There are 2 groups: one has an average of 305.15 and standard deviation of 241.83 while the second group has an average of 198.1 and a standard deviation of 98.1. Given the large standard deviation of the first group, its mean should not be significantly different from the second group. But when I conducted the independent sample t-test, it was which doesn't make sense. Is there any another test I can conduct to analyze the date (quantitative)?
The data is about solid waste generation on a monthly basis (averages). And I am comparing March and April data with that of other months. Also, the sample sizes are not equal i.e. less days in Feb as compared December for example.
It is difficult to give a comprehensive answer without more details on 1) the data, 2) the experimental design, 3) your question and 4) what exactly you did to perform the Student's T-test.
Note however that usual Student's T-test assumes equal theoretical variances between the two populations, which is probably wrong considering your two values. You should use corrected T-tests for this case, like the Aspin-Welch test (default in R / t.test).
Note also that the T-test compares the means, that can be very precise even if the sample SD is very high if your sample size is big enough, so you cannot conclude just « by eye » based on means and sample SD. You should look at sample SEM to do that and better understand what's going on.
Never forget that T-test assumes Gaussian distributions, which may well be wrong also in your samples, and this all the more matter, than sample sizes are small and different (just like the hypothesis of variance equality is more important for unequal sized samples).
Last, don't confuse « statically significant » and « of practical interest », results of your test should be interpreted along with effect size / confidence intervals for the difference and so on.
Technically speaking, the t-Test operates through the standard error, not the standard deviation. Check Levene's test to see if you have violated the assumption of equal standard errors, and if so, use a corrected version of the t-Test.
First, you say you are using means for the months. Why? Given the difference in SDs I am curious if there are any really high measurements in that month (March?). Are you using the individual measurements, or are you using means? And the two most important questions:
1. You say you want to compare these months of many. Why? If you are choosing these because they have a big difference, that is not how this works.
2. You say you are comparing across months. I assume the sites from where the data are drawn are matched (or can be based on spatial location, or were they all at the same site, at different times?) and that you want to model any changes across several months. Time series methods are usually used for this. If there is something that happened on March 31st, you can incorporate this into the models.
But, as Emmanuel Curis and Bruce Weaver say, more information is needed.
Just on your specific question, Student's version of the independent group t-test estimates the pooled standard deviation as a weighted average of the group standard deviations, but if you want to be more conservative you could use the larger SD in those calculations. Welch's version is often used when you do not want to make the assumption that the groups, in their respective populations, have the same spread. And there are lots of robust alternatives (e.g., use least absolute value rather than least squares, but we would need to know more about the data to speculate on this).
BUT, you should also be interested in changes in SD.
The prerequisite for a t-test is homoscedasticity and that also means an (almost) equal standard deviation. You should use the Welsh test instead.
"Given the large standard deviation of the first group, its mean should not be significantly different from the second group" . I don't see why this should be true!
What strikes me is that the SD is larger in the group with larger mean. Is there a correlation between mean and variance? This should probably be addressed in an analysis.
Further "about solid waste generation" should not be a normal distributed variable: it cannot be negative. This may be negligible when the mean ≫ SD, but here mean ≈ SD, so the distributional assumption as such is likely not adequate/reasonable.
If you still consider the normal distribution as a reasonable model, Claudia Paladini is right: you should use Welsh's test instead.
Yes, the independent samples t-test takes into account the standard deviation of the two groups being compared. The t-test is a statistical hypothesis test used to determine whether there is a significant difference between the means of two independent groups. It considers both the means and the variability (standard deviation) within the groups to make this determination. The formula for the t-test involves calculating the difference between the means of the two groups and then normalizing it by dividing by the standard error of the difference. The standard error of the difference takes into account the standard deviations of both groups and the sample sizes. In other words, the t-test for both the mean difference and the variability within the groups, which is represented by their standard deviations. This helps to assess whether the observed mean difference between the groups is statistically significant or if it could have occurred due to random chance.
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