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Give the following table

                  Male    |   Female   |   Total
Likes              10     |     3      |     13
Indifferent        20     |     20     |     40
Dislikes           10     |     9      |     19
Total              40           32           72

Here the independent variable is gender which is nominal. The dependent variable which is ordinal is either likers, indifferent, or dislikes. I can use the Chi Square test or Exact Fisher to determine if a H0 such as men and women are equally likely to be likers should be rejected.

However, the question I would like to answer after this is: which gender is significantly more likely than the other to be a liker? And, by how much?

For example: Men are x% significantly more likely than Women to be likers.

I thought that the next step would be to perform a test between liker and each group: i.e. males and likers. But I got stumped because I thought I should be comparing males to females with respect to likers.

Is there any standard way of approaching this question?

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  • $\begingroup$ "Men are 30% significantly more likely" -- I don't know what this is intended to mean. $\endgroup$ – Glen_b Feb 7 '14 at 11:33
  • $\begingroup$ One of the very first steps in approaching any statistical question is to verify that the data are correct. Because I am unable to match your totals to your counts (they are way off in one case) this point, although trivial, seems relevant. $\endgroup$ – whuber Feb 9 '14 at 23:09
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I thought that the next step would be to perform a test between liker and each group: i.e. males and likers. But I got stumped because I thought I should be comparing males to females with respect to likers.

Chi-Square and generalized Fisher test do not address the ordinality of the second categorical variable. To test if either gender is more prone to higher degree od liking (association between dichotomous and ordered categorical) I would suggest applying contingecny table trend test by Cochran and Armitage.

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  • $\begingroup$ This is by far the most useful answer. Thanks for your help. Again most examples of these methods are done in SAS or SPSS, what I need is an algorithm to implement myself. $\endgroup$ – dominic Feb 10 '14 at 8:25
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If you just want to say what proportion of men vs. women are likers (or something similar) you can do it straight from the table: 10/40 men are likers, 3/31 women are.

You could then run a test of equality of proportions on these two proportions.

If, on the other hand, you want to answer all the questions you asked in your text, and treat one variable as independent, then you want ordinal logistic regression. This would also allow you to control for any other variables.

However, your title question isn't well posed: With a 2-group independent variable, the influences are, by necessity, equal (and opposite).

EDIT:

Here is how you would do it in SAS

data liking;
 input sex $ liking $ count;
 datalines;
 M 3 10
 F 3  3
 M 2 20
 F 2 20
 M 1 10
 F 1 9
;
run;

proc logistic data = liking;
 class sex;
 model liking = sex;
 freq count;
run;

and sex is not significant at p = 0.05. Its odds ratio is 1.69 with a 95% CI of 0.682 to 4.198.

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  • $\begingroup$ Hi Peter, when you say "and treat one variable as independent" do you mean treat 'male' as an independent variable? With that in mind, should the title become "which independent variable is more likely to influence the dependent variable?" ? $\endgroup$ – dominic Feb 7 '14 at 13:19
  • $\begingroup$ No. You have two variables: "Degree of liking" and "gender". If you want to treat gender as an independent variable, then you need some form of regression, but you would then have ONE independent variable. It makes no sense to say maleness influenced the DV more than femaleness. One is the negative of the other $\endgroup$ – Peter Flom Feb 7 '14 at 22:04
  • $\begingroup$ Sorry Peter I don't understand how this allows me to test if either gender is more prone to higher degree of liking... ? I have not been able to find a working example of ordinal logistic regression - all the logic is done in SPSS or SAS with little explanation. Could you give me a more detailed answer? $\endgroup$ – dominic Feb 9 '14 at 20:08

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