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Solution:-Given;\begin{array}{ll}n_{1}=32, & n_{2}=25 \\\bar{x}_{1}=103.5, & x_{2}=114.2 \\s_{1}=12.3 & s_{2}=10.3\end{array}Test \mu_{1} \mathrm{cu}_{2}Hypothesis are \alpha=0.05\begin{array}{l}H_{0}:-\mu_{1}=\mu_{2} \\H_{1}:-\mu_{1}<\mu_{2}\end{array}Test statistics are,\begin{array}{l}t c a=\frac{\bar{x}_{1}-\bar{x}_{2}}{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)} \\=32-25103 \cdot 5-114.2 \\\frac{(32-1) 12 \cdot 3^{2}+(25-1) 13 \cdot 3^{2}}{32+25-2} \\\end{array}\begin{array}{c}t c a l=-3.144973 \\t c a l=-3.145\end{array}\begin{aligned}P \text {-value } & =0.00133932-\because(P(x \leq T)=0.00133932) \\= & =0.001 \\P \text {.value } & =0.001\end{aligned}Here P-valuz <\alpha then we reject H 0 If p value c \alpha then we reject H 0 Here 0.001<0.05 then we reject to.D] Reject Ho. There is sufficient evidence at \alpha=0.05 level of significance to conclude that \mu_{11}-u_{2} ...