stata考试复习题库

stata考试复习题库 例2-1 从某单位1999年的职工体检资料中获得101名正常成年女子的血清总胆固醇()mmol/L的测量结果如下，试编制频数分布 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 。 4.21 3.32 5.35 4.17 4.13 2.78 4.26 3.58 4.34 4.84 4.41 2.35 4.78 3.95 3.92 3.58 3.66 4.28 3.26 3.50 2.70 4.61 4.75 2.91 3.91 4.59 4.19 2.68 4.52 4.91 3.18 3.68 4.83 3.87 3.95 3.91 4.15 4.55 4.80 3.41 4.12 3.95 5.08 4.53 3.92 3.58 5.35 3.84 3.60 3.51 4.06 3.07 3.55 4.23 3.57 4.83 3.52 3.84 4.50 3.96 4.50 3.27 4.52 3.19 4.59 3.75 3.98 4.13 4.26 3.63 3.87 5.71 3.30 4.73 4.17 5.13 3.78 4.57 3.80 3.93 3.78 3.99 4.48 4.28 4.06 5.26 5.25 3.98 5.03 3.51 3.86 3.02 3.70 4.33 3.29 3.25 4.15 4.36 4.95 3.00 3.26 例题2-1(EX2-1.dta): . sum x Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------- x | 101 4.029505 .6592183 2.35 5.71 . di r(max)-r(min) 3.3600001 . gen group=int((x-2.30)/0.30)*0.30+2.3 . tab group group | Freq. Percent Cum. ------------+----------------------------------- 2.3 | 1 0.99 0.99 2.6 | 3 2.97 3.96 2.9 | 6 5.94 9.90 3.2 | 8 7.92 17.82 3.5 | 18 17.82 35.64 3.8 | 19 18.81 54.46 4.1 | 17 16.83 71.29 4.4 | 12 11.88 83.17 4.7 | 9 8.91 92.08 5 | 5 4.95 97.03 5.3 | 2 1.98 99.01 5.6 | 1 0.99 100.00 ------------+----------------------------------- Total | 101 100.00 例2-2 用直接法计算例2-1某单位名正常成年女子的血清总胆固醇的均数 101 例2-3 利用表2-1计算101名正常成年女子的血清总胆固醇均数。 第2、3题了解即可，可以用最简单的 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 求出:比如 su，tabstat等 例2-4 某地5例微丝蚴血症患者治疗7年后用间接荧光抗体试验测得其抗体滴度倒数分别为 ，求几何均数。 10,20,40,40,160 例题2-4(EX2-4.dta): . means x Variable | Type Obs Mean [95% Conf. Interval] -------------+---------------------------------------------------------- x | Arithmetic 5 54 -21.32307 129.3231 | Geometric 5 34.82202 9.715194 124.8121 | Harmonic 5 24.24242 . . ------------------------------------------------------------------------ Missing values in confidence interval(s) for harmonic mean indicate that confidence interval is undefined for corresponding variable(s). Consult Reference Manual for details. -5 69例类风湿关节炎(RA)患者血清EBV-VCA-lgG抗体滴度的分布见表2-5第(1)、(2)栏,求其平均例2 抗体滴度。 表2-5 69例RA患者血清EBV-VCA-lgG抗体测定结果 fX,lg 抗体滴度 lgX滴度倒数fX人数 ? ? ? ? ? 1: 10 4 10 1.0000 4.0000 1: 20 3 20 1.3010 3.9030 1: 40 10 40 1.6021 16.0210 1: 80 10 80 1.9031 19.0310 1: 160 11 160 2.2041 24.2451 1: 320 15 320 2.5051 37.5765 1: 640 14 640 2.8062 39.2868 1: 1280 2 1280 3.1072 6.2144 合 计 69 — — 150.2778 例题2-5(EX2-5.dta): . means x [fw=f] Variable | Type Obs Mean [95% Conf. Interval] -------------+---------------------------------------------------------- x | Arithmetic 69 280.8696 212.754 348.9851 | Geometric 69 150.6411 110.9425 204.545 | Harmonic 69 64.84581 47.14289 103.8391 ------------------------------------------------------------------------ 例2-6 7名病人患某病的潜伏期分别为天，求其中位数。 2,3,4,5,6,9,16 例题2-6(EX2-6.dta): . centile x -- Binom. Interp. -- Variable | Obs Percentile Centile [95% Conf. Interval] -------------+------------------------------------------------------------- x | 7 50 5 2.314286 13.8 例2-7 8名患者食物中毒的潜伏期分别为小时，求其中位数。 1,2,2,3,5,8,15,24 例题2-7(EX2-7.dta): . centile x -- Binom. Interp. -- Variable | Obs Percentile Centile [95% Conf. Interval] -------------+------------------------------------------------------------- x | 8 50 4 1.675 17.925 例2-8 试计算表2-2某医院1123名产后出血孕妇人工流产次数的中位数。 