Hi Ann,

If you are using categorical data you can use the Kruskal-Wallis test (the non-parametric equivalent of the one-way ANOVA) to determine group differences. If the test shows there are differences between the 3 groups. You can use the Mann-Whitney test to do pairwise comparisons as a post hoc or follow up analysis. Since you're only doing a few comparisons (i.e. X vs. Y, X, vs Z, and Y vs. Z) you wouldn't have to worry about family wise error rates.

An alternative solution is to run a normality test for the data (you can use the Shapiro–Wilk test) if the results come back as non-significant, you could use a parametric test to do the analysis (i.e. ANOVA) and run post-hocs if the ANOVA is significant.

Some people might argue that categorical data should be analyzed using non-parametric tests no matter what, while others would say that as long as the data is normal, a parametric test is the best way to go as those tests are more powerful.

I hope this helps. Good luck! 

Hi Ann,

If you are using categorical data you can use the Kruskal-Wallis test (the non-parametric equivalent of the one-way ANOVA) to determine group differences. If the test shows there are differences between the 3 groups. You can use the Mann-Whitney test to do pairwise comparisons as a post hoc or follow up analysis. Since you're only doing a few comparisons (i.e. X vs. Y, X, vs Z, and Y vs. Z) you wouldn't have to worry about family wise error rates.

An alternative solution is to run a normality test for the data (you can use the Shapiro–Wilk test) if the results come back as non-significant, you could use a parametric test to do the analysis (i.e. ANOVA) and run post-hocs if the ANOVA is significant.

Some people might argue that categorical data should be analyzed using non-parametric tests no matter what, while others would say that as long as the data is normal, a parametric test is the best way to go as those tests are more powerful.

I hope this helps. Good luck! 

Ann, I'm not sure I completely understand your question. But if you mean you have the proportion who ride the tube for each group and the sample size for each group, see the link below for advice on how to proceed.  HTH.

http://www.theanalysisfactor.com/chi-square-test-proportions/

you can also try correspondance analysis. e.g pl follow the following link:

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3718710/

Dear Ann

I think chi-square test between each two groups represent good chose because the chi-square test of three categories give a general decision for not equality between groups and not determine the pair comparisons.

Good Luck

The answer of DR Khalid Hassan is the most appropriate

Maybe you should at least consider adjusting your criterion for per comparison statistical significance by Bonferroni's or Sidak's correction even though you will only make a very few comparisons. The difference is substantial. If you were to use a control you would be making 2 comparisons. More than a 5% overall cutoff might not impress, So, under this limitation you would be able to use no more than a 2.5% per comparison cutoff using the Bonferroni correction.

However, there is an awkward philosophical problem here in that if you did two (or even three) little experiments everyone would accept 5% each. This is one of the many things we, as the research community need to think out. 

And, finally, a lot of us would say that simply finding confidence intervals is closer to ideal than testing. In any case, as some mathematical statisticians will tell you,  non overlap of confidence intervals is not a valid statistical test.

With SPSS, you can run a chi-square, and test for pair-wise differences between the pair of groups. Put your 3 groups in columns, and "ride the tube: yes/no in the rows.  Use the raw numbers in the 6 cells.  Select the option to use Bonferroni corrections for the pairwise comparisons. Using sub-scripted (super-scripted?) letters, SPSS will tell you which of the pairwise differences are significant, while preserving the overall (experiment-wise) 0.05 significance level.   

Looking back at your question, I see that you seem to be thinking that you cannot do multiple comparisons with a Pearson's chi square. Black boxes such as  SPSS or SAS will do it for you automatically, But we should have given the simple answer that after the 2X3 test you can run a 2X2 Pearson's chi square for each comparison. Percentages usually do not fit the assumptions of the ANOVA, unless the sample size is huge even under the popular conversions. This is because the ANOVA is rather sensitive to skew,  But if the sample is above, say 400 per comparison, then Pearson's chi square will possibly  be hyper sensitive. So, maybe that might be a sufficient definition of "huge" enough for the ANOVA. And at some point in your choice of a sample size the central limit theorem (CLT) will protect you from all but the most extreme non normality (ie. if one of the samples is less than a few percent.) The CLT will cause the sums and means within a test to approach normality as the sample size grows very large. Therefore the distribution of the raw data does not need to pass a normality test if the sample is very large (greater than 30 per comparison and, in the ANOVA additionally that the df for error>20). I might as well point out that assuming your data is counts, you will not have mush sensitivity (power) to detect a difference of, say a few percent, unless your total sample size is greater than say 1500. 

I imagined the original question to be about a 3x2 contingency table, which is why I directed the OP to the resource I did.  If it is a 3x2 contingency table, and if it makes sense to partition the overall Chi-square with df = 2 into two orthogonal components, the likelihood ratio Chi-square test (which SPSS reports automatically) has the nice property that the LR Chi-square values for the two orthogonal components sum exactly the the LR Chi-square value for the overall test.  Here is an example, with the frequencies taken from a 3x2 example I have in some of my notes.  The OP can replace my frequencies with their own, if they wish.  (If you are a point-and-click only SPSS user, click on File > New > Syntax to open a new syntax window.  Then copy the code below and paste it into that syntax window.  Then select Run > All to run it.)  HTH.

* Read in cell counts for a 3x2 contingency table.

DATA LIST LIST / Group Tube N (3F5.0).

BEGIN DATA

1 1 9

1 2 21

2 1 17

2 2 13

3 1 22

3 2 8

END DATA.

VARIABLE LABELS Tube "Takes tube to work".

VALUE LABELS Tube 1 "1-Yes" 2 "2-No".

WEIGHT by N.

CROSSTABS Group by Tube

  /CELLS = COUNT COL

  /STATISTICS = CHISQ.

* The overall Chi-square for this table has df=2.

* The likelihood ratio Chi-square can be partitioned into

* two orthogonal components that sum to the overall

* likelihood ratio Chi-square. E.g., we could make

* the following two contrasts:

* 1) Group 1 vs groups 2 & 3 combined

* 2) Group 2 vs Group 3.

* Contrast 1: 1 v [2,3].

TEMPORARY.

RECODE Group (3=2) (ELSE=COPY).

CROSSTABS Group by Tube

  /CELLS = COUNT COL

  /STATISTICS = CHISQ.

* Contrast 2: 2 v 3.

TEMPORARY.

SELECT IF Group GT 1.

CROSSTABS Group by Tube

  /CELLS = COUNT COL

  /STATISTICS = CHISQ.

IF $CASENUM EQ 1 ChiSqSum = SUM(10.020741, 1.844780).

FORMATS ChiSqSum(F8.4).

LIST ChisqSum /CASES=FROM 1 to 1.

* The sum of the LR Chi-square values from the two

* orthogonal contrasts = the overall LR Chi-square

* value (with df = 2).

I would do this as logistic regression. Generally speaking, it is not a good idea to convert categorical data into percentages; percentages look like a continuous variable but they generally violate the assumption of most continuous-variable tests (like ANOVA and t-tests).

From as an a priori design viewpoint, I agree with Stephen Politzer-Ahles.  Good answer. 

I agree with Dr. Khalid Hassan., Lewis Maccarter and Emet D Schneiderman.