a 'stress on the workfloor' factor, which was measured on a Likert scale (1 to 7, total 4 questions) and is now transformed into a factor (factor analysis). The gender variable is in my data set where 0 means a male and 1 means a female. The other variable (stress on the workfloor) has many numbers behind the comma, but it ranges from -1.792 to 1.854.

My question:
Now, I want to measure whether stress on the workfloor depends on gender. I also need to formulate the null hypothesis, why I select the appropriate test and if the test meets the assumptions. I also want to interpret these results.

I honestly have no clue which test to pick and why... if any one can help me out with it, it would be really appreciated!

thanks in advance!

[Weiwen Ng]

We don't necessarily shy away from basic questions on the forum, but does this one relate to a homework assignment? Our FAQ does state not to ask homework questions, as you would be better off asking your instructor if so.

One point is that your title refers to ordinal variables. However, your question asked about "stress on the workfloor" - which you had described as a factor score derived from four ordinal variables. Thus, that variable is actually continuous.

[Nico Dekker]

We don't necessarily shy away from basic questions on the forum, but does this one relate to a homework assignment? Our FAQ does state not to ask homework questions, as you would be better off asking your instructor if so.

One point is that your title refers to ordinal variables. However, your question asked about "stress on the workfloor" - which you had described as a factor score derived from four ordinal variables. Thus, that variable is actually continuous.

Thanks for your response! Excuse me, I do not completely understand the code delimiters thing. I did find it, it creates than, excuse me!
Indeed true, it were four ordinal factors, which were transformed into a continuous variable, namely the factor score. I will rewrite the title (if I can) to that.

It is not part of a homework assignment per se. I am writing a thesis and asked a questionaire for that thesis. Now I want to interprete these results better. My lecturerer knows little to nothing about this topic, hence I am trying Google to find out more about it. Personally I thought I would use the Kruskall-Wallis test to find more about it, with the H0 that both variables are independent.

Since the further writing of my paper depends on these findings, I really want to get this right.

[Weiwen Ng]

Thanks for your response! Excuse me, I do not completely understand the code delimiters thing. I did find it, it creates than, excuse me!
Indeed true, it were four ordinal factors, which were transformed into a continuous variable, namely the factor score. I will rewrite the title (if I can) to that.

It is not part of a homework assignment per se. I am writing a thesis and asked a questionaire for that thesis. Now I want to interprete these results better. My lecturerer knows little to nothing about this topic, hence I am trying Google to find out more about it. Personally I thought I would use the Kruskall-Wallis test to find more about it, with the H0 that both variables are independent.

Since the further writing of my paper depends on these findings, I really want to get this right.

In that case, there's a class of basic tests to determine if two (or more) groups have different average levels of some variable. You could treat the dependent variable as each question individually, or you could use the factor score. One-way ANOVA and Kruskal-Wallis are in this family of tests. You have only two groups, so the null for both tests is that both groups have identical means (ANOVA) or medians (K-W). The alternative is merely that at least one group has a different mean or median, I believe. If you had normally distributed data, you would be better off using ANOVA. If your data are skewed left or right, you'd be better off with K-W. In my experience, a number of subjective constructs like stress may be skewed, e.g. most people report zero, some report low levels, and progressively fewer report moderate and high levels of stress. Basically, you're on the right track for a simple analysis.