表2-2 某医院1123名产后出血孕妇人流次数的分布 产后出血人数 人流次数 累计频数 累计频率(%) (1) (2) (3) (4) 0 402 402 35.80 1 330 732 65.18 2 232 964 85.84 3 118 1082 96.35 4 27 1109 98.75 5 11 1120 99.73 6 3 1123 100.00 合计 1123 — — 例题2-8(EX2-8.dta): . tab x [fw=f] x | Freq. Percent Cum. ------------+----------------------------------- 0 | 402 35.80 35.80 1 | 330 29.39 65.18 2 | 232 20.66 85.84 3 | 118 10.51 96.35 4 | 27 2.40 98.75 5 | 11 0.98 99.73 6 | 3 0.27 100.00 ------------+----------------------------------- Total | 1123 100.00 . expand f (1116 observations created) . centile x -- Binom. Interp. -- Variable | Obs Percentile Centile [95% Conf. Interval] -------------+------------------------------------------------------------- x | 1123 50 1 1 1 例2-10 某地118名链球菌咽喉炎患者的潜伏期频数表见表2-5第(1)、(2)栏，试分别求中位数及第、25第百分位数。表2-5 118名链球菌咽喉炎患者的潜伏期 75 人数， f天 数 累计频数 累计频率(%) (1) (2) (3) (4) 12, 4 4 3.4 24, 17 21 17.8 36, 32 53 44.9 48, 24 77 65.3 60, 18 95 80.5 72, 12 107 90.7 84, 5 112 94.9 96, 4 116 98.3 ?108 2 118 100.0 例题2-10(EX2-10.dta): . tab x [fw=f] x | Freq. Percent Cum. ------------+----------------------------------- 12 | 4 3.39 3.39 24 | 17 14.41 17.80 36 | 32 27.12 44.92 48 | 24 20.34 65.25 60 | 18 15.25 80.51 72 | 12 10.17 90.68 84 | 5 4.24 94.92 96 | 4 3.39 98.31 108 | 2 1.69 100.00 ------------+----------------------------------- Total | 118 100.00 (cm)例2-11 试计算下面三组同龄男孩的身高均数和极差。 甲组: ，; X,100cm90 95 100 105 110R,,,1109020cm甲甲 乙组: ，; 96 98 100 102 104R,,,104968cmX,100cm乙乙 丙组: ， 。 96 99 100 101 104R,,,104968cmX,100cm丙丙 例2-13 续例2-11，计算三组资料的 标准 excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载 差。 例题2-11、2-13(EX2-11.dta): . sum x1-x3 Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------- x1 | 5 100 7.905694 90 110 x2 | 5 100 3.162278 96 104 x3 | 5 100 2.915476 96 104 PP例2-12 续例2-10。已知=，=，计算名链球菌咽喉炎患者潜伏期的四分位39.267.71182575 数间距。 例题2-12(EX2-10.dta): Tabstat x [fw=f], st(iqr) 直接求出四分位间距 或者 Tabstat x [fw=f], st(q) 求出p25 p50 p75 例2-15 对例2-1，已计算出名正常成年女子的血清总胆固醇均数，标准X,4.06101mmol/L 差。试估计该单位:?正常女子血清总胆固醇在以下者占正常女S,0.654mmol/L4.00mmol/L 子总人数的百分比;?在4.00~5.00之间者占正常女子总人数的百分比;?在以mmol/L5.00mmol/L 上者各占正常女子总人数的百分比。 例题2-15(EX2-1.dta): . sum x Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------- x | 101 4.029505 .6592183 2.35 5.71 recode x (min/4=1) (4.01/5=2) (5.01/max=3), gen(group) tab group 同时求出三个比例 例2-16 由例2-1资料估计正常女子血清总胆固醇的95%的参考值范围 例题2-16(EX2-1.dta): . sum x Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------- x | 101 4.029505 .6592183 2.35 5.71 . di r(mean)-1.96*r(sd) 2.7374371 . di r(mean)+.96*r(sd) 4.6623546 例3-1 若某市1999年18岁男生身高服从均数为167.7cm，标准差为5.3cm的正态分布。从该正态 2n分布N(167.7, 5.3)cm总体中随机抽样100次即共抽取样本g=100个，，每次样本含量=10人，得到每个j S样本均数及标准差如图3-1和下表3-1所示。 Xjj 167.41, 2.74 165.56, 6.57 168.20, 5.36 ,=167.7cm 100个 ,=5.3cm , x,x,x,？x,？ 123i, n=10 j 2S-1 N(167.7, 5.3表3)总体中100个随机样本的、和95%CL(n=10) Xjjj SS XX样本号 95%CL 样本号 95%CL jjjj 1 167.41 2.74 165.45 169.37 51 168.47 3.91 165.67 171.27 2 165.56 6.57 160.86 170.26 52 165.95 3.76 163.26 168.64 3 168.20 5.36 164.37 172.03 53 168.87 5.77 164.74 173.00 4 166.67 4.81 163.24 170.11 *54 169.53 2.07 168.05 171.00 5 164.89 5.41 161.02 168.76 55 166.10 5.58 162.11 170.10 6 166.36 4.50 163.14 169.58 56 167.20 4.56 163.94 170.47 7 166.16 4.04 163.27 169.05 57 170.50 7.66 165.02 175.98 8 169.11 5.71 165.02 173.19 58 166.44 4.93 162.91 169.97 9 167.17 8.26 161.27 173.08 59 168.68 4.52 165.45 171.91 10 166.13 5.24 162.38 169.87 60 168.40 6.95 163.43 173.37 11 167.71 6.42 163.12 172.31 61 171.21 6.30 166.70 175.72 12 168.68 5.93 164.44 172.92 62 170.33 4.34 167.23 173.44 13 166.83 3.69 164.19 169.47 63 169.03 7.38 163.75 174.31 14 169.62 4.81 166.18 173.06 64 167.63 4.58 164.36 170.90 15 166.95 3.64 164.35 169.55 65 168.66 3.33 166.27 171.04 16 170.29 4.91 166.78 173.80 66 168.84 2.78 166.85 170.83 17 169.20 5.72 165.11 173.30 67 169.31 5.31 165.51 173.11 18 167.65 2.79 165.65 169.65 68 168.46 4.81 165.02 171.90 19 166.51 5.39 162.65 170.36 69 168.60 5.48 