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st: comparing different means using ttest


From   David Lempert <david.lempert@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   st: comparing different means using ttest
Date   Thu, 16 Dec 2010 14:54:19 +0100

I apologize beforehand if this question is a bit trivial, but I have
run into a problem and I can't seem to find the solution in any
previous statalist correspondence I have come across.

I am currently using a two-sample t-test with equal variances to
compare the % GDP-growth between 2 periods and I am trying to figure
out if my method has been correct. The following is part of the code I've
used:

reg gdp_ year
predict res, r

sum res (I check and see if the mean is close to zero, which it is -
hence homoscedastic as I see it)

pnorm res (I see if the residuals are normally distributed, which it
looks to me as if they are - they are all relatively close to the
trend-line)

The gdp data I have is quarterly, due to the fact that I am only
investigating 8 years and need it to be normally distributed. The
pnorm test seems to confirm that this is the case. I after this
generate a code for the percentual gdp-growth for period 1 and for
period 2 - the main thing I want to examine.

gen percdiff1 = (gdp_[_n] - gdp_[_n-4]) / gdp_[_n-4] if year<=20034
gen percdiff2 = (gdp_[_n] - gdp_[_n-4]) / gdp_[_n-4] if year>=20041

sum percdiff1
sum percdiff2
sdtest.... (this is to check and see that they have equal variances,
degrees of freedom is 15, 15 and f-value is 1.2392 which is smaller
than all critical f-values in any table I have access to, so I can not
reject the null hypothesis that the ratio of their standard deviations
is 1, hence equal variances)

ttest percdiff1=percdiff2, unpaired

I get the following printout:

------------------------------------------------------------------------------
        |    Obs    Mean    Std. Err.   Std. Dev.   [95% Conf. Interval]
---------+--------------------------------------------------------------------
percdi~1 |    16   .0714719    .004791    .019164    .0612601    .0816837
percdi~2 |    16   .1033814   .0043038   .0172151    .0942081    .1125547
---------+--------------------------------------------------------------------
combined |    32   .0874267   .0042715   .0241634    .0787148    .0961385
---------+--------------------------------------------------------------------
   diff |        -.0319094   .0064402              -.0450621   -.0187568
------------------------------------------------------------------------------
   diff = mean(percdiff1) - mean(percdiff2)                 t =  -4.9547
Ho: diff = 0                                degrees of freedom =       30

    Ha: diff < 0               Ha: diff != 0              Ha: diff > 0
  Pr(T<t) = 0.0000         Pr(|T|>|t|) =  0.0000       Pr(T>t) = 1.0000

And, I can reject the null-hypothesis that the means for each period
is equal on an alfa=0.05 significance level.

My first question is: Have I done this correctly? This is for my
thesis paper and I will have to defend this in the beginning of
January.
My second question is: Can I say that the difference is negative from
these numbers? I ask because I am trying to prove that the mean for
percdiff2 is greater than the mean for percdiff1.

Thanks a lot,

David
Student at Stockholm University

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