设男、女大学生月饮食费支出（单位：元）分别服从(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)，为研究男、女大学生在月饮食费支出上的差异，在某大学随机抽取(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)名 男生和(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)名女生，得到以下结果：男生：(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)；女生：(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)。试求均值差(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)的置信度为(mu (1)_(1),({sigma )_(1)}^2), ((mu )_(2),({sigma )_(2)}^2)的置信区间。

步骤 1：确定样本信息
根据题目所给信息，已知${v}_{1}={n}_{2}=30$，${c}_{1}=520$ ${{S}_{1}}^{2}=200$ ${S}_{1}=\sqrt {200}$，${c}_{2}=480$ ${{S}_{2}}^{2}=280$ ${S}_{2}=\sqrt {280}$。$I'O=r$，因为本题总体方差未知，样本方差已知，故可以用样本方差代替总体方差。

步骤 2：确定置信区间边界公式
置信区间边界公式用$(\overline {X}-{t}_{\dfrac {\alpha }{2}}(n-1)\dfrac {S}{\sqrt {n}}$ $\overrightarrow {X}+{t}_{\dfrac {a}{2}}(n-1)\dfrac {S}{\sqrt {n}})$，其中$\overline {X}$是样本均值，$S$是样本标准差，$n$是样本大小，${t}_{\dfrac {\alpha }{2}}(n-1)$是自由度为$n-1$的t分布的临界值。

步骤 3：计算置信区间
查表可知$\dfrac {a}{2}(n-1)={t}_{0.05}(29)$，计算可得：
${c}_{1}-{t}_{\dfrac {\alpha }{2}}({n}_{1}-1)\dfrac {{S}_{1}}{\sqrt {{n}_{1}}}$$=520-{t}_{0.05}(29$ $\dfrac {\sqrt {200}}{\sqrt {30}}=514.7198$
${c}_{2}-t\dfrac {\alpha }{2}({n}_{2}-1)\dfrac {{S}_{2}}{\sqrt {{n}_{2}}}$$=480-{t}_{0.05}(29$ $\dfrac {\sqrt {280}}{\sqrt {30}}=473.7524$
${c}_{1}+{t}_{\dfrac {\alpha }{2}}({n}_{1}-1)\dfrac {{S}_{1}}{\sqrt {{n}_{1}}}$$=520+{t}_{0.05}(29$ $\dfrac {\sqrt {200}}{\sqrt {30}}=525.2802$
${x}_{2}+t\dfrac {\alpha }{2}({n}_{2}-1)\dfrac {{S}_{2}}{\sqrt {{n}_{2}}}$$=480+{t}_{0.05}(29$ $\dfrac {\sqrt {280}}{\sqrt {30}}=486.2476$
根据计算结果可得出均值声的置信度为90%的置信区间为：514.7198,525.2802)；均值12的置信度为90%的置信区间为：473.7524,486.2476)。

步骤 4：计算均值差的置信区间
均值差${L}_{1}-{\mu }_{2}$的置信度为90%的置信区间为：39.0326,40.9674)。