164.68 172.52 *20 163.28 3.19 161.00 165.57 70 168.47 5.05 164.86 172.09 21 166.29 4.95 162.75 169.84 71 165.68 5.19 161.97 169.40 22 167.65 5.27 163.88 171.42 72 165.68 8.22 159.80 171.56 23 167.64 4.61 164.35 170.94 73 168.03 4.89 164.53 171.53 24 172.61 7.74 167.07 178.15 74 169.37 5.00 165.79 172.94 25 166.65 4.12 163.70 169.59 75 169.16 8.36 163.18 175.14 26 165.19 4.41 162.04 168.34 *76 171.27 4.99 167.71 174.84 27 168.80 7.68 163.31 174.30 77 168.36 4.50 165.14 171.58 28 167.99 2.58 166.14 169.83 78 168.50 3.55 165.96 171.04 29 168.41 3.43 165.95 170.86 79 168.08 5.33 164.27 171.90 30 167.75 7.53 162.36 173.13 80 165.51 4.71 162.14 168.88 *31 164.25 4.30 161.17 167.33 81 167.59 3.73 164.93 170.26 32 166.42 5.19 162.71 170.13 *82 171.12 4.40 167.98 174.27 33 166.90 4.41 163.74 170.05 83 165.92 5.11 162.26 169.58 34 166.77 4.34 163.66 169.88 84 167.86 4.44 164.69 171.04 35 165.77 5.34 161.95 169.59 85 167.43 6.15 163.03 171.83 36 164.12 6.63 159.38 168.86 86 167.90 6.13 163.51 172.28 37 169.83 4.20 166.82 172.84 87 167.59 6.33 163.06 172.12 38 165.16 4.01 162.29 168.02 88 167.74 4.60 164.45 171.03 39 166.59 6.20 162.15 171.03 89 167.40 8.27 161.49 173.32 40 165.65 3.56 163.10 168.20 90 167.18 6.00 162.89 171.48 41 165.72 4.17 162.74 168.71 91 166.43 3.87 163.66 169.21 42 166.22 7.44 160.90 171.54 92 166.62 4.08 163.70 169.54 43 167.71 6.12 163.33 172.09 93 166.30 4.84 162.83 169.76 44 167.25 5.24 163.50 170.99 94 169.70 5.26 165.94 173.45 45 165.69 5.91 161.46 169.92 95 169.17 6.32 164.65 173.69 46 169.06 5.65 165.03 173.10 96 167.89 6.07 163.54 172.23 47 168.76 6.14 164.36 173.15 97 167.48 6.03 163.16 171.79 48 168.64 4.54 165.39 171.89 98 169.93 4.80 166.50 173.37 49 167.72 3.82 164.99 170.45 99 169.40 5.57 165.42 173.39 50 170.39 4.15 167.42 173.35 100 165.69 5.09 162.06 169.33 *表示该样本资料算得的可信区间未包含已知总体均数167.7cm 例题3-1: . drop _all . set seed 050322 . set obs 10 obs was 0, now 10 . quietly for num 1/100:gen varX=invnorm(uniform())*5.3+167.7 . format var* %9.2f . ci var* Variable | Obs Mean Std. Err. [95% Conf. Interval] -------------+------------------------------------------------------------- var1 | 10 167.23 1.53 163.78 170.69 var2 | 10 164.70 1.97 160.24 169.17 var3 | 10 165.39 1.67 161.61 169.16 var4 | 10 168.49 2.12 163.70 173.27 var5 | 10 169.38 1.37 166.28 172.48 var6 | 10 166.84 1.66 163.07 170.60 var7 | 10 167.51 1.21 164.79 170.24 var8 | 10 167.09 1.84 162.93 171.24 var9 | 10 167.06 1.64 163.36 170.77 var10 | 10 167.21 1.82 163.11 171.32 var11 | 10 167.01 1.85 162.84 171.18 var12 | 10 167.19 1.99 162.67 171.70 var13 | 10 168.04 1.76 164.07 172.02 var14 | 10 168.68 1.28 165.80 171.57 var15 | 10 167.12 1.95 162.72 171.52 var16 | 10 168.29 1.52 164.86 171.72 var17 | 10 165.97 2.11 161.19 170.75 var18 | 10 164.30 1.72 160.42 168.18 var19 | 10 168.97 0.92 166.89 171.05 var20 | 10 167.53 1.31 164.57 170.49 var21 | 10 165.98 1.21 163.24 168.71 var22 | 10 166.99 1.77 162.98 170.99 var23 | 10 166.98 1.25 164.15 169.80 var24 | 10 167.64 1.86 163.43 171.85 var25 | 10 163.26 1.03 160.93 165.59 var26 | 10 167.96 1.17 165.30 170.61 var27 | 10 166.20 2.64 160.23 172.17 var28 | 10 168.23 1.62 164.57 171.89 var29 | 10 167.10 1.81 163.00 171.21 var30 | 10 165.99 1.76 162.01 169.97 var31 | 10 169.32 2.59 163.46 175.17 var32 | 10 168.70 1.84 164.54 172.85 var33 | 10 165.80 1.18 163.14 168.47 var34 | 10 169.07 1.12 166.55 171.60 var35 | 10 166.59 1.65 162.87 170.32 var36 | 10 165.32 2.35 160.01 170.63 var37 | 10 169.29 1.36 166.22 172.36 var38 | 10 167.59 1.15 164.99 170.18 var39 | 10 167.71 1.79 163.65 171.77 