中国石油大学-应用统计方法作业2表格推荐下载.xlsx

中国石油大学-应用统计方法作业2表格推荐下载.xlsx

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p139例例3-73-7实实测测数数据据x2345781011y106.42108.2109.58109.5110109.93110.49110.591415161819散散点点图图如如下下图图3-13-1所所示示：

@#@110.6110.9110.76111111.2由散点图看出，以下四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）（4）1.1.选选取取曲曲线线回回归归（11）求求解解。

@#@令令，应应用用EXCELEXCEL可可算算得得数数据据，列列入入表表3-13-1中中：

@#@表表3-13-1例例3-73-7数数据据预预处处理理计计算算序号xy121.414214106.42-1.62822-3.516155.72506231.732051108.2-1.31038-1.736152.27502342109.58-1.04243-0.356150.371266452.236068109.5-0.80636-0.436150.351698572.645751110-0.396680.063846-0.02533682.828427109.93-0.214-0.006150.0013177103.162278110.490.1198470.5538460.0663778113.316625110.590.2741950.6538460.1792819143.741657110.60.6992270.6638460.46417910153.872983110.90.8305530.9638460.80052511164110.760.957570.8238460.7888912184.2426411111.200211.0638461.27683913194.358899111.21.3164691.2638461.66381413239.551591429.17-平均10.153853.04243109.9362-平方和-11.6670321.2105144.128由由表表3-13-1得得lxx=11.66703lyy=21.21051lxy=13.93894b=1.194729a=106.3013故故所所求求回回归归方方程程为为y=106.3013+1.1947检验假设H01：

@#@1=0y20181614121086420104105106107108109110111112x图3-1散点图xxxbayxbaylgxbay/xxyyi）（yyxxiixxix2bxayS回2=16.65325S总2=21.21051S残2=4.557255F=40.19652对于给定的=0.01，查F（1，11）表得临界值lambda=9.64603由于F,检验效果显著，所以拒绝H01，即回归方程有意义。

@#@2.2.选选取取曲曲线线回回归归（22）求求解解。

@#@令令，应应用用EXCELEXCEL可可算算得得数数据据，列列入入表表3-23-2中中：

@#@表表3-23-2例例3-73-7数数据据预预处处理理计计算算序号xy120.30103106.42-0.6166-3.516152.168044230.477121108.2-0.4405-1.736150.764783340.60206109.58-0.31557-0.356150.11239450.69897109.5-0.21866-0.436150.095367570.845098110-0.072530.063846-0.00463680.90309109.93-0.01454-0.006158.94E-057101110.490.0823750.5538460.0456238111.041393110.590.1237680.6538460.0809259141.146128110.60.2285030.6638460.15169110151.176091110.90.2584660.9638460.24912211161.20412110.760.2864950.8238460.23602812181.2552731110.3376471.0638460.35920513191.278754111.20.3611281.2638460.45641113211.929131429.17-平均10.153850.917625109.9362-平方和-1.19471721.210515.793803由由表表3-23-2得得lxx=1.194717lyy=21.21051lxy=4.715045b=3.946578a=106.3147故故所所求求回回归归方方程程为为y=106.3147+3.9466lgx检验假设H01：

@#@1=0S回2=18.6083S总2=21.21051S残2=2.602211F=78.66052对于给定的=0.01，查F（1，11）表得临界值lambda=9.64603由于F,检验效果显著，所以拒绝H01，即回归方程有意义。

@#@3.3.选选取取曲曲线线回回归归（33）求求解解。

@#@令令，应应用用EXCELEXCEL可可算算得得数数据据，列列入入表表3-33-3中中：

@#@表表3-33-3例例3-73-7数数据据预预处处理理计计算算序号xy120.5106.420.34224-3.51615-1.20337230.333333108.20.175573-1.73615-0.30482340.25109.580.09224-0.35615-0.03285450.2109.50.04224-0.43615-0.01842570.142857110-0.01490.063846-0.00095680.125109.93-0.03276-0.006150.000202xxlglgxxyyixxi）（yyxxiixx/11/xxyyixxi）（yyxxii7100.1110.49-0.057760.553846-0.031998110.090909110.59-0.066850.653846-0.043719140.071429110.6-0.086330.663846-0.0573110150.066667110.9-0.091090.963846-0.087811160.0625110.76-0.095260.823846-0.0784812180.055556111-0.10221.063846-0.1087313190.052632111.2-0.105131.263846-0.132871322.0508821429.17-平均10.153850.15776109.9362-平方和-0.21367121.210511.591993由由表表3-33-3得得lxx=0.213671lyy=21.21051lxy=-2.1011b=-9.83337a=111.4875故故所所求求回回归归方方程程为为y=111.4875-9.8333/x检验假设H01：

@#@1=0S回2=20.66092S总2=21.21051S残2=0.549586F=413.5295对于给定的=0.01，查F（1，11）表得临界值lambda=9.64603由于F,检验效果显著，所以拒绝H01，即回归方程有意义。

@#@4.4.选选取取曲曲线线回回归归（44）求求解解。

@#@令令，应应用用EXCELEXCEL可可算算得得数数据据，列列入入表表3-13-1中中：

@#@表表3-43-4例例3-73-7数数据据预预处处理理计计算算序号xy124106.42-130.615-3.51615459.2638239108.2-125.615-1.73615218.08763416109.58-118.615-0.3561542.245334525109.5-109.615-0.4361547.809175749110-85.61540.063846-5.466216864109.93-70.6154-0.006150.434556710100110.49-34.61540.553846-19.1716811121110.59-13.61540.653846-8.90237914196110.661.384620.66384640.749941015225110.990.384620.96384687.116861116256110.76121.38460.823846100.00221218324111189.38461.063846201.47611319361111.2226.38461.263846286.115313217501429.17-平均10.15385134.6154109.9362-平方和-186413.121.21051404737.6由由表表3-43-4得得lxx=186413.1lyy=21.21051lxy=1449.761b=0.007777a=108.8892故故所所求求回回归归方方程程为为y=108.8892+0.0078检验假设H01：

@#@1=02xxyyi）（yyxxiixxi2x2xxS回2=11.27499S总2=21.21051S残2=9.935514F=12.48299对于给定的=0.01，查F（1，11）表得临界值lambda=9.64603由于F,检验效果显著，所以拒绝H01，即回归方程有意义。

@#@结结论论：

@#@通过对这一问题用以上四种曲线作回归，我们发现S残2最小者为（3），即方案（3）回归方程较优1415161819110.6110.9110.76111111.2xbaylg