#### Answer

$\mu_1-\mu_2$ is between -6.3462 and 8.189, yes.

#### Work Step by Step

Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1\ne\mu_2$. The degree of freedom: $min(n_1-1,n_2-1)=min(40-1,40-1)=39.$
The corresponding critical value using the table: $t_{\alpha/2}=t_{0.025}=2.023.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.023\sqrt{\frac{16.5201^2}{40}+\frac{15.5987^2}{40}}=7.2676.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(16.5201-15.5987)-7.2676=-6.3462 and$\overline{x_1}-\overline{x_2}+E$=(16.5201-15.5987)+7.2676=8.189. THis interval contains 0, hence there don't seem to be different sample means, hence the samples seem to originate from the same sample.