var40 | 10 169.70 2.06 165.05 174.36 var41 | 10 167.52 1.90 163.23 171.82 var42 | 10 166.11 1.74 162.16 170.05 var43 | 10 166.45 1.48 163.09 169.80 var44 | 10 168.09 0.98 165.87 170.30 var45 | 10 167.14 1.46 163.85 170.43 var46 | 10 169.69 1.63 165.99 173.38 var47 | 10 165.67 1.55 162.17 169.18 var48 | 10 165.29 1.99 160.79 169.78 var49 | 10 167.99 1.88 163.73 172.24 var50 | 10 167.58 2.18 162.65 172.51 var51 | 10 165.19 1.44 161.93 168.45 var52 | 10 169.19 1.51 165.76 172.62 var53 | 10 168.37 1.66 164.62 172.12 var54 | 10 167.13 1.75 163.16 171.10 var55 | 10 167.01 1.72 163.12 170.90 var56 | 10 166.02 1.89 161.74 170.30 var57 | 10 169.41 1.89 165.13 173.69 var58 | 10 169.13 2.55 163.35 174.90 var59 | 10 165.51 1.41 162.32 168.70 var60 | 10 165.71 2.17 160.81 170.61 var61 | 10 167.46 1.62 163.80 171.12 var62 | 10 168.41 1.70 164.56 172.27 var63 | 10 168.06 1.98 163.59 172.54 var64 | 10 167.63 2.38 162.25 173.00 var65 | 10 168.22 2.38 162.83 173.61 var66 | 10 171.20 1.69 167.39 175.02 var67 | 10 166.51 1.61 162.88 170.14 var68 | 10 169.31 1.31 166.34 172.29 var69 | 10 164.97 1.53 161.52 168.42 var70 | 10 169.08 1.44 165.82 172.35 var71 | 10 168.69 2.03 164.10 173.28 var72 | 10 168.44 1.72 164.54 172.34 var73 | 10 167.77 2.15 162.91 172.64 var74 | 10 171.81 1.07 169.38 174.24 var75 | 10 168.29 1.37 165.20 171.38 var76 | 10 166.58 1.83 162.44 170.72 var77 | 10 169.60 0.97 167.42 171.79 var78 | 10 167.82 1.59 164.23 171.41 var79 | 10 166.57 1.97 162.12 171.02 var80 | 10 166.03 1.54 162.55 169.51 var81 | 10 167.85 1.36 164.77 170.93 var82 | 10 167.95 1.23 165.17 170.73 var83 | 10 169.40 1.20 166.68 172.13 var84 | 10 170.53 1.59 166.93 174.14 var85 | 10 170.45 1.32 167.48 173.43 var86 | 10 167.09 2.34 161.79 172.39 var87 | 10 166.03 1.70 162.19 169.87 var88 | 10 168.89 1.41 165.70 172.09 var89 | 10 169.31 1.86 165.11 173.50 var90 | 10 168.25 1.62 164.59 171.91 var91 | 10 166.20 1.20 163.48 168.93 var92 | 10 167.00 1.95 162.58 171.42 var93 | 10 168.28 1.60 164.65 171.90 var94 | 10 170.93 1.87 166.70 175.15 var95 | 10 163.68 0.87 161.70 165.65 var96 | 10 165.29 1.38 162.17 168.42 var97 | 10 169.60 1.64 165.88 173.32 var98 | 10 170.40 1.73 166.48 174.31 var99 | 10 166.90 2.03 162.31 171.50 var100 | 10 168.60 1.56 165.06 172.14 . xpose,clear . egen mean=rmean(v1-v10) . sum mean Variable | Obs Mean Std. Dev. Min Max -------------+----------------------------------------------------- mean | 100 167.5869 1.628082 163.259 171.808 . di 5.3/sqrt(10) 1.6760072 . histogram mean,bin(9) xlab(163(1)172) ylab freq gap(4) title("样本均数的抽样分布图") 30 20 Frequency 10 0 163164165166167168169170171172mean 样本均数的抽样分布图 例3-3 某地抽得正常成人200名，测得其血清胆固醇的均数为3.64 mmol/L，标准差为1.20mmol/L，试 估计该地正常成人血清胆固醇均数的95%可信区间。 例题3-3: . cii 200 3.64 1.20 Variable | Obs Mean Std. Err. [95% Conf. Interval] -------------+------------------------------------------------------------- | 200 3.64 .0848528 3.472674 3.807326 例3-4 为了解甲氨蝶呤(MTX)对外周血IL-2水平的影响，某医生将61名哮喘患者随机分为两组。其中 nn对照组29例()，采用安慰剂;试验组32例()，采用小剂量甲氨蝶呤(MTX)进行治疗。测得12 XS对照组治疗前IL-2的均数为20.10 IU/ml ()，标准差为7.02 IU/ml ();试验组治疗前IL-211 XS的均数为16.89 IU/ml ()，标准差为8.46 IU/ml ()。