Testing the significant difference in Proportions

[Jims More]

I have two categorical variables.  One is the response variable, a yes/no variable.  The other is a categorical predictor variable which has 5 categories.  The proportion table of the gathered data (n=5000) looks like as follows:

 

                         Row % of Yes Response           Row % of No Response

Category1                .40                                    .60

Category2                .30                                    .70

Category3                .80                                     20

Category4                .40                                    .60

Category5                .34                                    .66

 

How to test using SPSS if the proportions of Yes and No are significantly equal for each category?

 

Thank you for any help.

Jims

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[Bruce Weaver]

Jims More wrote

I have two categorical variables.  One is the response variable, a yes/no variable.  The other is a categorical predictor variable which has 5 categories.  The proportion table of the gathered data (n=5000) looks like as follows:
 
                         Row % of Yes Response           Row % of No Response
Category1                .40                                    .60
Category2                .30                                    .70
Category3                .80                                     20
Category4                .40                                    .60
Category5                .34                                    .66
 
How to test using SPSS if the proportions of Yes and No are significantly equal for each category?
 
Thank you for any help.
Jims

You can always roll your own, like this:

data list list / cat (f2.0) x N (2f6.0).
begin data
1 40 100
2 30 100
3 80 100
4 40 100
5 34 100
end data.

compute p = x/N.
compute se = SQRT( p*(1-p)/N).
compute z = (p - .5)/se.
compute pvalue = 2*(1-(CDF.NORMAL(abs(z),0,1))) .
format se pvalue (f5.3).
list.

You'll have to fill in the correct values of X and N, of course.

[Peck, Jon]

Note also that there is a PROPOR extension command available from SPSS Developer Central (www.spss.com) that will construct various confidence intervals for proportions.  It requires at least version 16 and the Python plugin, but no Python knowledge is required to use it.

-----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of Bruce Weaver
Sent: Monday, July 06, 2009 5:27 AM
To: [hidden email]
Subject: Re: [SPSSX-L] Testing the significant difference in Proportions

Jims More wrote:

>
> I have two categorical variables.  One is the response variable, a yes/no
> variable.  The other is a categorical predictor variable which has 5
> categories.  The proportion table of the gathered data (n=5000) looks like
> as follows:
>
>                          Row % of Yes Response           Row % of No
> Response
> Category1                .40                                    .60
> Category2                .30                                    .70
> Category3                .80                                     20
> Category4                .40                                    .60
> Category5                .34                                    .66
>
> How to test using SPSS if the proportions of Yes and No are significantly
> equal for each category?
>
> Thank you for any help.
> Jims
>

You can always roll your own, like this:

data list list / cat (f2.0) x N (2f6.0).
begin data
1 40 100
2 30 100
3 80 100
4 40 100
5 34 100
end data.

compute p = x/N.
compute se = SQRT( p*(1-p)/N).
compute z = (p - .5)/se.
compute pvalue = 2*(1-(CDF.NORMAL(abs(z),0,1))) .
format se pvalue (f5.3).
list.

You'll have to fill in the correct values of X and N, of course.


-----
--
Bruce Weaver
[hidden email]
http://sites.google.com/a/lakeheadu.ca/bweaver/
"When all else fails, RTFM."

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[E. Bernardo]

How to open SPSS16 outputs in SPSS 15?

Thanks for any help!
Eins

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[Lemon, John S.]

As far as I know, and I’m willing to be corrected by someone else, it is not possible to open output generated by 16 in any earlier version. The way I get round this problem is to export the output to a Word file and then send that to the people who need it. This obviously works for people who do not have SPSS

 

Best Wishes

 

John S. Lemon

Student Liaison Officer

Directorate of Information Technology (DIT) - University of Aberdeen

Edward Wright Building: Room G51

 

Tel:  [phone]

Fax: [phone]

 

Diary ( Free / Busy )

 

 

From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of Eins Bernardo
Sent: 07 July 2009 02:58
To: [hidden email]
Subject: How to open SPSS16 outputs in SPSS15?

 

How to open SPSS16 outputs in SPSS 15?

Thanks for any help!
Eins

 

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[Jianfeng Gu]

hi,

in spsstools.net, i found a interesting scripts.
http://www.spsstools.net/Scripts/PivotTables/CropAndRetain.txt

'This Script allows you to delete all rows in a Pivot Table beyond a cut
'point that you can specify (intNumberRowsToKeep).
'It also allows you to chose a particular row label (strSearchRowToKeep)
'to keep, even if it is below the cut point intNumberRowsToKeep.

but i am newbie of spss scripts. i think this one can help me, but i do not know how to change some places to fit my requirment.

anyway, thanks for lemon's rapid reply.

if anyone can help me about that script, please continue....

thanks.

ashura ku

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[Allan Lundy, PhD]

This is a real drag, and can cause a lot of problems, especially with someone like me who deals with a lot of different people.  I used to boast about SPSS that, unlike the jerks at Microsoft and many other software purveyors, the good folks at SPSS were careful to keep versions compatible.  But even MS Word allows you to save files in a whole range of different (outmoded) formats.  I have also noticed that the modern assumption of user stupidity has started creeping into SPSS/PASW.  Anyone else annoyed at the overly complex, slow to load syntax program?  With the name change, has there been a major shift toward assuming less competence on the part of customers?

