Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: st: sign test output
From
Nahla Betelmal <[email protected]>
To
[email protected]
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 <[email protected]> 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 <[email protected]> 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.
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/faqs/resources/statalist-faq/
> * http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/