B26002500240023002200210020001900180015202530354045505560品质指标支数4.解：将
22x?35.353 y?2211. 2 xy?76061.6 7 6mx 0 my?34527. 46?132.13代入得
????xy?xy76061.676?35.353?2211.2???15.982mx132.130???y??x?2211.2?15.98?35.353?2776.142??m??mx?34527.46???15.98??132.130?786.692y2?2?2
??*2为?的无偏估计量
?*2?2n?220??786.69?874.10 ??n?2185. 解：将 代入得
22x?6 y?210. 4 xy?1558 mx?8 my?10929. 84????xy?xy1558?6?210.4??36.952mx8???y??x?210.4?36.95?6??11.3n?25???10929.84?36.952?8??12.37 ??n?23?*2??3.517假设
?*H0:??38 H1:??38
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用T检验法 拒绝域为
???0?*???x?x?ii?1n2?t?2?n?2?
查表得
t0.025?3??3.1824
将上面的数据代入得
t?1.89?t0.025?3?
所以 接受H0 即认为?为38
6. 解：（1）由散点图看，x的回归函数具有线性函数形式，认为长度对于质量的回归是线性的。
B121110长度98751015202530质量
（2）将
22x?17. 5 y?9.4 9 xy?179.3 7 mx my?2.45 ?72.9 2?代入得
???xy?xy179.37?17.5?9.49??0.182 2mx72.92?
??y??x?9.49?0.182?17.5?6.305
y????x?6.305?0.182x
???（3）当x?16时 由T分布定义
y0?a?16b??0
??T?Y0????x0??*11??n?x?x???x?x?20nii?1?t?n?2?
2 26
??????????Y0????x0?? P??tn?2????0.95 0.0252???x0?x1*??1??n?2n??x?x?i??i?1??????所以Y0的预测区间为
???1??*??x?tn?2?1???00.025??n??x?x?,???x?t??x?x?20??n20ii?10.025?n?2???*?x0?x?1? 1??n2nxi?x???i?1???2??查表得 将（2）的数据代入得
t0.025?4??2.776
?*2n?262?????2.4?50.1?82n?24
7?2.?92
0.0075??0.0866计算得Y0的预测区间为
?*?8.9521,9.4? 7212x?21 y?141. 2 xy?3138 mx?90
9. 解：利用第八题得到的公式 将 代入得
??
??xy?xy3138?21?141.2??1.922mx90
???y??x?141.2?1.92?21?100.8810.。解：二元线性回归模型为Yi离差平方和为
??1xi1??2xi2??i,i?1,2,???,n
nQ???yi??ixi1??2xi2?
2i?1对Q求?1,?2的偏导数并令其为0
?n?yi??1xi1??2xi2?xi1?0???i?1 ?
n??y??x??x?x?0?i1i12i2i2??i?1可变换为
27
?n? ??xyn2?ni1i??1?xi1?2?xi1xi2?0?i?1i?1i?1?n???yixi2??nn
21?xi1xi2??2?xi2?0i?1i?1i?1
正规方程为
? ??x2??1?1?x1x2?2?x1y
?x???21x2?1?x2?2?x2y最小二乘估计为
?
?x21?2yx1x2?x1yx2x222
1x2?x1x2??x22?1yx1x2?x2yx1x221x2?x21x2xn其中
1n11n1n1y?n?xi1yi x2y?n?xi2yi x1x2??xix1i 2 x2j?i?1i?1ni?1n?x2iji?111解：（1）
p?2 n?15
采用线性回归模型
Y????1?x1?x???2?x2?x???
15
?yi?248.25 y?16.55
i?1151515
?y22i?4148.3125
i1?56734
i?1?xi1?920
i?1?xi?115
x1?61.33
?xi2?7257 x2?483. 8i?11515
?x2i2?3524489
?445366
i?1?xi1xi2i?11515
?xi1yi?15170 i?1?xi2yi?12063925
i?1
28
j?1,2
1?15?2???xi?1?56734?56426.66?307.34 L11??xi115?i?1?i?11521?15?2???xi? L22??xi22?3524489?3510936.6?13552.4 15i?1?i?1?1521?15??15 L12?L2?xi??1?xi?1?xixi1?2?15?i?1??i?1i?11515??6445?096 270??244536?1?15??15? L1y??xi1yi???xi10152?2?6 56???yi??1517?15i?1?i?1??i?1?1?15??15? L2y??xi2yi???xi2 5yi??120639.?251201?03.2???15?i?1??i?1?i?115536于是
??y?16.55
?307.34270? L????27013552.4???L1y???56? ?????
L536???2y???????1?1??56?可得 ????L? ????536???2?所以 12.解
y?10.504?0.216x1?0.04x2 p?3 n?18
采用线性回归模型
18Y????1x1?x??2x2?x??3x3?x??
18i1??????
?x?y?1463 y?81.27 7ii?1?215
i?1
x1?11.944
?xi?118i2?758 x2?42.11
?xi?118i?118i3?2214 x3?123 ?35076
?xi?1182i1?4321.02
?x2i2?xi?1182i3?307864
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