Can you find the variability accounted for using only the mean and s.d I am comparing the variability accounted for from two different studies. One gives me F(obs) and the two df's right off the bat so I can calculate that easily. The other, however, only gives me the M and s.d. of the treatments. Can I use those somehow to find the explained variability? It gives me no other statistics, even a t would be nice. 
 A: 1. Test of equal means: $\mu_1=\mu_2$
Assuming the two treatment's mean are normally distributed, you can correspondingly run a two sample $t$-test:
$$
t=\frac{\bar{X}-\bar{Y}}{\sqrt{\hat{\sigma}_X^2/n_x+\hat{\sigma}_Y^2/n_y}}
$$
here $\bar{X},\;\bar{Y}$ would be your two treatments' mean, while $\hat{\sigma}^2_X,\;\hat{\sigma}^2_Y$ would be the estimated variances from your sample. $n_x,\;n_y$ would be your sample size.
You can also get a two-sided $p$-value from this $t$, 
$$
p=2\Phi(-|t|).
$$
2. Test of equal variances: $\sigma_X^2=\sigma_Y^2$
Your degree of freedom is for these two treatments are just respectively $n_x-1$ and $n_y-1$, assuming the samples are i.i.d.
You can use the given mean and sds to estimate the sample means' variance ,just like in the aforementioned $t$-test:
$$
F=\frac{\sum_{i=1}^{n_x}(X_i-\bar{X})^2/(n_x-1)}{\sum_{j=1}^{n_y}(Y_j-\bar{Y})^2/(n_y-1)}
$$
this $F$-statistic has an $F$-distribution with $n_x − 1$ and $n_y − 1$ degrees of freedom if the null hypothesis of equality of variances is true.
