Both cohen's D and hedges' correction are negative . Does it means that after intervention there was negative effect in the population
No. The t test tells you if there is a difference between treatments according to the the p value. You use the mean values to determine if the effect by intervention was negative or positive. Those corrections merely indicate the size of the difference :)
Andrew Paul McKenzie Pegman , no, please look up the definition of g: it is the mean difference, divided by the standard deviation (via the pooled variance estimate). Hedge's modification is only concerning the standard deviation in cases where the sample sized are different. Thus, for both statistics the sign is that of the mean difference.
And regarding your statement "Those corrections merely indicate the size of the difference :)": I don't subscribe her either. The size, directly and simply, is the mean difference. d (and g) give standardized sizes, a rather convoluted measure.
Sudeepa Kumari , you cannot know what the effect is in the population. That's why you do inference. The sample will either have a positive of a negative mean difference. You see this by comparing (i.e. subtracting) the sample means directly, or via d or g (as you did). This may or may not be informative for the population. The larger the sample, the more informative is it (given it's a random sample). The p-value from a test about the hypothesis that the mean difference is zero (usually a t-test) will tell you if the sample size may (or may not) be considered large enough to conclude that the sign of your sample mean difference is the same as the population mean difference (aka the expected difference).
The term "Hedges' correction" may be misleading, since the literature is ambiguous. Some refer to the use of the pooled standard deviation as Hedges' correction, other mean the small sample correction formula (where I would use it for the latter one). Of course, the small sample correction is only a factor to shrink the d-value and cannot change the sign.
Therefore, Andrew, your answer "no" is not correct, since it does not matter what Hedges' correction you meant, the resulting standardized effect sizes will have the same sign as the original mean difference and hence, it of course implies that the effect is negative.
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