问两组治疗前基线的IL-2总体均 22 例题3-4: . ttesti 29 20.10 7.02 32 16.89 8.46 Two-sample t test with equal variances ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 29 20.1 1.303581 7.02 17.42973 22.77027 y | 32 16.89 1.495531 8.46 13.83984 19.94016 ---------+-------------------------------------------------------------------- combined | 61 18.41607 1.012939 7.911309 16.38989 20.44225 ---------+-------------------------------------------------------------------- diff | 3.21 2.002304 -.7966004 7.2166 ------------------------------------------------------------------------------ Degrees of freedom: 59 Ho: mean(x) - mean(y) = diff = 0 Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0 t = 1.6032 t = 1.6032 t = 1.6032 P < t = 0.9429 P > |t| = 0.1142 P > t = 0.0571数相差有多大, 例3-5 某医生测量了36名从事铅作业男性工人的血红蛋白含量，算得其均数为130.83g/L，标准差为 25.74g/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140g/L, 例题3-5(EX3-5.dta): . ttest x=140 One-sample t test ------------------------------------------------------------------------------ Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 36 130.8333 4.29017 25.74102 122.1238 139.5428 ------------------------------------------------------------------------------ Degrees of freedom: 35 Ho: mean(x) = 140 Ha: mean < 140 Ha: mean ~= 140 Ha: mean > 140 t = -2.1367 t = -2.1367 t = -2.1367 P < t = 0.0198 P > |t| = 0.0397 P > t = 0.9802 . ttesti 36 130.8333 25.74102 140 One-sample t test ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 36 130.8333 4.29017 25.74102 122.1238 139.5428 ------------------------------------------------------------------------------ Degrees of freedom: 35 Ho: mean(x) = 140 Ha: mean < 140 Ha: mean ~= 140 Ha: mean > 140 t = -2.1367 t = -2.1367 t = -2.1367 P < t = 0.0198 P > |t| = 0.0397 P > t = 0.9802 例3-6 为比较两种方法对乳酸饮料中脂肪含量测定结果是否不同，随机抽取了10份乳酸饮料制品，分别 用脂肪酸水解法和哥特里,罗紫法测定其结果如表3-5第(1)~(3)栏。问两法测定结果是否不同, 表3-5 两种方法对乳酸饮料中脂肪含量的测定结果(%) 编号 哥特里,罗紫法 脂肪酸水解法 差值d (1) (2) (3) (4)=(2),(3) 1 0.840 0.580 0.260 2 0.591 0.509 0.082 3 0.674 0.500 0.174 4 0.632 0.316 0.316 5 0.687 0.337 0.350 6 0.978 0.517 0.461 7 0.750 0.454 0.296 8 0.730 0.512 0.218 9 1.200 0.997 0.203 10 0.870 0.506 0.364 2.724 例题3-6(EX3-6.dta): . ttest x1=x2 Paired t test ------------------------------------------------------------------------------ Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x1 | 10 .7952 .0583003 .1843618 .6633155 .9270845 x2 | 10 .5228 .0588125 .1859814 .389757 .655843 ---------+-------------------------------------------------------------------- diff | 10 .2724 .034368 .1086812 .1946542 .3501458 ------------------------------------------------------------------------------ Ho: mean(x1 - x2) = mean(diff) = 0 Ha: mean(diff) < 0 Ha: mean(diff) ~= 0 Ha: mean(diff) > 0 t = 7.9260 t = 7.9260 t = 7.9260 P < t = 1.0000 P > |t| = 0.0000 P > t = 0.0000 例3-7 为研究国产四类新药阿卡波糖胶囊的降血糖效果，某医院用40名2型糖尿病病人进行同期随机对 照试验。试验者将这些病人随机等分到试验组(用阿卡波糖胶囊)和对照组(用拜糖平胶囊)，分别测得试验开 始前和8周后的空腹血糖，算得空腹血糖下降值见表3-6，能否认为该国产四类新药阿卡波糖胶囊与拜糖平 胶囊对空腹血糖的降糖效果不同, 表3-6 试验组和对照组空腹血糖下降值(mmol/L) 试验组X -0.70 -5.60 2.00 2.80 0.70 3.50 4.00 5.80 7.10 -0.50 1 (n=20) 2.50 -1.60 1.70 3.00 0.40 4.50 4.60 2.50 6.00 -1.40 1 对照组X 3.70 6.50 5.00 5.20 0.80 0.20 0.60 3.40 6.60 -1.10 2 (n=20) 6.00 3.80 2.00 1.60 2.00 2.20 1.20 3.10 1.70 -2.00 2 例题3-7(EX3-7.dta): . ttest x,by(group) Two-sample t test with equal variances ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- 1 | 20 2.065 .6842697 3.060147 .6328071 3.497193 2 | 20 2.625 .54124 2.420499 1.492172 3.757828 ---------+-------------------------------------------------------------------- combined | 40 2.345 .4329231 2.738046 1.46933 3.22067 ---------+-------------------------------------------------------------------- diff | -.56 .8724482 -2.326179 1.206179 ------------------------------------------------------------------------------ Degrees of freedom: 38 Ho: mean(1) - mean(2) = diff = 0 Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0 t = -0.6419 t = -0.6419 t = -0.6419 P < t = 0.2624 P > |t| = 0.5248 P > t = 0.7376 例3-8 在上述例3-7国产四类新药阿卡波糖胶囊的降血糖效果研究中，测得用拜糖平胶囊的对照组20例 病人和用阿卡波糖胶囊的试验组20例病人，其8周时糖化血红蛋白HbAc(%)下降值如表3-7。