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Re: st: sign test output


From   Nahla Betelmal <[email protected]>
To   [email protected]
Subject   Re: st: sign test output
Date   Fri, 18 Jan 2013 12:34:42 +0000

Dear Paul,

Thank you for your kind feedback.

Briefly I am looking if firms manage earnings upwardly via
discretionary accruals prior certain event. The literature suggests
that the discretionary accruals (DA) should be positive (i.e.. greater
than zero).

In order to calculate DA for my sample (346 firms), I must first
perform a cross sectional regression on the whole population available
from the database (thousands firms). Then generate the estimated
(normal level of accrual) substitute the coefficient values generated
into my sample' variables. Finally subtract the normal from the actual
(reported level) to get DA (abnormal level) of accruals. Again
According to the literature DA should be positive.

So, I performed t-test on DA to see if there is sufficient evidence
that mean DA >0, but as you can see the test shows that there is no
sufficient evidence.

 ttest wDA_T_1== 0

One-sample t test
------------------------------------------------------------------------------
Variable |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf. Interval]
---------+--------------------------------------------------------------------
 wDA_T_1 |     346   -.0137077    .0645203    1.200146   -.1406103    .1131949
------------------------------------------------------------------------------
    mean = mean(wDA_T_1)                                          t =  -0.2125
Ho: mean = 0                                     degrees of freedom =      345

    Ha: mean < 0                 Ha: mean != 0                 Ha: mean > 0
 Pr(T < t) = 0.4159         Pr(|T| > |t|) = 0.8319          Pr(T > t) = 0.5841


On the other hand when I performed the non-parametric sign test on DA,
the test shows that there is sufficient evidence that median DA >0

signtest wDA_T_1= 0


        sign |    observed    expected
-------------+------------------------
    positive |         221         173
    negative |         125         173
        zero |           0           0
-------------+------------------------
         all |         346         346

One-sided tests:
  Ho: median of wDA_T_1 = 0 vs.
  Ha: median of wDA_T_1 > 0
      Pr(#positive >= 221) =
         Binomial(n = 346, x >= 221, p = 0.5) =  0.0000

  Ho: median of wDA_T_1 = 0 vs.
  Ha: median of wDA_T_1 < 0
      Pr(#negative >= 125) =
         Binomial(n = 346, x >= 125, p = 0.5) =  1.0000

Two-sided test:
  Ho: median of wDA_T_1 = 0 vs.
  Ha: median of wDA_T_1 != 0
      Pr(#positive >= 221 or #negative >= 221) =
         min(1, 2*Binomial(n = 346, x >= 221, p = 0.5)) =  0.0000

I got the same conclusion even after I winsorized DA, Also, the
normality test show rejection of normality both before and after I
winsorized the data


                    Skewness/Kurtosis tests for Normality
                                                         ------- joint ------
    Variable |    Obs   Pr(Skewness)   Pr(Kurtosis)  adj chi2(2)    Prob>chi2
-------------+---------------------------------------------------------------
     wDA_T_1 |    346       0.000          0.000            .         0.0000

 Shapiro-Wilk W test for normal data

    Variable |    Obs       W           V         z       Prob>z
-------------+--------------------------------------------------
     wDA_T_1 |    346    0.41873    140.579    11.689    0.00000

This is what initially made me go for the sign test instead of t-test
as I first thought that if the data is not normally distributed, then
use non-parametric test, but from previous discussions in statalist ,
no agreement on that!

So, now I am confused how to interpret the results. Using the sign
test: there is sufficient evidence that earnings are managed upwardly.
Using the t-test: earnings are not managed at all!! and this is what I
meant by irrationality in previous email.

Also, As I am not using regression in my 346 firm sample, robust or
bootstrapped regression  would not solve the normality issue.


Thank you all again, I do appreciate all the feedback and time you gave me,

Nahla



On 18 January 2013 10:48, Seed, Paul <[email protected]> wrote:
> Dear Statalist,
>
> While the discussion of Nahla Betelmal's query has been interesting and
> informative, one point seems to have been missed: the question is ill-defined.
>
> It appears that Nahla Betelmal has a variable that she wants/expects
> for good theoretical reasons to have an average of 0; and wants to test
> if this is true.  We are not told any more.
>
> If (s)he came to me for statistical advice, I would instantly want to know
>         - what the theoretical reasons were
>         - which average (the mean or the median) was expected to be 0
>         - how large a tolerance was acceptable
>         - what the implications would be if the average was not 0.
> Until I had a clear understanding, I would not want start analysing data.
>
> The second question is crucial.  For a seriously non-normal distribution,
> the mean and the median can be quite different, and it is possible
> to construct examples where the mean is significantly > 0, while
> the median is significantly < 0.
>
> Normality checks would be mainly graphical, for the reasons discussed;
> but I might look at measures of skewness, kurtosis and in particular compare
> whether the mean and median were sufficiently close for it not to matter which
> I used.  (Estimates of the mean are usually more robust, so with low skew and
> mean close to median, I might prefer to use the mean even if the median were
> the main object of interest.)
>
> Assuming interest was in the mean, I would advise one or more of
>         one-sample t-test  (quick simple, and usually sufficient)
>         linear regression with robust standard errors (a basic correction for non-Normality)
>         bootstrapped linear regression with BCa confidence intervals, (a fuller correction,
>                 that can give asymmetrical CI where appropriate, e.g. in cases of extreme non-Normality).
>
> All methods are well described in the Stata manual, and usually give very similar answers
> (except for extreme cases of non-Normality).
>
> If interest was in the median, and I didn't trust the Normal approximation,
> I would use the -centile- command with the -cci- option to get a confidence interval
> for the median.
>
> In each case I would direct attention to the confidence interval, and to the question of whether
> the answer was sufficiently close to 0  (As defined by the third question.)
>
> All this assumes that the ultimate interest is in the answer to this question.
> If it was just a preliminary to another analysis, or the answer was wanted for
> some deduction that could be made from it, I would also look for other
> ways of addressing the real question, whatever it might be.
>
> On Jan 17, 2013, at 5:13, Nahla Betelmal <[email protected]> wrote:
>
>> Again, thank you both for your comments.
>>
>> However, if normality test is proved to be useful only for huge sample
>> as Maarten mentioned. How can we determine which test (i.e. parametric
>> or non-parametric ) to be used for smaller sample size in hundreds?!
>>
>> I personally think it is irrational to run both t-test and sign test
>> on the same sample and hope they both produce the same conclusion! and
>> what if they don't!
>>
>> I will follow Nick's advise to look deeper in the data, but I still
>> believe that there must be another way to give obvious solution to
>> this situation.
>>
>> Thank you both again, I highly appreciate your kind help and time,
>>
>> Nahla
>>
>>
>
>
> Paul T Seed, Senior Lecturer in Medical Statistics,
> Division of Women's Health, King's College London
> Women's Health Academic Centre, King's Health Partners
> (+44) (0) 20 7188 3642.
>
>
>
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