发布时间 : 星期一 文章统计学（第五版）贾俊平等著 - 课后习题答案更新完毕开始阅读9d45e67469eae009581bec78
???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 ）