问用两种不1 同药物的病人其HbAc下降值是否不同, 1 表3-7 对照组和试验组HbAc下降值(%) 1 分 组 n S X 对照组 20 1.46 1.36 试验组 20 1.13 0.70 例题3-8: . ttesti 20 1.46 1.36 20 1.13 0.70,une Two-sample t test with unequal variances ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 20 1.46 .3041052 1.36 .8235004 2.0965 y | 20 1.13 .1565248 .7 .8023899 1.45761 ---------+-------------------------------------------------------------------- combined | 40 1.295 .1708602 1.080615 .9494026 1.640597 ---------+-------------------------------------------------------------------- diff | .33 .3420234 -.3701514 1.030151 ------------------------------------------------------------------------------ Satterthwaite's degrees of freedom: 28.4068 Ho: mean(x) - mean(y) = diff = 0 Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0 t = 0.9648 t = 0.9648 t = 0.9648 P < t = 0.8286 P > |t| = 0.3428 P > t = 0.1714 . ttesti 20 1.46 1.36 20 1.13 0.70,une w Two-sample t test with unequal variances ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 20 1.46 .3041052 1.36 .8235004 2.0965 y | 20 1.13 .1565248 .7 .8023899 1.45761 ---------+-------------------------------------------------------------------- combined | 40 1.295 .1708602 1.080615 .9494026 1.640597 ---------+-------------------------------------------------------------------- diff | .33 .3420234 -.3691062 1.029106 ------------------------------------------------------------------------------ Welch's degrees of freedom: 29.397 Ho: mean(x) - mean(y) = diff = 0 Ha: diff < 0 Ha: diff ~= 0 Ha: diff > 0 t = 0.9648 t = 0.9648 t = 0.9648 P < t = 0.8288 P > |t| = 0.3425 P > t = 0.1712 . 例3-9 试用矩法对表3-1中计算机模拟抽样所得100个样本均数进行正态性检验 (数据见前表3-1) 。 例题3-9: . drop _all . set seed 050322 . set obs 10 obs was 0, now 10 . quietly for num 1/100:gen varX=invnorm(uniform())*5.3+167.7 . xpose,clear . egen mean=rmean(v1-v10) . sktest mean Skewness/Kurtosis tests for Normality ------- joint ------ Variable | Pr(Skewness) Pr(Kurtosis) adj chi2(2) Prob>chi2 -------------+------------------------------------------------------- mean | 0.881 0.755 0.12 0.9418 -10 对例3-7，请用F检验判断两总体空腹血糖下降值的方差是否不等。 例3 例题3-10(EX3-7.dta): . sdtest x,by(group) Variance ratio test ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- 1 | 20 2.065 .6842697 3.060147 .6328071 3.497193 2 | 20 2.625 .54124 2.420499 1.492172 3.757828 ---------+-------------------------------------------------------------------- combined | 40 2.345 .4329231 2.738046 1.46933 3.22067 ------------------------------------------------------------------------------ Ho: sd(1) = sd(2) F(19,19) observed = F_obs = 1.598 F(19,19) lower tail = F_L = 1/F_obs = 0.626 F(19,19) upper tail = F_U = F_obs = 1.598 Ha: sd(1) < sd(2) Ha: sd(1) ~= sd(2) Ha: sd(1) > sd(2) P < F_obs = 0.8424 P < F_L + P > F_U = 0.3153 P > F_obs = 0.1576 例3-11 对例3-8，请用F检验判断对照组和试验组病人HbAc(%)下降值的总体方差是否不等。 1 例题3-11: . sdtesti 20 1.46 1.36 20 1.13 0.70 Variance ratio test ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 20 1.46 .3041052 1.36 .8235004 2.0965 y | 20 1.13 .1565248 .7 .8023899 1.45761 ---------+-------------------------------------------------------------------- combined | 40 1.295 .1708602 1.080615 .9494026 1.640597 ------------------------------------------------------------------------------ Ho: sd(x) = sd(y) F(19,19) observed = F_obs = 3.775 F(19,19) lower tail = F_L = 1/F_obs = 0.265 F(19,19) upper tail = F_U = F_obs = 3.775 Ha: sd(x) < sd(y) Ha: sd(x) ~= sd(y) Ha: sd(x) > sd(y) P < F_obs = 0.9971 P < F_L + P > F_U = 0.0057 P > F_obs = 0.0029 例4-1 某医生为了研究一种降血脂新药的临床疗效，按统一纳入标准选择120名患者，采用完全随机设计方法将患者等分为4组进行双盲试验。