Re: st: sign test output

[Nick Cox <njcoxstata@gmail.com>]

Your t-test is testing a quite different hypothesis. If the two states
0 and 1 of a binary variable have equal frequencies, then its mean is
0.5, not 0.

That aside, the t-test can not be more appropriate for a binary
variable than what you have done already, and this is predictable in
advance, as a distribution with two distinct states is not a normal
distribution. You do not need a Kolmogorov-Smirnov test to tell you
that.

For the record, what I suggested is best not described as a robust
test. It was calculating a confidence interval, and I showed that for
your data the result was robust to the method of calculation, meaning
merely not sensitive. The word "robust" was used informallly.

You never define what you mean by u, so I am not commenting on any
details about u.

I recommend that you read (or re-read) a good introductory text on
statistics, as you appear confused on some basic matters.

Nick

On Thu, Jan 17, 2013 at 7:52 AM, Nahla Betelmal <nahlaib@gmail.com> wrote:

> 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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