t-test
I am trying to test if there has been a significant change in the weight of two groups, one without exercise and one with exercise.
The two samples have large spread, which creates a standard deviation that is larger than the mean for each group.
My question is, is it still valid to use the unpaired t-test to compare means when the standard deviation is larger than the mean.
If not, what statistical test can I perform?
Let's use $C$ for 'control' and $T$ for treatment (i.e. 'intervention')
It seems like you have the following situation.
There are four quantities you have measurements on:
1) Control, baseline - from which you can estimate $\mu_{C1}$
2) Control, endline - from which you can estimate $\mu_{C2}$
3) Intervention, baseline - from which you can estimate $\mu_{T1}$
4) Intervention, endline - from which you can estimate $\mu_{T2}$
Let $\delta_i=\mu_{i2}-\mu_{i1}$ be the after-before difference.
You're interested in testing against the alternative $\delta_T\neq\delta_C$ (or maybe $\delta_T>\delta_C$ in a one-tailed test).
This is straightforward; it may be done in a number of ways.
Since you say you'd normally use a t-test, you can do that here, but you'll need some assumptions; the suitability of those assumptions will be a question you'll need to consider carefully.
The numerator of your t-statistic will be
$\hat{\delta}_T-\hat{\delta}_C \,= \hat{\mu}_{T2}-\hat{\mu}_{T1}-(\hat{\mu}_{C2}-\hat{\mu}_{C1}) $
$\quad\quad\quad\quad= \bar{x}_{T2}-\bar{x}_{T1}-(\bar{x}_{C2}-\bar{x}_{C1}) $
The questions come down to whether you will assume independence of all four sets of measurements, and whether you will assume equality of variances (in the second case, if not, some Welch-type adjustment will be required).
a. If you assume independence and equal variances for all four measurements, your denominator would work similarly to the way it does in an ordinary two-sample t-test, so you'd get:
$$t = \frac{\bar{x}_{T2}-\bar{x}_{T1}-(\bar{x}_{C2}-\bar{x}_{C1})}{ s_{p} \cdot \sqrt{\frac{1}{n_{T1}}+\frac{1}{n_{T2}}+\frac{1}{n_{C1}}+\frac{1}{n_{C2}}}}$$
where
$$s_p^2= \frac{(n_{T1}-1)s_{{T1}}^2+(n_{T2}-1)s_{{T2}}^2+(n_{C1}-1)s_{{C1}}^2+(n_{C2}-1)s_{{C2}}^2}{n_{T1}+n_{T2}+n_{C1}+n_{C2}-4}$$
and you have $n_{T1}+n_{T2}+n_{C1}+n_{C2}-4$ d.f. for the $t$ distribution.
b. if they don't all necessarily have the same variance, the Welch-Satterthwaite approximation to the d.f. for the natural test statistic can be used:
$$t = {\overline{x}_{T2} - \overline{x}_{T1}-(\overline{x}_{C2} - \overline{x}_{C1})} \over s_{d}$$
$$s_{d} = \sqrt{{s_{T1}^2 \over n_{T1}} + {s_{T2}^2 \over n_{T2}}+{s_{C1}^2 \over n_{C1}} + {s_{C2}^2 \over n_{C2}}}$$
with d.f:
$$ \frac{(\frac{s_{T1}^2}{n_{T1}} + \frac{s_{T2}^2}{n_{T2}}+\frac{s_{C1}^2}{n_{C1}} + \frac{s_{C2}^2}{n_{C2}})^2}{\frac{(s_{T1}^2/n_{T1})^2}{n_{T1}-1} + \frac{(s_{T2}^2/n_{T2})^2}{n_{T2}-1}+\frac{(s_{C1}^2/n_{C1})^2}{n_{C1}-1} + \frac{(s_{C2}^2/n_{C2})^2}{n_{C2}-1}}$$
It might even make sense to assume that the two baseline measurements have equal variance and that the two endline measurements have equal variance, but that the baseline and endline might not have the same variance; that, too, could be done fairly readily (as could the assumption that the two control measurements had equal variance and the two intervention measurements had equal variance but control and treatment variances might differ).
Alternatively, the equality of that linear combination can be tested as a contrast in a one-way ANOVA. Many packages make this simple. Again, some packages will support the option of doing this with unequal variances.
Another approach is to do it as a two-way ANOVA type model; you can fit a model like:
response ~ group + time + group:time
where group is control or treatement, and time is the baseline/endline ('1' or '2'). In this case the group:time interaction is the thing you're interested in testing for.
You could also construct a permutation test of this hypothesis, or a variety of other tests can be similarly adapted in the way I did above.
If you had paired data (before and after), many people would suggest a different approach to the differences-in-differences style of test here (such as repeated measures, or at least putting the baseline in the model as a control variable), but I don't think those sort of approaches are an option in this situation.
That said, if your sample sizes are equal, you perhaps shouldn't fear pairing up your samples artificially; simulations suggest that actually works surprisingly well.
The tl;dr version: I think it would be better to do a two-sample Kolmogorov–Smirnov test on the original (not binned) data.
Extended answer:
A direct reply to your last question is: you can use an F-test to compare variances (the square of the standard deviations). Better versions of this test are the Levene's test or the Bartlett's test.
However, there are other approaches that may be more accurate. In particular, you didn't say anything about the Normality. And the t-test and the F-test assume Normality of the distributions. For such large samples, taking a look at the skewness and kurtosis, together with a Q-Q plot might help to assess Normality.
Nevertheless, comparing means and standard deviations do not guarantee that the distributions are similar -- you may have two distributions with the same mean and standard deviation that, e.g., have different skewness and/or kurtosis. So, to compare distributions, you can use the two-sample Kolmogorov–Smirnov test. Again, a graphical method is also useful to see if the differences (even if significant) are relevant. For instance, you may plot the CDF of both samples to see how large are the differences between them.
Regarding binning the data, I usually advise against it. I think you can compute the mean and standard deviation of $N$ for group A and for group B and these will be indicative of each sample characteristics.
By the way, if you had different type of data, where cells would either be on or off, then the more appropriate test for this dichotomous case would be Fisher's exact test.
Best Answer
If you can, you should show us the plots of the two distributions of people.
However, for weight loss among people who span a large range of weights, it may be preferable to use log scale regardless of whether the distributions are skewed. That is because weight loss is (I think) better thought of on a ratio basis than an additive basis. For a 100 pound person to lose 10 pounds is not the same as for a 300 pound person to lose 10 pounds.
There are also questions of what covariates you should include, whether the people are randomized and so on.