第二章習(xí)題答案Prob2-2:Prob2-3別捷爾斯法：極差法：其中；最大誤差法：其中；Prob2-4；；；或然誤差平均誤差：Prob2-5；根據(jù)t分布，測量結(jié)果為：Prob2-7查表得到滿足上述條件的最小N為5，所以至少測量5次Prob2-9（1）（2）;（3）先去除粗大誤差26.202526.202526.202526.202826.202826.202826.202826.202826.202820.202520.202526.202626.202626.202626.202226.202226.202220.202326.202526.202526.202526.202626.202626.202626.202226.202226.2022ave=25.002525.5358626.20253stdev2.5298752.0000080.000231；Prob2-122025602535202050154530.033.018.3711713.50926337.5182.5P1：182.5P2:337.5;Prob2-13證明：由于是等精度測量，設(shè)每個(gè)測量值的標(biāo)準(zhǔn)差為，權(quán)為P。那么測量結(jié)果的平均值的標(biāo)準(zhǔn)差為，權(quán)重為由于：，所以：Prob2-16使用計(jì)算數(shù)據(jù)比較法判別前一組數(shù)據(jù)和后一組數(shù)據(jù)之間是否存在系統(tǒng)誤差；；結(jié)論，不存在系統(tǒng)誤差。Prob2-18xvv220.06-0.0650.00422520.07-0.0550.00302520.06-0.0650.00422520.08-0.0450.00202520.1-0.0250.00062520.12-0.0050.00002520.11-0.0150.00022520.140.0150.00022520.180.0550.00302520.180.0550.00302520.210.0850.00722520.190.0650.0042250.03210020.125stdev10.05402stdev20.05998281（1）先使用殘余誤差校核法結(jié)論：明顯不為零，所以存在線性系統(tǒng)誤差。（2）使用不同公式計(jì)算標(biāo)準(zhǔn)差比較法；結(jié)論：不存在系統(tǒng)誤差。Prob2-20xv28.53-0.04128.530.02728.5328.52-0.05128.520.01728.5228.5-0.07128.5-0.00328.529.520.94928.530.02728.5328.53-0.04128.530.02728.5328.53-0.04128.5-0.00328.528.5-0.07128.49-0.01328.4928.49-0.08128.49-0.01328.4928.49-0.08128.510.00728.5128.51-0.06128.530.02728.5328.53-0.04128.520.01728.5228.52-0.05128.49-0.01328.4928.49-0.08128.4-0.10328.528.4-0.17128.5-0.00328.5-0.071ave28.57128.50328.51077stdev0.2650.0340.0165643sigma0.7943sigma