问如何进行分组, 例题4-1: . drop _all . set obs 120 obs was 0, now 120 . gen id =_n . set seed 050323 . gen r=uniform() . sort r . gen treat=group(4) . list id r treat 1. 25 .0097145 1 2. 49 .0115311 1 3. 77 .0149384 1 4. 23 .0373307 1 5. 3 .0525425 1 6. 7 .0916106 1 7. 101 .0992928 1 8. 11 .099738 1 9. 102 .1009754 1 10. 78 .1012358 1 11. 12 .1049003 1 12. 117 .1056932 1 13. 19 .1135953 1 14. 9 .1182419 1 15. 87 .1236629 1 16. 56 .1280095 1 17. 66 .128401 1 18. 31 .1398393 1 19. 40 .1411199 1 20. 92 .1436865 1 21. 36 .1561403 1 22. 108 .1726452 1 23. 26 .172652 1 24. 8 .1749743 1 25. 114 .1861417 1 26. 55 .1996692 1 27. 38 .2088119 1 28. 17 .2416117 1 29. 79 .2514814 1 30. 29 .265756 1 31. 18 .2705794 2 32. 32 .2962818 2 33. 71 .297382 2 34. 97 .3019295 2 35. 28 .3051004 2 36. 86 .3072144 2 37. 16 .310899 2 38. 99 .3155963 2 39. 44 .3252935 2 40. 20 .3308867 2 41. 57 .334819 2 42. 27 .3732665 2 43. 107 .3835843 2 44. 109 .3836386 2 45. 83 .3853779 2 46. 80 .3899993 2 47. 45 .3995801 2 48. 111 .4147781 2 49. 98 .4163906 2 50. 46 .4204507 2 51. 96 .4225546 2 52. 90 .432752 2 53. 22 .4387289 2 54. 39 .4402532 2 55. 13 .4532539 2 56. 88 .469486 2 57. 42 .4870647 2 58. 76 .4933496 2 59. 52 .5007715 2 60. 61 .5123829 2 61. 51 .51855 3 62. 68 .5363081 3 63. 73 .5425012 3 64. 43 .5569742 3 65. 82 .562066 3 66. 35 .5621918 3 67. 62 .5644192 3 68. 104 .5692632 3 69. 75 .5735795 3 70. 84 .5802925 3 71. 113 .5853601 3 72. 118 .5888116 3 73. 65 .5891058 3 74. 15 .5920883 3 75. 116 .5928208 3 76. 10 .5994575 3 77. 24 .6161131 3 78. 93 .6194515 3 79. 60 .6268067 3 80. 115 .6550769 3 81. 64 .6562381 3 82. 5 .6615584 3 83. 106 .6622884 3 84. 95 .6650478 3 85. 94 .6652233 3 86. 70 .6655526 3 87. 34 .672056 3 88. 21 .698502 3 89. 69 .6995025 3 90. 1 .7129369 3 91. 119 .7130946 4 92. 110 .7300173 4 93. 48 .7426195 4 94. 37 .7506956 4 95. 105 .751996 4 96. 74 .7546608 4 97. 85 .7708155 4 98. 112 .7745966 4 99. 41 .7877266 4 100. 81 .8045244 4 101. 33 .8229868 4 102. 120 .8457658 4 103. 50 .8584244 4 104. 4 .8790787 4 105. 2 .8856059 4 106. 30 .8878745 4 107. 63 .8897925 4 108. 53 .8912752 4 109. 6 .8947296 4 110. 100 .9076214 4 111. 72 .9080909 4 112. 91 .9109184 4 113. 89 .9205835 4 114. 67 .9230537 4 115. 103 .9371982 4 116. 47 .9607754 4 117. 58 .9620151 4 118. 54 .9672734 4 119. 14 .9880295 4 120. 59 .9993106 4 例4-2 某医生为了研究一种降血脂新药的临床疗效，按统一纳入标准选择120名高血脂患者，采用完全随机设计方法将患者等分为4组(具体分组方法见例4-1)，进行双盲试验。6周后测得低密度脂蛋白作为试验结果，见表4-3。问4个处理组患者的低密度脂蛋白含量总体均数有无差别? 表4-3 4个处理组低密度脂蛋白测量值(mmol/L) 统计量 分 组 测量值 2 n XXX,,i 3.53 4.59 4.34 2.66 3.59 3.13 2.64 2.56 3.50 3.25 3.30 4.04 3.53 3.56 3.85 4.07 3.52 3.93 4.19 2.96 安慰剂组 30 3.43 102.91 367.85 1.37 3.93 2.33 2.98 4.00 3.55 2.96 4.3 4.16 2.59 降血脂新药 2.42 3.36 4.32 2.34 2.68 2.95 1.56 3.11 1.81 1.77 30 2.72 81.46 233.00 2.4g组 1.98 2.63 2.86 2.93 2.17 2.72 2.65 2.22 2.90 2.97 2.36 2.56 2.52 2.27 2.98 3.72 2.80 3.57 4.02 2.31 2.86 2.28 2.39 2.28 2.48 2.28 3.21 2.23 2.32 2.68 2.66 2.32 2.61 3.64 2.58 3.65 2.66 3.68 2.65 3.02 4.8g组 30 2.70 80.94 225.54 3.48 2.42 2.41 2.66 3.29 2.70 3.04 2.81 1.97 1.68 0.89 1.06 1.08 1.27 1.63 1.89 1.19 2.17 2.28 1.72 1.98 1.74 2.16 3.37 2.97 1.69 0.94 2.11 2.81 2.52 7.2g组 30 1.97 58.99 132.13 1.31 2.51 1.88 1.41 3.19 1.92 2.47 1.02 2.10 3.71 例题4-2(EX4-2.dta): . one x group Analysis of Variance Source SS df MS F Prob > F ------------------------------------------------------------------------ Between groups 32.1560309 3 10.718677 24.88 0.0000 Within groups 49.967022 116 .43075019 ------------------------------------------------------------------------ Total 82.1230529 119 .690109688 Bartlett's test for equal variances: chi2(3) = 5.2192 Prob>chi2 = 0.156 例4-3 如何按随机区组设计，分配5个区组的15只小白鼠接受甲、乙、丙三种抗癌药物, 例题4-3: . drop _all . set obs 15 obs was 0, now 15 . egen id=seq() . egen b=seq(),block(3) . set seed 050323 . gen r=uniform() . sort b r . gen group=mod(_n-1,3)+1 . sort id b . l id b g id b group 1. 