At 02:55 AM 7/7/2009, you wrote:

As far as I know, and I’m willing to be corrected by someone else, it is not possible to open output generated by 16 in any earlier version. The way I get round this problem is to export the output to a Word file and then send that to the people who need it. This obviously works for people who do not have SPSS
 
Best Wishes
 
John S. Lemon
Student Liaison Officer
Directorate of Information Technology (DIT) - University of Aberdeen
Edward Wright Building: Room G51
 
Tel:  [phone]
Fax: [phone]
 
Diary ( Free / Busy )
 
 
From: SPSSX(r) Discussion [[hidden email]] On Behalf Of Eins Bernardo
Sent: 07 July 2009 02:58
To: [hidden email]
Subject: How to open SPSS16 outputs in SPSS15?
 
How to open SPSS16 outputs in SPSS 15?

Thanks for any help!
Eins
 

Start chatting with friends on the all-new Yahoo! Pingbox today!!
It's easy to create your personal chat space on your blogs


The University of Aberdeen is a charity registered in Scotland, No SC013683.

Allan Lundy, PhD
Research Consultant      [hidden email]

Business & Cell (any time): 215 820-8100   
Home: Voice and fax (8am - 10pm,  7 days/week):  215 885-5313
Address:  108 Cliff Terrace,  Wyncote, PA 19095
Visit my Web site at   www.dissertationconsulting.net
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[E. Bernardo]

Hi Jims, Bruce, et al,

 

Jims More wrote:

>
> I have two categorical variables.  One is the response variable, a yes/no
> variable.  The other is a categorical predictor variable which has 5
> categories.  The proportion table of the gathered data (n=5000) looks like
> as follows:
>
>                          Row % of Yes Response           Row % of No
> Response
> Category1                .40                                    .60
> Category2                .30                                    .70
> Category3                .80                                     20
> Category4                .40                                    .60
> Category5                .34                                    .66
>
> How to test using SPSS if the proportions of Yes and No are significantly
> equal for each category?
>
> Thank you for any help.
> Jims
>

Bruce replied:

You can always roll your own, like this:

data list list / cat (f2.0) x N (2f6.0).
begin data
1 40 100
2 30 100
3 80 100
4 40 100
5 34 100
end data.

compute p = x/N.
compute se = SQRT( p*(1-p)/N).
compute z = (p - .5)/se.
compute pvalue = 2*(1-(CDF.NORMAL(abs(z),0,1))) .
format se pvalue (f5.3).
list.

You'll have to fill in the correct values of X and N, of course.

Can someone translate the syntax of Bruce into syntax that do the same but for SPSS data file with variables(predictor and dependent) in the columns.

 

Thank you.

Eins

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[Marta Garcia-Granero]

Hi Eins:

You don't need a special syntax to achieve that task. You have to split
your file by the categorical predictor and then run a goodness of fit
test (ANALYZE -> NON PARAMETRIC -> CHI-SQUARE) for the Yes/No variable
with equal expected values.

That's it.

Marta

Eins Bernardo wrote:


> Hi Jims, Bruce, et al,
>
>
Can someone translate the syntax of Bruce into syntax that do the same
but for SPSS data file with variables(predictor and dependent) in the
columns

> Jims More wrote:
> >
> > I have two categorical variables.  One is the response variable, a
> yes/no
> > variable.  The other is a categorical predictor variable which has 5
> > categories.  The proportion table of the gathered data (n=5000)
> looks like
> > as follows:
> >
> >                          Row % of Yes Response           Row % of No
> > Response
> > Category1                .40                                    .60
> > Category2                .30                                    .70
> > Category3                .80                                     20
> > Category4                .40                                    .60
> > Category5                .34                                    .66
> >
> > How to test using SPSS if the proportions of Yes and No are
> significantly
> > equal for each category?
> >
> > Thank you for any help.
> > Jims
> >
> Bruce replied:
>
> You can always roll your own, like this:
>
> data list list / cat (f2.0) x N (2f6.0).
> begin data
> 1 40 100
> 2 30 100
> 3 80 100
> 4 40 100
> 5 34 100
> end data.
>
> compute p = x/N.
> compute se = SQRT( p*(1-p)/N).
> compute z = (p - .5)/se.
> compute pvalue = 2*(1-(CDF.NORMAL(abs(z),0,1))) .
> format se pvalue (f5.3).
> list.
>
> You'll have to fill in the correct values of X and N, of course.
> .
>
> Thank you.
> Eins
>
>
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