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1、第五节正态总体均值与方差的区间估计一、单个总体的情况二、两个总体的情况三、小结一、单个总体的情况由上节例1可知:1.推导过程如下:解例1有一大批糖果,现从中随机地取16袋,重量(克)如下:称得设袋装糖果的重量服从正态分布,试求总体均值这个估计的可信程度为95%.就是说估计袋装糖果重量的均值在500.4克与507.1克之间,这个误差的可信度为95%.解代入公式得标准差的置信区间(续例)求补充1中总体标准差的置信度为例20.95的置信区间.附表2-1附表2-2推导过程如下:2.根据第六章第二节定理二知进一步可得:在密度函数不对称时,注意:习惯上仍取对称的分位点来确定置信区间(如图
2、).二、两个总体的情况1.推导过程如下:(2)例3为比较І,ІІ两种型号步枪子弹的枪口速度,随机地取ІІ得枪口速度平均值为假设两总体都可认为近似且由生产过程可认为它们的方差信区间.随机地取І型子弹10发,得到枪口速度的平均值为型子弹20发,地服从正态分布,相等,求两总体均值差解两总体样本是相互独立的.由题意,但数值未知.又因由假设两总体的方差相等,由于例4试图采用为提高某一化学生产过程的得率,一种新的催化剂,为慎重起见,在试验工厂先进行试验.体都可认为近似地服从正态分布,且方差相等,求解现在2.由第六章第三节定理四即例5设两样本相互独研究由机器A和机器B生产的钢管内径,随机抽取
3、机器A生产的管子18只,测得样本方差抽取机器B生产的管子13只,区间.且设由机器A和机器B生产的钢管内径分别服立,均未知,解补充例题三、小结附表2-1=0.250.100.050.0250.010.005123456789101112131415161.3232.7734.1085.3856.6267.8419.03710.21911.38912.54913.70114.84515.98417.11718.24519.3692.7064.6056.2517.7799.23610.64512.01713.36214.68415.98717.27518.54919.81220.06
4、422.30723.5423.8415.9917.8159.48811.07112.59214.06715.50716.91918.30719.67521.02622.36223.68524.99626.2965.0247.3789.34811.14312.83314.44916.01317.53519.02320.48321.92023.33724.73626.11927.48828.8456.6359.21011.34513.27715.08616.81218.47520.09021.66623.20924.72526.21727.68829.14130.57832.000
5、7.87910.59712.83814.86016.75018.54820.27821.95523.58925.18826.75728.29929.89131.31932.80134.267分布表27.488返回附表2-2=0.9950.990.9750.950.900.75123456789101112131415160.0100.0720.2070.4120.6760.9891.3441.7352.1562.6033.0743.5654.0754.6015.1420.0200.1150.2970.5540.8721.2391.6462.0882.5583.0533.5714
6、.1074.6605.2295.8120.0010.0510.2160.4840.8311.2371.6902.1802.7003.2473.8164.4045.0095.6296.2626.9080.0040.1030.3520.7111.1451.6352.1672.7333.3253.9404.5755.2265.8926.5717.2617.9620.0160.2110.5841.0641.6102.2042.8333.4904.1684.8655.5786.3047.0427.7908.5479.3120.1020.5751.2131.9232.6753.4554.2
7、555.0715.8996.7377.5848.4389.29910.16511.03711.912分布表6.262返回