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Re: st: Median and CI with predict


From   Carla Guerriero <[email protected]>
To   [email protected]
Subject   Re: st: Median and CI with predict
Date   Tue, 11 Feb 2014 11:18:56 +0100

Hi Nick
I used your coding meta:... and the proportion come out ..
I eventually apply  the ic command to my wtp dependent variable and it
runs without error  but the output is blank ..with both the approaches
..(Wilson and Jeffreys)
also another quesiton I need to test that the WTP values for different
health risk redcution are the same or they statistically different ..
usually I do the test command on coefficient but in this case I need
to compare the values the come from different regression with
intercpet only model .. there is a way to do that in stata ?
Kind Regards
Carla

On Fri, Feb 7, 2014 at 5:00 PM, Carla Guerriero
<[email protected]> wrote:
> Thank you so much Nick that's great!!!
> Kind  Regards
> Carla Guerriero
>
> On Fri, Feb 7, 2014 at 4:56 PM, Nick Cox <[email protected]> wrote:
>> I'd apply -ci- directly; indeed you have a choice of ways to do it.
>>
>> But as for -glm-, my answer is the same answer as before:
>>
>> 1. -glm- gives you confidence intervals in its main output. The only
>> indirectness is that you need to invert the link.
>>
>> 2. -predict- is not needed.
>>
>> Examples:
>>
>> . sysuse auto
>> (1978 Automobile Data)
>>
>> . glm foreign, link(logit)
>>
>> Iteration 0:   log likelihood = -53.942063
>> Iteration 1:   log likelihood = -47.679133
>> Iteration 2:   log likelihood = -47.065235
>> Iteration 3:   log likelihood = -47.065223
>> Iteration 4:   log likelihood = -47.065223
>>
>> Generalized linear models                          No. of obs      =        74
>> Optimization     : ML                              Residual df     =        73
>>                                                    Scale parameter =  .2117734
>> Deviance         =  15.45945946                    (1/df) Deviance =  .2117734
>> Pearson          =  15.45945946                    (1/df) Pearson  =  .2117734
>>
>> Variance function: V(u) = 1                        [Gaussian]
>> Link function    : g(u) = ln(u/(1-u))              [Logit]
>>
>>                                                    AIC             =   1.29906
>> Log likelihood   = -47.06522292                    BIC             = -298.7373
>>
>> ------------------------------------------------------------------------------
>>              |                 OIM
>>      foreign |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
>> -------------+----------------------------------------------------------------
>>        _cons |  -.8602013   .2560692    -3.36   0.001    -1.362088   -.3583149
>> ------------------------------------------------------------------------------
>>
>> . mata: invlogit((-.8602013, -1.362088, -.3583149))
>>                  1             2             3
>>     +-------------------------------------------+
>>   1 |    .29729729   .2039011571   .4113675423  |
>>     +-------------------------------------------+
>>
>> . ci foreign, jeffreys binomial
>>
>>                                                          ----- Jeffreys -----
>>     Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
>> -------------+---------------------------------------------------------------
>>      foreign |         74    .2972973    .0531331        .2024107    .4076909
>>
>> . ci foreign, wilson binomial
>>
>>                                                          ------ Wilson ------
>>     Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
>> -------------+---------------------------------------------------------------
>>      foreign |         74    .2972973    .0531331        .2052722    .4093291
>>
>>
>> Nick
>> [email protected]
>>
>>
>> On 7 February 2014 15:45, Carla Guerriero <[email protected]> wrote:
>>> Hi Nick my dependent variable is a proportion (of the budget that
>>> given a budget constraint individuals are willing to give up)
>>> so I used  logit link function to ensure linearity and binomial family
>>> distribution.. For example for 19 in 100 risk reduction I get a
>>> coefficent of -.657211*** and If i use predict the mean WTP is 0.20
>>> which makes sense .. but the SD is 0 .. I want to get CI for the mean
>>> .. maybe boostrapping is an option? I know how to do for DCE where you
>>> have a ratio of the coefficent (delta or boostrapping or parametric
>>> boostrapping) but I have no clue how to make CI for eman WTP estimate
>>> from regression ..
>>>
>>>
>>> On Fri, Feb 7, 2014 at 4:26 PM, Nick Cox <[email protected]> wrote:
>>>> -glm- with no covariates gives you confidence intervals for mean
>>>> response, directly or indirectly, depending on the link. No need to
>>>> use -predict- at all. I don't think you can get confidence  intervals
>>>> for the median that way.
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