Stata基本操作和数据分析入门第08章多组独立定量资料统计分析Stata实现.doc

第八章 多组独立定量资料的统计分析本章应用的STATA命令为：单因素方差分析oneway response_var factor_var, bonferroni 非参数多组比较kwallis varname, by(groupvar)例8-1 为研究茶多酚保健饮料对急性缺氧的影响，某研究者将60只小白鼠随机分为低、中、高三个剂量组和一个对照组，每组15只小白鼠。对照组给予蒸馏水0.25ml灌胃，低、中、高剂量组分别给予2.0gkg、4.0gkg、8.0gkg的饮料溶于0.20.3ml蒸馏水后灌胃。每天一次，40天后，对小白鼠进行耐缺氧存活时间试验，结果如表8-1。试比较不同剂量的茶多酚保健饮料对延长小白鼠的平均耐缺氧存活时间有无差别。表8-1 各组小白鼠耐缺氧时间组别耐缺氧时间（min）对照组（）20.7922.9127.2119.3417.8523.7922.6018.531521.323.4023.2320.1426.7119.3617.2224.1315.85低剂量组（）22.2224.7421.5319.6625.8929.1018.9318.641523.233.5226.3925.4920.4322.6929.6720.3622.74中剂量组（）28.5628.6725.2830.3823.1323.4728.8829.621528.144.0024.8234.6422.2929.2225.6335.1232.32高剂量组（）31.9337.9439.7627.9429.6534.2332.6329.131532.844.6639.6236.1528.8524.0729.2935.2436.13合计6026.385.92解：STATA数据为：gx120.79122.91127.21119.34117.85123.79122.6118.53123.23120.14126.71119.36117.22124.13115.85222.22224.74221.53219.66225.89229.1218.93218.64226.39225.49220.43222.69229.67220.36222.74328.56328.67325.28330.38323.13323.47328.88329.62324.82334.64322.29329.22325.63335.12332.32431.93437.94439.76427.94429.65434.23432.63429.13439.62436.15428.85424.07429.29435.24436.13 ：四个总体均数相等：四个总体均数不等或不全相等 STATA命令为：oneway x g结果为： Analysis of Variance Source SS df MS F Prob F-Between groups 1205.79582 3 401.931941 26.09 0.0000 Within groups 862.796912 56 15.4070877- Total 2068.59273 59 35.0608938Bartletts test for equal variances: chi2(3) = 1.7256 Probchi2 = 0.631检验统计量F26.09，p F-Between groups .173146778 2 .086573389 5.90 0.0075 Within groups .396399936 27 .014681479- Total .569546715 29 .019639542Bartletts test for equal variances: chi2(2) = 1.9926 Probchi2 = 0.369 Comparison of x by g (Bonferroni)Row Mean-|Col Mean | 1 2-+- 2 | .098 | 0.245 | 3 | .186 .088 | 0.006 0.3483个组比较F=5.9，p0.0075，差别有统计意义，然后用Bonferroni法进行两两比较，其中1比2的p0.245，1比3为0.006，2比3为0.348。例8-4 对按完全随机设计分组的四组大白鼠，给予不同剂量的某种激素后，测量耻骨间隙宽度的增加量（），结果如表8-9所示，试分析给予不同剂量的某种激素后大白鼠耻骨间隙宽度的平均增加量有无差异？表8-9 耻骨间隙宽度的增加量（）1组0.150.300.400.502组1.201.351.401.501.902.303组0.501.201.402.002.202.204组1.501.502.502.50 解：对于表8-9数据，如果对四组数据进行Levene方差齐性检验，P0.025，按水准，表明该资料不满足方差分析要求方差齐性的基本条件，所以这里采用Kruskal-Wallis检验。1） 建立检验假设：接受不同剂量（4种）激素的大白鼠耻骨间隙宽度的增加量总体分布相同 ：接受不同剂量（4种）激素的大白鼠耻骨间隙宽度的增加量总体分布不全相同 =0.05解：STATA数据为gx110.15210.3310.4410.5521.2621.35721.4821.5921.91022.31130.51231.21331.414321532.21632.21741.51841.51942.52042.5STATA命令为：kwallis x,by( g)结果为：Test: Equality of populations (Kruskal-Wallis test) +-+ | g | Obs | Rank Sum | |-+-+-| | 1 | 4 | 10.50 | | 2 | 6 | 68.00 | | 3 | 6 | 68.50 | | 4 | 4 | 63.00 | +-+chi-squared = 10.501 with 3 d.f.probability = 0.0148chi-squared with ties = 10.572 with 3 d.f.probability = 0.0143按=0.05水准，拒绝，接受，故可认为接受不同剂量（4种）激素的大白鼠平均耻骨间隙宽度的增加量总体分布不全相同。