This is the Case 1.3'. The two samples are:

$n_{1}=8,\bar{x}=16.5,s_{x}=2.8;\hskip 56.9055ptn_{2}=10,\bar{y}=14.2,s_{y}=2.2.$

The one-sided CI is:

$I_{\mu_{1}-\mu_{2}}$ $=((\bar{x}-\bar{y})-t_{\alpha}(n_{1}+n_{2}-2)\cdot s\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}},~{}~{}+\infty)$
$=((16.5-14.2)-t_{0.05}(8+10-2)\cdot s\cdot\sqrt{\frac{1}{8}+\frac{1}{10}},~{}~{}+\infty)$
$=(2.3-1.75\cdot 2.48\cdot 0.4743,~{}~{}+\infty)=(0.242,~{}~{}+\infty),$

where $s^{2}=\frac{(n_{1}-1)s_{x}^{2}+(n_{2}-1)s_{y}^{2}}{n_{1}+n_{2}-2}=6.1525.$ As $0.242>0$ , it follows that $\mu_{1}-\mu_{2}>0$ , that is $\mu_{1}>\mu_{2}.$ Yes, the samples do provide evidence that the perch in Roxen are bigger than the perch in Glan. ∎