I asked 14 participants "what is your preference among six keyboard designs?" The data resulted in the following picture. For that case, which statistical method should I use? I thought that fisher's exact test is suitable for that case, but I could not be sure.
The basic approach would be a chi-square goodness-of-fit-test. (Against a null of each cell having 1/6 of the total counts.) But since you have such small counts, you would use something like a multi-nomial exact test or Monte Carlo simulation for a chi-square goodness-of-fit-test. That being said, having 14 observations across six categories isn't going to tell you much.
Salvatore S. Mangiafico thank you for answer. My sample size is the same with this research (Conference Paper Effects of Key Size, Gap and the Location of Key Characters ...
). Please see user preference section. How did the researcher apply chi-square test at that research?
In that manuscript, they give you enough information to re-create what they did for those user preferences chi-square tests. I'm getting somewhat different results than they did for some analyses; I don't have an explanation for that.
For example, the counts for the preferences for Key size were (0, 9, 5, 0). They used a chi-square goodness-of-fit test, which yields ( X-squared = 16.195, df = 3, p-value = 0.001034 ), which is almost the same as is reported in the paper. With those zero counts, the straight chi-square test may not be appropriate. It's possible that SPSS applies some correction or alternate method automatically, and that could be why the manuscript reports somewhat different results than I'm getting in R. Results from Monte Carlo simulation are more-or-less similar.
Code in R:
N = 14
pkey = c(0, 0.64, 0.36, 0)
pgap = c(0.43, 0.57)
ploc = c(0.29, 0, 0.71, 0, 0)
Key = pkey * N
Gap = pgap * N
Loc = ploc * N
Key = c(0, 9, 5, 0)
Gap = c(6, 8)
Loc = c(4, 0, 10, 0, 0)
chisq.test(Key)
### X-squared = 16.195, df = 3, p-value = 0.001034
chisq.test(Key)$expected
chisq.test(Gap)
### X-squared = 0.2744, df = 1, p-value = 0.6004
chisq.test(Loc)
### X-squared = 27.174, df = 4, p-value = 1.833e-05
chisq.test(Key, simulate.p.value=T, B=10000)
### X-squared = 16.195, df = NA, p-value = 0.0016
chisq.test(Loc, simulate.p.value=T, B=10000)
### X-squared = 27.174, df = NA, p-value = 9.999e-05
As a follow-up to my comments above. I will note that in the section on user preference, the authors of that manuscript are a little ambivalent about if their statistical analyses support their statements of the results. For example, they say
The majority of participants preferred 15 mm key size (64%, Χ2 = 16.286, p = 0.010)...
However, their data do not support the idea that the 15 mm key size is statistically preferred when compared to the 20 mm key size. This is pretty sloppy, blending results from statistical analysis with conclusions that aren't supported by the statistical analysis.
Thank Salvatore S. Mangiafico for clarifying me. Regarding my question, I am going to prefer using Monte Carlo simulation for a chi-square goodness-of-fit-test.
Salvatore S. Mangiafico I also have one more question.
I applied that R code:
Key = c(1, 2, 4, 6, 0, 1)
chisq.test(Key, simulate.p.value = T, B=10000)
I found: X-squared = 10.857, df = NA, p-value = 0.06039
No significant result found.
Can I do it over proportion like:
pkey = c(7.1, 14.3, 28.6,42.9,0.0,7.1)
I found: X-squared = 77.821, df = NA, p-value = 9.999e-05
Significant result found.
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