1?18?2L11??xi1???xi1??4321.02?2568.05?1752.9718?i?1?i?11821?18?2 L22??xi2???xi2??35076?31920.22?3155.78
18?i?1?i?11821?18?2L33??xi3???xi3??307894?272322?3557218?i?1?i?1182
?xxi?118i1i2?10139.5
1?18??18? L12?L21??xi1xi2???xi1???xi2??10139.5?9053.88?1085.62
18?i?1??i?1?i?118?xxi?118i1i3?96598
1?18??18?L?L31??xi1xi3???xi1???xi3??27645?26445?1200 13 18?i?1??i?1?i?11818
?xi?1i2i3x?96598
181?18??18? L32?L23??xi2xi3???xi2???xi3??96598?93234?3364
18?i?1??i?1?i?1
?xy?20706.2
i1ii?1181?18??18? L1y??xi1yi???xi1???yi??20706.2?17474.7?3231.5
18?i?1??i?1?i?118
?xi?118i2iy?63825
1?18??18? L2y??xi2yi???xi2???yi??63825?61608.5?2216.5
18?i?1??i?1?i?118
?xi?118i3iy?187542
1?18??18?L?xi3yi???xi3???yi??187542?179949?7593 3y? 18?i?1??i?1?i?118 30
于是
??y?4.582
?1200???1752.93 1085.62?? L?1085.62 3155.78 3364 ???35572??1200 3364 ?
?L1y??3231.5??????L2y???2216.5? ?L??7593???3y??可得
?????1??3231.5?????1?? ??2??L2216.5
??????7593?????3???所以 第三章
y?43.65?1.78x1?0.08x2?0.16x3
?1.解: 假设: H0:??26,H1:??26 由于??5.2已知,故用统计量u?x????~N(0,1)
n???x??27.56?26??1.2 P?u?u???? u的拒绝域u?u? u??5.222??4n因显著水平??0.05,则u?1.2?u??u0.025?1.96
2这时,就接受H0 2. 解: (1) ?已知,故u?x????0n??~N(0,1) P?u?u???? u的拒绝域u?u?
22??u?x????0?n25.32?5?3.2因显著水平??0.01,则 110u?3.2?u??u0.005?2.576 故此时拒绝H0:u?5
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(2) 检验u?4.8时犯第二类错误的概率?
?????0??12??0?0??1?u??0n22nn???????e0?????x??????2?2?0n2dx 令t??x????0则上式变为
2nn1??2??0??0??1?u?21edt?2?t22?4.58??0.58edtt22n??(4.58)??(?0.58)??(4.58)??(0.58)?1?0.9999979?0.71990?1?0.7180
3. 解:假设H0:??3.25,H1:??3.25
?x??用t检验法拒绝域T?*?t?(n?1) ??0.01, x?3.252
s2n?查表t0.0112(14)?4.6041 s*2?0.00017,s?0.0130 代入计算T?0.344?t0.0112(14) 故接受H0,认为矿砂的镍含量为3.25
4解:改变加工工艺后电器元件的电阻构成一个母体,则在此母体上作假设
H0:??2.64,用大子样检验 x??0u?snx??0s?n??~N(0,1) 拒绝域为u?u? 由n?200,x?2.62,s?0.06,??0.01
2?查表得u??2.575 u?20.02?3.33?2.575?u? 0.06210故新加工工艺对元件电阻有显著影响.
x??0近似5 .解：用大子样作检验，假设H0:???0 u?~N(0,1) sn? 32
拒绝域为u?u?由n?200,?0?0.973,x?0.994,s?0.162,??0.05,u0.025?1.96
2?x??00.021??1.833?1.96故接收H0，认为新工艺与旧工艺无显著差异。 s0.162n200?6.解:由题意知，母体X的分布为二点分布B(1,p)，作假设H0:p?p0(p0?0.17) 此时x??m(m为n个产品中废品数) n?因n?400很大，故由中心极限定理知x近似服从正态分布。 故u?mm?p0?p0nn~N(0,1)即P{?u?}??
p0(1?p0)p0(1?p0)2nnm?p0?u?n22计算得拒绝域为
p0(1?p0) n把m?56,n?400,u??u0.025?1.96,p0?0.17代入
m?p0?0.14?0.17?0.03?1.96?0.0188?0.037 n即接受H0，认为新工艺不显著影响产品质量。
7解：金属棒长度服从正态分布原假设H0:???0?10.5，备择假设H1:???0
x??t?x???s~t(n?1) 拒绝域为t?t? n21(10.4?10.6???10.7)?10.48 15样本均方差s?x??0s?n?1(10.4?10.48)2???(10.7?10.48)2?0.237 140.02?0.327而t0.025(14)?2.144 因0.327?2.144
0.23715于是t?故接受H0，认为该机工作正常。
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8．解：原假设H0:???0?12100，备择假设H1:???0
???x??0sn?n?,s?323,??0.05代入计算 ~t(n?1)，拒绝域为T?t? 将x?119582?x??0s142?2.153?t0.025(13)?2.068 故拒绝原假设即认为期望。
323249． 假设H0:????20.8,H1:???0?20.8 使用新安眠药睡眠平均时间
1s2?[(26.7?24.2)2???(23.4?24.2)2]16x?(26.7?22.0???23.4)?24.2
7s?s2?2.296?x??0t?s?n?3.42.296?4.046 所以拒绝域为t?t0.05(n?1) 7查表t0.05(6)?1.943?4.046?t 故否定H0
又因为x?24.2?20.8?3 故认为新安眠药已达到新疗效。 10． 原假设H0:?甲??乙，H1:?甲??乙u?解得拒绝域u?u?
2?x甲?x乙ss?n1n22122??近似~N(0,1)
x1?2805,x2?2680x1?x2125??8.03 s1?120.41,s2?105.00代入计算2222s1s2120.41105n1?140,n2?100??110100n1n2????查表u??u0.025?1.96 因8.03?1.96
2故拒绝原假设即两种枪弹速度有显著差异。
211．解：因两种作物产量分别服从正态分布且?12??2
假设H0:?1??2,H1:?1??2 故统计量T?X?Y~t(n1?n2?2) 11Sw?n1n2?? 34