A new medicine was invented to threat a serious disease. An experiment on $5000$ women and $4000$ men showed that the mean of effect is $10.2$ in women group and $10$ in men group. If samples were drawn down from the normal population with variance $3$, test on $10$% level significance a hypothesis that the effect for women is significantly more than for men.
We assume that samples are independent. Denote $\xi_1 \sim N(a_1,3)$ - women results, $\xi_2 \sim N(a_1,3)$- men results. I used t-test with $$ H_0: a_1=a_2,$$ $$ H_1: a_1> a_2.$$
Test statistic: $$t= \frac{10.2-10}{\sqrt{3}\sqrt{\frac{1}{5000}+\frac{1}{4000}}} \approx 5.44$$
As samples drawn from normal distribution, $t \sim N(0,1),$ then $$A_{crit} = [z_{1-\alpha},+\infty) \approx [1.282, +\infty)$$
So we reject a null-hypothesis in favour to alternative. Is this a correct solution?