Re: st: mean and SD by categories

[Nick Cox <njcoxstata@gmail.com>]

You need to take square roots of the sample sizes in calculating
standard errors.
Nick
njcoxstata@gmail.com


On 16 July 2013 13:56, Federico Belotti <f.belotti@gmail.com> wrote:
> This could be a strategy to get a table using Ben Jann -estout- command
>
> qui {
> forval a=1/2 {
>         forval b=1/7 {
>                 forval c=1/5  {
>                         sum bmi if sex==`a' & agegp==`b' & ses==`c'
>                         mat mean = nullmat(mean) \ r(mean)
>                         mat sd = nullmat(sd) \ r(sd)
>                         mat up = nullmat(up) \ r(mean)+(1.96*r(sd)/r(N))
>                         mat lo = nullmat(lo) \ r(mean)-(1.96*r(sd)/r(N))
>                         local lab `"`lab' "sex=`a' agegp=`b' ses=`c'""'
>                 }
>         }
> }
> mat all = (mean,sd,lo,up)
> mat rown all = `lab'
> mat coln all = mean sd lb ub
> noi estout m(all), varw(22) mlab(,none)
> }
>
> Federico
>
> On Jul 16, 2013, at 2:23 PM, Nick Cox wrote:
>
>> There are various problems with your approach.
>>
>> 1. The immediate problem is that you cycle over combinations of
>> categories creating local macros with names like mean111, sd111, and
>> so forth.
>>
>> But a command like
>>
>> display mean111
>>
>> is understood by Stata to mean that you want to see a value of a
>> variable -mean111- or a scalar -mean111-, but you have no such
>> variable or scalar, hence the message you received.
>>
>> To see the value of a local macro, you would need to do something like
>>
>> display `mean111'
>>
>> 2. Adding +/- 1.96 SD to the mean is an attempt to get 95% confidence
>> intervals. But there are at least two problems here. 1.96 is an
>> optimistic multiplier and a more accurate multiplier would be based on
>> the t-distribution. However, that's secondary compared with using SDs
>> when you should be using standard errors.
>>
>> 3. A strategic difficulty here is that creating lots of local macros
>> first is not a good way to create graphs. In fact if you are a novice
>> in programming trying to program graphs from scratch is not advisable.
>> In your case 70 cross-combinations seems likely to produce only a very
>> complicated graph. Much depends on whether each cross-combination is
>> well enough represented to show stable patterns.
>>
>> -stripplot- from SSC may be of help. It has a -ci- option that means
>> that confidence intervals are shown for you.
>>
>> Nick
>> njcoxstata@gmail.com
>>
>>
>> On 16 July 2013 12:43, Dherani, Mukesh <M.K.Dherani@liverpool.ac.uk> wrote:
>>> Dear all,
>>> I am very novice in programming.
>>> I have a dataset with BMI by three categorical variables age (7 groups), sex (2 groups) and SES (5 groups).  I was to generate a line graph showing mean BMI with 95%CI by age , sex  and SES.  Below is my code to create local that I want to use in twoway line graph:
>>>
>>> forval a=1/2 {
>>> forval b=1/7 {
>>> forval c=1/5  {
>>> sum bmi if sex==`a' & agegp==`b' & ses==`c'
>>> local mean`a'`b'`c'=r(mean)
>>> local sd`a'`b'`c'=r(sd)
>>> local up`a'`b'`c'=r(mean)+(1.96*r(sd))
>>> local lo`a'`b'`c'=r(mean)-(1.96*r(sd))
>>> di mean`a'`b'`c' sd`a'`b'`c' up`a'`b'`c' lo`a'`b'`c'
>>> }
>>> }
>>> }
>>>
>>>
>>> At the end of this command stata says mean111 was not found!
>>> Secondly, I want graph to show me mean bmi (y axis) and age (x axis) for each ses for each sex separately. Should I carry this out in the loop or outside the loop?
>>>
>>> Any help is highly appreciated.
>>>
>>> BW,m
>>>
>>>
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>>
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>
> --
> Federico Belotti, PhD
> Research Fellow
> Centre for Economics and International Studies
> University of Rome Tor Vergata
> tel/fax: +39 06 7259 5627
> e-mail: federico.belotti@uniroma2.it
> web: http://www.econometrics.it
>
>
> *
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