医学统计学第四版各章例题SAS与STATA实现第三章.docx

1、医学统计学第四版各章例题SAS与STATA实现第三章医学统计学(第四版)各章例题SAS/STATA实现(第三章)例3J若某市1999年18岁男生身高服从均数为167.7cm,标准差为53cm的正态分布。从该正态分布NU677,532)cm总体中随机抽样100次即共抽取样本=100个，每次样本含量川)=10人.得到每个样本均数Xj及标准差S/如图3-1和下表3-1所示。167.41, 2.74165.56, 6.57168.20, 536图3-1 1999年某市18岁男生身aN(1677 5.32)抽样示意图表3-1 N(1677 5.32)总体中100个随机样本的乂j. Sj和95%CI(j=

2、10)样木号乂 jSj9S%CZ样本号乂 JSj95%CZ1167.412.74165.45169.375116&473.91165,6717L27165.566.57160.86170.2652165.953.76163.26168.64316S.205.36164.37172.0353168.875.77164.74173,004166.674.81163.24170.11*54169.532.07168.05171.005164.895.41161.02168.7655166.105.58162.11170,106166.364.5016344169.5856167.204.56163.

3、94170477166.164.04163.27169.0557170.507.66165.02175.98S169.115.71165.02173.1958166.444.93162.91169.979167.178.2616L27173.085916&684.521654517L9110166.135.241623S169.876016&406.9516343173.3711167.716.42163.12172.3161171.21630166.70175.721216&685.93164.44172.9262170.334.34167.23173,4413166.833.6916449

4、169.4763169.037.38163.751743114169.624.81166.18173.0664167634.58164.36170,9015166.953.6416435169.566516&663.33166.27171.0416170.294.91166.78173.806616&842.78166.85170,8317169.205.72165.11173.3067169.315.31165.51173.111S167.652.79165.65169.6568168.464.8116302171.9019166.515.39162.65170.3669168.605.4S

5、164,68172.5220163.283.1916L00165.5770168.475.05164.86172,0921166.294.95162.75169.8471165.6S5.19161.97169.40167.655.27163$8171.4272165.688.22159.80171.5623167.644.6116435170.9473168.034.89164.53171.5324172.617.74167.07178.1574169.375.0016579172.9425166.654.12163.70169.5975169.168.36163,18175,1426165.

6、194.41162.04168.34*76171.274.99167.71174,8427168.807.6816331174.307716&364.50165,14171.5828167如2.58166.14169.8378168.503.55165,96171,0429168.413.43165.95170.8679168.085.33164.27171.9030167.757.53162.36173.1380165.514.71162.14168.8831164.254.30161.17167.33S1167.593.73164.931702632166.425.1916271170.1

7、3*82171.124.40167.98174,2733166.904.41163.74170.0583165.925.11162.26169.5834166.774.34163&169&84167864.44164.69171,0435165.775.34161.95169.5985167.436.15163.03171.8336164426.6315938168.8686167906.13163.51172.2837169.834.20166.82172.8487167.59633163.06172.1238165.164.01162.29168.0288167.744.601644517

8、L0339166.596.201623171.0389167.408.27161.49173.3240165.653.56163.10168.2090167.1S6.00162.89171.4841165.724.17162.74168.7191166433.87163.66169,2142166.227.441603171.5492166.624.08163.70169.5443167.716.12163.33172.0993166.304.84162.83169.7644167255.24163.50170.9994169.705.26165.941734545165.695.91161.

9、46169.9295169.17632164.65173.6946169.065.65165.03173.1096167.896.07163.54172.234716&766.1416436173.1597167.486.03163.16171.794S16&644.5416539171.8998169.934.80166.50173.3749167.723.82164.99170.4599169405.57163421733950170.394.1516742173.35100165.695.09162,0616933*：表该样本资料算得的可信区间未包含已知总体均数167.7cm例35某医生

10、测量了 36洛从事铅作业男性工人的血红蛋白含量，算得加均数为130.83或L,标准差为2574g/L.问从事铅作业工人的血红蛋白是否不同于正常成年男性平均值140gzL?(1)建立检验假设，确定检验水准Ho：严3=140g/L,即从事铅作业的男性工人平均血红蛋白含量与正常成年男性平均值相等Hl： /岸MF140弓L 即从事铅作业的男性工人平均血红蛋白含量与正常成年男性平均值 不等0=0.05讣算检验统il嗤本例 =36 乂 =13083g/L, 42574g/L, “o=140g/L。按公式(3-15)仟361=35J30.83一斗3825.74/后(3)确定P值，做出推断结论以仟35、f =

