Answer
$\mu_1-\mu_2$ is between 0.0291 and 0.0349, yes.
Work Step by Step
Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1\ne\mu_2$. The degree of freedom: $min(n_1-1,n_2-1)=min(36-1,36-1)=35.$
The corresponding critical value using the table: $t_{\alpha/2}=t_{0.025}=2.03.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.03\sqrt{\frac{0.0075^2}{36}+\frac{0.0044^2}{36}}=0.0029.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(0.8168-0.7848)-0.0029=0.0291 and$\overline{x_1}-\overline{x_2}+E$=(0.8168-0.7848)+0.0029=0.0349. This interval doesn't contain 0, hence there seem to be different sample means, hence the samples seem to originate from not the same sample.
