应用统计方法(大作业)

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1、例3-7在某项实验中，测得变量y与因素x数据如表1所示。试建立适当的y与因素x的回归方程（）。表3-14例3-7实测数据x23457810111415161819y106.42108.2109.58109.5110109.93110.49110.59110.6110.9110.76111111.2解：绘制散点图，如图1所示：图1从图1中可看出，以下三种曲线方程的曲线图都与散点图接近，因此都可以作为曲线回归的选择对象。（1）.（2）.（3）.现对方案1和方案2方案3进行求解分析，通过对S2残作比较，S2残小这则回归方程交优

2、1.方案1选取曲线回归（1）进行求解。令，应用EXCEL进行相应处理算得数据，列入表1。表1方案1数据处理计算xx=xyx'-x'yi-y(x'-x')(yi-y)21.414213562106.42-1.628216714-3.5161538465.72506046131.732050808108.2-1.310379469-1.7361538462.27502035542109.58-1.042430277-0.3561538460.37126555252.236067977109.5-0.806362299-0.4

3、361538460.35169801872.645751311110-0.3966789660.063846154-0.02532642682.828427125109.93-0.214003152-0.0061538460.001316942103.16227766110.490.1198473830.5538461540.066377012113.31662479110.590.2741945140.6538461540.179281028143.741657387110.60.699227110.663846154

4、0.464179228153.872983346110.90.8305530690.9638461540.800525381164110.760.9575697230.8238461540.788890133184.2426406871111.200210411.0638461541.276839229194.358898944111.21.3164686671.2638461541.663813862∑39.55159361429.1713.93894078平均3.042430277109.9362平方和1321571

5、38.711.6670341421.2105076944.12800113由表2得：由此得：故所求的回归方程为：进行变量还原得回归方程：检验假设H01：.对给定的，查F(1,11)表（附表5）得临界值。由于F>，检验效果显著，所以拒绝H01，即回归方程有意义。2.方案2表2方案2数据处理计算xxˊ=lgxYx'-x'yi-y(x'-x')(yi-y)20.301030106.420000-0.616595-3.5162002.16807130.477121108.200000-0.440504-1.7362000.764

6、80340.602060109.580000-0.315565-0.3562000.11240450.698970109.500000-0.218655-0.4362000.09537770.845098110.000000-0.0725270.063800-0.00462780.903090109.930000-0.014535-0.0062000.000090101.000000110.4900000.0823750.5538000.045619111.041393110.5900000.1237680.653800

7、0.080919141.146128110.6000000.2285030.6638000.151680151.176091110.9000000.2584660.9638000.249110161.204120110.7600000.2864950.8238000.236015181.255273111.0000000.3376481.0638000.359189191.278754111.2000000.3611291.2638000.456394总和11.9291271429.170000—————————平均0.

8、917625109.936154—————————平方和——————1.19471721.21.5104.715045由此得：故所求回归方程：进行变量还原得回归方程：检验假设：对给定的α=0.01，查F(1，11)表得到临界值λ=9.65.由于F>λ,检验效果显著，所以拒绝,即回归方程有意义。3.方案3选取曲线回归（