11、 2.138 =2.138 査附表 2 的 r 界值表，因 0,05/235 2.138 vo02/235 故双尾概率0.02Ple t testmean = meaii(hb)Ho: mean = 140-2.1367例36为比较两种方法对乳酸饮料中脂肪含量测；4结果是否不同，随机抽取了 10份乳酸饮料制品，分別用脂肪酸水解法和哥特里一罗紫法测定貝结果如表35第(1X3)栏。问两法测崔 结果是否不同？表3-5两种方法对乳酸饮料中脂肪含量的测崔结果()編号1)哥特里一罗紫法脂肪酸水解法差值d(4X2)73)10.84005800.26020.59105090.0S230.67405000.17

12、440.63203160.31650.68703370.3506097805170.46170.7500,4540.2968073005120.21S9L20009970.203100.87005060.3642.724(1)建立检验假设，确定检验水准Hz “a，即两种方法的测定结果相同血耳工0即两种方法的测企结果不同0=0.05汁算检验统il量本例 =10. 加2724, N=08483. 7 = ZJ/n = 2724/10 = 02724。0.8483-22空L 叫一=0.108710-1按公式(3-16)v = i0-l=9尸 2724 7.925,0J 087/710(3)确企P值，

13、做出推断结论査附表2的r界值表得P|t!7.930001SAS系统UNIVARIATE PROCEDURE 变& 矩N 均值 标准差 偏度 未校平方和 变异系数100.27240.10868119-0.03375030.84832239.8976451权重总和 观测总和 方差 峰度 校正平方和 标准误差均值1027240,01181160.06565577010630440.03436801基本统计测度 位置 均值 中位数 众数0.2724000.278000变异性 标准追 方差 极差0.108680.01181037900四分位极差 0.14700位置检验:Mu0=0 检验Student t

14、 符号 符号秩统计量-P值一 |t|Pr=|M| 0.0020Pr = |S| 0.0020分位数（是义5） 分位数 100%最大值 99%95%90%75% Q350%中位数25% QI10%5%1%0%最小值27.5估il值0.46100.46100.46100.4125035000.27800.20300.12800.08200.08200.0820极值观测 最小值.最大值值观测值观测0.08220.29670.17430.31640.20390.35050.21880364100.26010.4616(2) STATA 实现.ttest xl=x2anable | ObsMeanStd

15、. ErrStd De95% Conf IniemlX11 10.7952.0583003.1843618.6633155 .9270845x2| 10.5228.0588125.1859814.389757 .655843diff| 10.2724.034368.1086812,1946542 .3501458niean(diff) = mean(xlx2)t= 7,9260Ho: meaa(di3) = 0degrees of freedom = 9Ha: 0Pr(T|t|) = 0,0000PrCT 1) = 0.0000.gen d=xl-x2.ttest d=0Oiie-sanqle

16、 t testVariable | ObsMeanStd. Err.Std. De*.95% Conf. Inienald| 10.2724.034368.1086812.1946542 .3501458mean = meaii(d)t= 7,9260Paired ttestHo: mean = 0degrees of freedom =9Pr(T t) = 0.0000Pr(T 0.50.按a=(M)5水准,不拒绝Hq,无统计学意义。还不能认为阿卡波糖胶囊与拜腐苹胶囊对空腹血糖的降糖效果 不同。(1) SAS实现SAS代码如下:data ex3_7:input X ; if_n_|tlPoo

17、ledEqual38-0,640.5248Unequal36.0860640.5250Equality ofXnancesMetiiodMumDFDenDFFyue PrFFolded F19191.60 0.3153(2) STATA 实现.nest xl=x2jinplvo-sample ttest wich equal variancesVariable |ObsMeanStd. Err.Std. Dev95% Conf Imervalxl|202.065.68426973,06014763280713.49 刀 93x2|202.625.5412424204991.4921723.757828combmed |402.34543292312.7380461.469333.22067&ff|568724482-2326179120617906419diff = mean(xl) - inean(x2).stack xl x2,into(x) .rename .stack group .ttest x.by(group)lvo-sample ttest uirh equal variancesGroup 1ObsMeanStd, ErrSuJ. Dev,(95% Conf. Intervalll202.065.68426973,060147632807134971932|2