统计学(第五版)贾俊平等著 - 课后习题答案

???Z????1?????1.962?0.5?0.52????150.0625 所以n为151 (2)n?E20.042???Z????1?????1.6452?0.55?0.452????267.89 所以n为268 (3)n?22E0.057.18(1)p?2232?0.64 502???Z????1?????1.962?0.8?0.22????61.46 所以n为62 (2)n?22E0.127.19(1)???n?1???02.05?50?1??66.339

22?2??n?1???0.95?50?1??33.930 1?2?n?1?s2??2??n?1?s2

2???2?1?22所以

?n?1?s??2??2??n?1?s??2?1?24949?2????2?1.72???2.40

66.33933.932?2??n?1???015?1??6.5706 .95?1?2(2)???n?1???0.05?15?1??23.6848

222?n?1?s??2??2??n?1?s??2?1?21414?0.02???706?0.02?0.015???0.04323.68486.52 ?2??n?1???0.95?22?1??11.59131?2 (3)???n?1???0.05?22?1??32.6706

222?n?1?s??2??2??n?1?s??2?1?22121?31????31?24.85???41.72536.70611.5913 7.20(1)x?1xi?7.15 s??n12??x?x?0.4767 ?in?11?2 ?n?1???02.025?10?1??19.0228 ?2??n?1???02.975?10?1??2.7004??22?n?1?s??2??2??n?1?s??2?1?299?0.4767????0.4767?0.328???0.87

19.02282.7004(2)

?n?1?s??2??2??n?1?s??2?1?299?1.822????1.822?1.253???3.326

19.02282.70042(n1?1)s12?(n2?1)s213?96.8?6?1027.21 s?=?98.442

n1?n2?2192p(1)?1??2的90%置信区间为: (x1?x2)?t?2(n1?n2?2)sp1111?? =9.8?1.729?98.442?147n1n2 =9.8?7.9411 (2)?1??2的95%置信区间为: (x1?x2)?t?2(n1?n2?2)sp1111?? =9.8?2.093?98.442?147n1n2 =9.8?9.613 (3)?1??2的99%置信区间为: 9.8?2.8609?98.442?11?=9.8?13.140 1477.22(1)(x1?x2)?z?22s12s2?=2?1.96?0.36=2?1.176 n1n212(n1?1)s12?(n2?1)s29?16?9?20(2)s?==18

n1?n2?2182p(x1?x2)?t?2(n1?n2?2)sp111?=2?2.1?18?=2?3.98

5n1n22s12s2(?)2n1n21(3)??2=17.78 222(s1n1)(sn)?22n1?1n2?12s12s2(x1?x2)?t?2(?)?=2?2.1?3.6=2?3.98

n1n21(4)t0.025(28)?2.048

2(n1?1)s12?(n2?1)s2s?=18.714

n1?n2?22p(x1?x2)?t?2(n1?n2?2)sp1111?=2?2.048?18.714??

1020n1n2 =2?3.43

2s12s21620(?)2(?)2n1n211020=20.05 (5)??2222?(s1n1)(s2n2)1.6212??n1?1n2?1919t?2(?)?2.086

2s12s2? (x1?x2)?t?2(?)=2?2.086?1.6?1=2?3.364 n1n2177.23(1)d? sd?4(2)d?t?2(n?1)?(di?d)2n?1=

332=6.917 48sd7=?4.185 n47.24 t?2(n?1)?2.6216 d?11,sd?6.53197 ?d的置信区间为:

d?t?2(n?1)sd6.53197=11?2.6216?=11?5.4152

10np1(1?p1)p2(1?p2)? n1n27.25(1)(p1?p2)?z?2=0.1?1.645?0.4?0.60.3?0.76?=0.1?0.0698

250250p1(1?p1)p2(1?p2)? n1n2(2)(p1?p2)?z?2=0.1?1.96?0.4?0.60.3?0.76?=0.1?0.0831

2502507.26 s1?0.241609 s2?0.076457

F?2(n1?1,n2?1)=F0.025(20,20)=2.464

F0.975(20,20)=0.40576

22s12s2?12s12s2?2? F?2?2F1??29.986?129.986?2? 2.446?20.40576?124.0528?2?24.611

?2(z?2)2?(1??)1.962?0.02?0.987.27 n?==47.06 22E0.04所以 n =48

(z?2)2?21.962?12027.28n?==138.30 22E20所以 n =139

第8章 假设检验

二、练习题

(说明:为了便于查找书后正态分布表,本答案中,正态分布的分位点均采用了下侧分位点。其他分位点也可。为了便于查找书后t分布表方便,本答案中,正态分布的分位点均采用了上侧分位点。)

8.1解:根据题意,这是双侧检验问题。

H0:??4.55H1:??4.55

已知:总体方差? ?0?4.55, z?2?0.1052 x?4.484,n?9,显著水平??0.05

x??0?/n?4.484?4.550.108/9??1.8333

当??0.05,查表得z1??/2?1.96。 拒绝域W={z?z1??/2}

因为z?z1??/2,所以不能拒绝H0,认为现在生产的铁水平均含碳量为4.55。

(注:z1??/2为正态分布的1-α/2下侧分位点 )

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