5.解 ?22?0.108,n?9,x?4.484,1???0.95,即??0.05，
查标准正态分布表可得u??u0.025?1.96. 从而
2x??nu??4.484?20.10830.1083?1.96?4.41
x??nu??4.484?2?1.96?4.55，
故该厂铁水的平均含碳量的置信度为0.95的置信区间为（4.41，4.55）.
6.解 n1?n2?20,x?y??6,1???0.99,即??0.01， 查标准正态分布表可得u??u0.005?2.57. 从而
2x?y?u?x?y?u??122n1???2n22??6?2.57???6?2.57?0.05202??0.05202??6.04 ??5.96，
?122?2n22n10.052020.05202故?1??2的置信度为0.99的置信区间为（-6.04，-5.96）.
227.解 s1?0.5419,s2?0.6065,n1?n2?10,1???0.95，即??0.05，查表得
F?(n1?1,n2?1)?F0.025(9,9)?4.03
2F1??(n1?1,n2?1)?1/F?(n2?1,n1?1)?1/F0.025(9,9)?1/4.03
22从而
s1s2222F?(n1?1,n2?1)?0.222，
s1s2222F1??(n1?1,n2?1)?3.601
故方差比的置信度为0.95的置信区间为（0.222，3.601）.
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