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Re: How can I calculate the P values in 6 X 6 contigency table

Posted by Bruce Weaver on May 07, 2013; 6:43pm
URL: http://spssx-discussion.165.s1.nabble.com/How-can-I-calculate-the-P-values-in-6-X-6-contigency-table-tp5719892p5720042.html

"However, the OP is asking for 36 p-values, apparently.  So I'm not sure what they want.  If they want p-values for 98.3% vs 1.7% (cell 1 in the original table), 97% vs 3% (cell 2, original table), etc, that seems rather pointless."

Oops...the OP appears to want 30 p-values, not 36.  Six rows x 5 columns in the original table = 30, not 36.  



Bruce Weaver wrote
It looks to me like six separate 5x2 (or 2x5) tables, as Rich Ulrich observed.  If that is correct, then one could do something like this to get a chi-square for each 5x2 table:

new file.
dataset close all.

data list list / row col (2f1) kount1 (f5.0) pct (f5.1).
begin data
1 1 237  98.3    
1 2 263  97.0    
1 3 257  94.1    
1 4 233  97.1    
1 5  83  92.2    
2 1 223  92.1    
2 2 237  87.1    
2 3 224  81.8    
2 4 201  83.8    
2 5  78  84.8    
3 1 222  91.4    
3 2 247  90.5    
3 3 229  83.0    
3 4 211  88.3    
3 5  72  78.3    
4 1 217  89.3    
4 2 234  85.7    
4 3 214  77.5    
4 4 207  86.6    
4 5  68  74.7    
5 1 224  92.2    
5 2 248  90.8    
5 3 236  85.5    
5 4 217  90.8    
5 5  77  83.7    
6 1 223  92.1    
6 2 247  90.5    
6 3 230  83.6    
6 4 214  89.5
6 5  76  82.6  
end data.

compute kount2 = rnd(kount1*100 / pct) - kount1.
formats kount2 (f5.0).
list.

VARSTOCASES
  /MAKE kount FROM kount1 kount2
  /INDEX=Y(2)
  /KEEP=row col
  /NULL=KEEP.

split file by row.
weight by kount.
crosstabs col by Y / cell = count row / stat = chisqr.
split file off.

However, the OP is asking for 36 p-values, apparently.  So I'm not sure what they want.  If they want p-values for 98.3% vs 1.7% (cell 1 in the original table), 97% vs 3% (cell 2, original table), etc, that seems rather pointless.


Mark Miller wrote
Salman,

This "table" may appear to have been constructed as a cross-tab but
it does not appear to me to represent a single 6x6 table.
Rather, it is  the conjunction of 6 separate 1-way tables where
the columns are instances of the nominal category Ed Level
and the rows are actually separate service items.

Otherwise, you need to explain what exactly is indicated by each row.

In any case, I know of no procedure which will give you a P-value
for the interior cells, although perhaps you could get a separate
Chi-squared value for each row as indicated by the P-value column
on the right.

... Mark Miller


On Tue, May 7, 2013 at 5:35 AM, David Marso <[hidden email]> wrote:

> What do you mean by "calculate the P-values of this table"?
> What do you wish to learn from this data?
> "What should I do for further calculation?"
> That is hardly a question which can be readily answered considering your
> question is ill defined at best.
> Back up a few paces, carefully define your question and perhaps describe
> the
> process by which this data table came about.
> Do you have access to the raw data or merely this summary table?
> --
>
> salman_stats wrote
> > If we are not considering the assumption of Chi-square test for
> proportion
> > and frequencies of the table. What should I do for further calculation?
> >
> > Could you please tell what will be most possible option or method to
> > calculate the P-values of this table.
>
>
>
>
>
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--
Bruce Weaver
bweaver@lakeheadu.ca
http://sites.google.com/a/lakeheadu.ca/bweaver/

"When all else fails, RTFM."

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