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Re: st: sign test output


From   Nahla Betelmal <nahlaib@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: sign test output
Date   Thu, 17 Jan 2013 07:52:21 +0000

Thank you Maarten and Nick  for the great help.

 So, in this case I would reject the null in favour of the alternative
u>0 as p value 0.000. However, using t-test on the same sample
provided the opposite (i.e. accept the null).

ttest DA_T_1 == 0

One-sample t test
------------------------------------------------------------------------------
Variable |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf. Interval]
---------+--------------------------------------------------------------------
  DA_T_1 |     346    1.564346     1.68628    31.36663   -1.752338     4.88103
------------------------------------------------------------------------------
    mean = mean(DA_T_1)                                           t =   0.9277
Ho: mean = 0                                     degrees of freedom =      345

    Ha: mean < 0                 Ha: mean != 0                 Ha: mean > 0
 Pr(T < t) = 0.8229         Pr(|T| > |t|) = 0.3542          Pr(T > t) = 0.1771


I think this is due to the distribution of the sample, so I performed
K-S normality test. It shows that data is not normally distributed,
hence I should use the non-parametric sign test instead of t-test. In
other words I would reject the null u=0 in favor of u>0 , right?


ksmirnov  DA_T_1 = normal((DA_T_1-DA_T_1_mu)/  DA_T_1_s)

One-sample Kolmogorov-Smirnov test against theoretical distribution
           normal((DA_T_1-DA_T_1_mu)/  DA_T_1_s)

 Smaller group       D       P-value  Corrected
 ----------------------------------------------
 DA_T_1:             0.4878    0.000
 Cumulative:        -0.4330    0.000
 Combined K-S:    0.4878    0.000      0.000


N.B. Thank you so much Nick for the robust test you mentioned, I will
use that as well)

Many thanks

Nahla

On 16 January 2013 09:33, Nick Cox <njcoxstata@gmail.com> wrote:
> In addition, it could be as or more useful to think in terms of
> confidence intervals. With this sample size and average, 0.5 lies well
> outside 95% intervals for the probability of being positive, and that
> is robust to method of calculation:
>
> . cii 346 221
>
>                                                          -- Binomial Exact --
>     Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
> -------------+---------------------------------------------------------------
>              |        346    .6387283    .0258248        .5856497    .6894096
>
> . cii 346 221, jeffreys
>
>                                                          ----- Jeffreys -----
>     Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
> -------------+---------------------------------------------------------------
>              |        346    .6387283    .0258248        .5871262    .6880204
>
> . cii 346 221, wilson
>
>                                                          ------ Wilson ------
>     Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
> -------------+---------------------------------------------------------------
>              |        346    .6387283    .0258248        .5868449    .6875651
>
> Nick
>
> On Wed, Jan 16, 2013 at 9:13 AM, Maarten Buis <maartenlbuis@gmail.com> wrote:
>> On Wed, Jan 16, 2013 at 9:38 AM, Nahla Betelmal wrote:
>>> I have generated this output using  non-parametric test "one sample
>>> sign test" with null: U=0 , & Ua > 0
>>>
>>> However, I do not understand the output. where is the p-value? is it
>>> 0.5 in all cases or the 0.000 ( as in the first and third cases) and
>>> 1.000 as in the second case?
>>>
>>>. signtest DA_T_1= 0
>>>
>>> Sign test
>>>
>>>         sign |    observed    expected
>>> -------------+------------------------
>>>     positive |         221         173
>>>     negative |         125         173
>>>         zero |           0           0
>>> -------------+------------------------
>>>          all |         346         346
>>>
>>> One-sided tests:
>>>   Ho: median of DA_T_1 = 0 vs.
>>>   Ha: median of DA_T_1 > 0
>>>       Pr(#positive >= 221) =
>>>          Binomial(n = 346, x >= 221, p = 0.5) =  0.0000
>>
>> The p-value is the last number, so in your case 0.0000. The stuff
>> before the p-value tells you how it is computed: it is based on the
>> binomial distribution, and in particular it is the chance of observing
>> 221 successes or more in 346 trials when the chance of success at each
>> trial is .5. For this tests this chance is the p-value, and it is very
>> small, less than 0.00005. If you type in Stata -di binomialtail(346,
>> 221, 0.5)- you will see that this chance is 1.381e-07, i.e.
>> 0.00000001381.
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