Hypothesis-test; test about equal sample means I'm asked to formulate and do a test of the hypothesis that the sample mean (average grade) of math students and economic students are equal. Data below:

I'm not sure how to "attack" this problem. I already did a exercise where I was asked to test the hypothesis that the variance of male and female math students were the same. 
But now, when the math-students are devided in two "subgroups"; male-students and female-students/, and the economic-students consists of only one subgroup (both males and females), I'm not sure how to begin. 
Can I somehow add the two sample means and use this "pooled" sample mean as a representative sample mean for math students and then just do a regular t-test, or how should I move forward? Any help will be much appreciated.
 A: Hint: Suppose $n = n_x + n_y$ is the number of economics students, $n_x$ being the number of male math students, $n_y$ being the number of math economics students.
Then the aggregate sample mean is$$\begin{align}\bar{X} = \dfrac{\sum\limits_{i=1}^{110}x_i+\sum\limits_{i=1}^{38}y_i}{n} &= \dfrac{\sum\limits_{i=1}^{110}x_i}{n}+\dfrac{\sum\limits_{i=1}^{38}y_i}{n} \\&= \dfrac{\sum\limits_{i=1}^{110}x_i}{n_x}\left(\dfrac{n_x}{n}\right) + \dfrac{\sum\limits_{i=1}^{38}y_i}{n_y}\left(\dfrac{n_y}{n}\right) \\&= \bar{X}_x\left(\dfrac{n_x}{n}\right)+\bar{X}_y\left(\dfrac{n_y}{n}\right)\text{.}\end{align}$$
The aggregate sample variance $S^2$ is
$$\begin{align}S^2 &= \dfrac{\sum\limits_{i=1}^{110}\left(x_i - \bar{X}\right)+\sum\limits_{i=1}^{38}\left(y_i - \bar{Y}\right)}{n-1} \\&= \dfrac{\sum\limits_{i=1}^{110}\left(x_i - \bar{X}_x\right)}{n-1} + \dfrac{\sum\limits_{i=1}^{38}\left(y_i - \bar{X}_y\right)}{n-1} \\
&= \dfrac{\sum\limits_{i=1}^{110}\left(x_i - \bar{X}_x\right)}{n_x-1}\left(\dfrac{n_x - 1}{n-1}\right) + \dfrac{\sum\limits_{i=1}^{38}\left(y_i - \bar{X}_y\right)}{n-1}\left(\dfrac{n_y - 1}{n-1}\right) \\
&= S_x^2\left(\dfrac{n_x - 1}{n-1}\right) + S_y^2\left(\dfrac{n_y - 1}{n-1}\right)\text{.}\end{align}$$
