According to some researchers , especially on research gate one can use fischer exact test if this is so can any one give me a reference or pdf which clearly mention that fischer exact test can be used if cell frequency is zero?
In top of that , in my thesis i have used fischer exact test where the cell frequency is zero , so now how do i justify this?
Kindly help as soon as possible
Thankyou
Fishers Exact Test is an alternative to chi square test. There's no lower bound on the amount of data that is needed for Fisher's Exact Test. You need to have at least one data value in each row and one data value in each column. If an entire row or column is zero, then you don't really have a 2 by 2 table. You can use Fisher's Exact Test when one of the cells in your table has a zero in it. Details about the test are available here: Statistical Methods for Rates and Proportions. Second Edition.
Joseph L. Fleiss. New York NY: John Wiley and Sons, Inc. (1981) (ISBN: 0-471-06428-9).
Fishers Exact Test is an alternative to chi square test. There's no lower bound on the amount of data that is needed for Fisher's Exact Test. You need to have at least one data value in each row and one data value in each column. If an entire row or column is zero, then you don't really have a 2 by 2 table. You can use Fisher's Exact Test when one of the cells in your table has a zero in it. Details about the test are available here: Statistical Methods for Rates and Proportions. Second Edition.
Joseph L. Fleiss. New York NY: John Wiley and Sons, Inc. (1981) (ISBN: 0-471-06428-9).
Is it that one entry in your data table has a frequency of 0 or that several (or all) of the cells in your table have a frequency of 0?.
Carmen sums it the response nicely. The presence of a single 0 or diagonal of 0 is not a problem for fisher's exact test, but for each group there must be valid observations and for each outcome there must be valid observations. It is not a prerequisite that each group exhibits each outcome or vice versa.
If your data fits the criteria for a valid table as described by Carmen you do not need a reference beyond the standard ones. If you data fails the criteria then the reference provided will justify why its not valid.
Fischer's exact test is used when cells have low expected values, but the only way a cell can have an expected value of zero is when one of the corresponding marginals is zero -- which makes no sense. In other words, an observed value of zero is not the issue here. One way to see that would be to think in terms of the residuals, so an observed value of zero will inherently have a negative residual, meaning that more observations would have been expected in that cell than actually occured.
I also have a similar question and wonder if I might ask it here; thanks for your patience in advance.
I'm asked to check association between 2 nominal variables; each variable has several categories, resulting in a 9X11 contingency table. Total N= approximately 2800. One of the cells has an observed count of 0. Is that a problem if I want to do a Chi-Squared test? (I ran the test in Stata to check if any of the expected frequencies was 20% of the cells had expected frequency
James Renwick Beattie or Carmen C W Lim I have a question related to this. Is it still valid to calculate an effect size (i.e. phi) if only a single cell in a 2x2 table is zero? It seems a bit off to me since it washes out the other cell it gets multiplied by in the calculation and as a result, I think I got a value of phi greater than 1.
Steve Bagienski , I don't think you can get a phi > 1. But if you end up with a zero in the denominator of the chi-square statistic, you might be unable to calculate phi. Do you want to share the table of counts and see what other software returns?
I don't think there's any reason why you can't calculate phi with zeros in the table. There's nothing inferential going on in the calculations. It's just adding and multiplying the numbers to get the chi-square statistic and dividing to get phi.
Ahh okay, I think it was something funky SPSS was doing. The table was:
12 0
6 15
And it was saying phi = 1.215
But I calculated it by hand using the formula here: http://www.pmean.com/definitions/phi.htm
And now I got 0.69... Any idea what SPSS is doing?
Hello Steve Bagienski. How is the data formatted in SPSS? Do you have 33 rows of raw data, or do you have 4 rows with row code, column code & cell count? If the latter, I gather you are using the WEIGHT command. It has trouble with weights = 0. Thanks for any additional detail you can offer.
Meanwhile, using Stata, I get:
Pearson chi2(1) = 15.7143 Pr = 0.000
Fisher's exact = 0.000
1-sided Fisher's exact = 0.000
And Pearson r for the two dichotomies (i.e., Phi-coefficient) = 0.6901
HTH.
In R (psych and vcd packages):
phi = 0.6901
P.S. All the expected counts for that table are > 5.
Output from the following SPSS code is in the attached Excel file.
NEW FILE.
DATASET CLOSE ALL.
DATA LIST LIST / row col n (3F2.0).
BEGIN DATA
0 0 12
0 1 0
1 0 6
1 1 15
END DATA.
WEIGHT by n.
CROSSTABS row BY col /STATISTICS=CHISQ PHI.
HTH.
The Fishers exact test is one of the Chi-2 family tests that examines the association or independence of the two quantities grouped by each other. Its use is also when the number of data in a group is at least equal to that of the other table cells. So you can use this test when faced with a zero value in a group.
I was formatting it with 33 rows: 2 columns, with each row having 2 categorical labels. I figured out what was going wrong...I had an empty row with no values in it that I forgot to delete. Once I deleted that row, I got the right value. Thanks so much for the help everyone! Much appreciated =)
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