Hi there,
I am currently investigating whether males nurses are more likely than female nurses to be brought before a fitness to practice hearing. I have tried to do a one sample binomial test in SPSS, however the expected results show a 50/50 split in terms of gender. The issue being that the actual proportion of male nurses is 11%. That being the case is it possible to discover whether the proportion in the sample is significant using a binomial approach or is another test needed?
Regards
TK
Why not use a z test to compare the two proportions? This approach takes sample size of each group into account.
https://online.stat.psu.edu/stat415/lesson/9/9.4
Thanks Blaine, I knew there would be a test that could do this!
Have a great weekend!
TK
Tony Kelly, IIRC, SPSS has no command to carry out the two-sample z-test for proportions. However, Pearson's Chi-square on a 2x2 table is equivalent to that z-test: z2 = Chi-square.
Also, if you run into problems with low expected frequencies, bear in mind that the test of linear-by-linear association in the CROSSTABS output is equivalent to Egon Pearson's N-1 Chi-square when you have a 2x2 table. See Ian Campbell's website for more info about the N-1 Chi-square. HTH.
- https://www.iancampbell.co.uk/twobytwo/n-1_theory.htm
Oh, my. Isn't there a binomial test that tests against a proportion other than 50/50 ? It's so simple in R, I'm feeling embarrassed for other software packages. :D
Salvatore S. Mangiafico, if you are responding to my post, and expressing shock and embarrassment about SPSS's lack of certain tests, note that the z-test comparing two independent proportions can be obtained as an equivalent Chi-square test; and one-sample z- and binomial tests (for a proportion) can be obtained via the NPAR TESTS command. Here is an example of the latter, where the observed proportion = 0.5 and the expected proportion = 0.4.
NEW FILE.
DATASET CLOSE ALL.
DATA LIST LIST / category (F1) N (F5.0).
BEGIN DATA
1 50
2 50
END DATA.
WEIGHT by N.
NPAR TESTS
/CHISQUARE=category
/EXPECTED=40 60
/MISSING ANALYSIS
/METHOD=EXACT TIMER(5).
RESULTS:
Chi-Square 4.167
Tony Kelly, can you give us the 4 cell counts?
a = females with fitness hearing
b = females with no fitness hearing
c = males with fitness hearing
d = males with no fitness hearing
Thanks.
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