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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 11:45:27 +0100
Sorry I will try to be more clear:
In my previous question I was asking about obtaining confidence
interval for my dependent variable (willigness to pay) wich in my case
is bounded between 0 and 1. After trying different models (zero
inflated beta and beta) I fund that glm with logit link function and
binomial family works better (test with AIC and BIC) . I previously
asked how to get the confidence intervals from constant only model and
I was (wrongly) using the predict command.
You suggested to run the regression and then to obtain the proportion
to use "meta: invlogit (constant, lower an upper confidece interval
values)" ..
In order to obtain the confidence interval for my dependent variable
you suggested to use the command: ci . I used "ci depvar, jeffreys
binomial" (I also tried Wilson" but the command show a blank results
(results with no number).. and I dont understand why.
My further question was: is there a test I can perform in stata to
test if the results from two different regressions are the same?
for example I have the willigness to pay for 1 in 100 risk reduciton
is equal to 0.21 and the willigness to pay for 19 in 100 is 0.50 I
want to test they are statisitcally different I can I do ?
Hope this makes sense ..
Kind Regards
Carla
On Tue, Feb 11, 2014 at 11:24 AM, Nick Cox <[email protected]> wrote:
> Sorry, but I don't understand almost any of this.
>
> meta: ?
>
> ic ?
>
> wtp ? WTP? (I think that means "willingness to pay", but please note
> that only some people here are economists)
>
> Note that -ci- is limited to single variables and that its -wilson-
> and -jeffreys- options don't travel to other commands.
>
> Whatever you did sounds at some considerable distance from your last
> question and my last answer. If someone else can't work out what you
> are saying, please read the FAQ advice again and give much more detail
> on your problem.
> Nick
> [email protected]
>
>
> On 11 February 2014 10:18, Carla Guerriero <[email protected]> wrote:
>> 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.
>>>>>> *
>>>>>> * For searches and help try:
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>>>>>> * http://www.ats.ucla.edu/stat/stata/
>>>>> *
>>>>> * For searches and help try:
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>>>>> * http://www.ats.ucla.edu/stat/stata/
>>>> *
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>> *
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> *
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