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# Re: st: chi2 on aggregate results

 From "Elizabeth Allred" To Subject Re: st: chi2 on aggregate results Date Mon, 10 May 2010 13:27:29 -0400

Exactly.

>>> On 5/10/2010 at  1:24 PM, in message
<i2ycde780461005101024uf1f12aftb067e77c8c0e2ede@mail.gmail.com>, mai7777
<mai7777@gmail.com> wrote:
> Thanks for the quick response. I guess it was the "col" option that I
> was looking for, but just to confirm: Below, is the Pearson chi2(1) =
>  5.9421   Pr =  testing the hypothesis that the proportion of 76.67 in
> a sample size of 30 equal to the proportion of 47.22 in a sample size
> of 36?
>
>
> . tabi 23 17 \ 7 19, chi col
>            |          col
>       row |         1          2 |     Total
> -----------+----------------------+----------
>         1 |        23         17 |        40
>            |     76.67      47.22 |     60.61
> -----------+----------------------+----------
>         2 |         7         19 |        26
>            |     23.33      52.78 |     39.39
> -----------+----------------------+----------
>    Total |        30         36 |        66
>            |    100.00     100.00 |    100.00
>
>           Pearson chi2(1) =   5.9421   Pr = 0.015
>
>
> On Mon, May 10, 2010 at 1:15 PM, Alan Neustadtl
>> Are you looking for something like this?
>>
>> - tabi 23 17  \ 7 19, chi col -
>>
>>
>>
>> On Mon, May 10, 2010 at 1:08 PM, mai7777 <mai7777@gmail.com> wrote:
>>> hi,
>>> I'm trying to conduct a chi2 on aggregate results to test that the
>>> proportions of incidence across two treatments are the same:
>>>
>>> Treatment 1 has a total of 30 subjects, 23 of which had incidence
>>> Treatment 2 has a total of 36 subjects, 17 of which had incidence.
>>>
>>> How do I do that?
>>> Thanks!
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>>
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>
> *
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