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Step 1SolutionSolution: Given data is Sandbe- Mean differane =\bar{d}=-1.06\begin{array}{r}S_{d}=1.18 \\r=16\end{array}Step 2Hypotheris:\begin{array}{l}H_{0}: H_{d}=0 \\H_{1}: \text { ud }_{d}<0\end{array}Test statistics.\begin{array}{c}t=\frac{\sqrt{d}}{S d / \sqrt{h}}=\frac{-1.06}{1.181 \sqrt{16}} \\t=-3.593\end{array}t=-3.593\begin{array}{c}d f=n-15 \\\text { Caitical value }=-t_{\alpha}(d t)=-t 0.05 \\\text { b-value }=P(t<-3.593)=0.0013\end{array}Since p-uclue \alpha \alpha \quad(\alpha=0.9) So, Reject HoConrlesion: Rgect H_{0}, data provide Sefticicut cvidence to conclude that anxicty levels decreased we the courre of semestas. ...