1 1 2 2. 2 1 3 3. 3 1 1 4. 4 2 2 5. 5 2 1 6. 6 2 3 7. 7 3 1 8. 8 3 3 9. 9 3 2 10. 10 4 3 11. 11 4 1 12. 12 4 2 13. 13 5 1 14. 14 5 3 15. 15 5 2 例4-4 某研究者采用随机区组设计进行实验，比较三种抗癌药物对小白鼠肉瘤抑瘤效果，先将15只 染有肉瘤小白鼠按体重大小配成5个区组，每个区组内3只小白鼠随机接受三种抗癌药物(具体分配方法 见例4-3)，以肉瘤的重量为指标，试验结果见表4-9。问三种不同的药物的抑瘤效果有无差别, 表4-9 不同药物作用后小白鼠肉瘤重量(g) g X区组 A药 B药 C药 ,iji,1 1 0.82 0.65 0.51 1.98 2 0.73 0.54 0.23 1.50 3 0.43 0.34 0.28 1.05 4 0.41 0.21 0.31 0.93 5 0.68 0.43 0.24 1.35 n ijX,()X,,3.07 2.17 1.57 6.81 ijj,1 0.614 0.434 0.314 0.454 ()XX i n 22X,2.0207 1.0587 0.5451 3.6245 ()X ,,ijij,j1 例题4-4(EX4-4.dta): . anova x group treat Number of obs = 15 R-squared = 0.8566 Root MSE = .097724 Adj R-squared = 0.7490 Source | Partial SS df MS F Prob > F -----------+---------------------------------------------------- Model | .456359996 6 .076059999 7.96 0.0050 | group | .228359987 4 .057089997 5.98 0.0158 treat | .228000009 2 .114000005 11.94 0.0040 | Residual | .076400007 8 .009550001 -----------+---------------------------------------------------- Total | .532760003 14 .038054286 2例4-5 某研究者为了比较甲、乙、丙、丁、戊、己 6种药物给家兔注射后产生的皮肤疱疹大小(mm)， 采用拉丁方设计，选用6只家兔、并在每只家兔的6个不同部位进行注射。试验结果见表4-11，试作方差 分析。 2表4-11 例4-5拉丁方设计与试验结果(皮肤疱疹大小，mm) 注射部位编号(列区组) 家兔编号 行区组 X i(行区组) 合计(R) 1 2 3 4 5 6 j 1 C(87) B(75) E(81) D(75)A (84) F(66) 468.0 78.0 2 B(73) A(81)D (87) C(85) F(64) E(79) 469.0 78.2 3 F(73) E(73) B(74) A(78)D (73) C(77) 448.0 74.7 4 A(77) F(68) C(69) B(74) E(76) D(73) 437.0 72.8 5 D(64) C(64) F(72) E(76) B(70) A(81) 427.0 71.2 6 E(75) D(77)A (82) F(61) C(82) B(61) 438.0 73.0 列区组 449.0 438.0 465.0 449.0 449.0 437.0 合计(C) i X74.8 73.0 77.5 74.8 74.8 72.8 j 药物 D E C A B F X=74.6 合计T449.0 483.0 464.0 483.0 427.0 404.0 k 74.8 80.5 77.3 80.5 71.2 67.3 X k 例题4-5(EX4-5.dta): . anova x treat row col Number of obs = 36 R-squared = 0.5829 Root MSE = 5.93025 Adj R-squared = 0.2701 Source | Partial SS df MS F Prob > F -----------+---------------------------------------------------- Model | 982.947661 15 65.5298441 1.86 0.0965 | treat | 657.336272 5 131.467254 3.74 0.0149 row | 251.662939 5 50.3325877 1.43 0.2561 col | 65.3365497 5 13.0673099 0.37 0.8621 | Residual | 703.357895 20 35.1678947 -----------+---------------------------------------------------- Total | 1686.30556 35 48.1801587 例4-7 对例4-2资料，问高血脂患者的降血脂新药2.4g组、4.8g组、7.2g组与安慰剂组的低密度脂 蛋白含量总体均数有无差别, 例题4-7(EX4-2.dta): . one x g ,t sch noanova 1='placebo' | ,2='newdrug | 2.4g',3='ne | wdrug | 4.8g,4='new | Summary of x drug 7.2g' | Mean Std. Dev. Freq. ------------+------------------------------------ 1 | 3.43 0.72 30 2 | 2.72 0.64 30 3 | 2.70 0.50 30 4 | 1.97 0.75 30 ------------+------------------------------------ Total | 2.70 0.83 120 Comparison of x by 1='placebo',2='newdrug 2.4g',3='newdrug 4.8g,4='newdrug 7.2g' (Scheffe) Row Mean-| Col Mean | 1 2 3 ---------+--------------------------------- 2 | -0.72 | 0.001 | 3 | -0.73 -0.02 | 0.001 1.000 | 4 | -1.46 -0.75 -0.73 | 0.000 0.000 0.001 例4-9 对例4-4资料，问三种不同药物的抑瘤效果两两之间是 否有差别, 例题4-9(EX4-4.dta): . anova x group treat Number of obs = 15 R-squared = 0.8566 Root MSE = .097724 Adj R-squared = 0.7490 Source | Partial SS df MS F Prob > F -----------+---------------------------------------------------- Model | .456359996 6 .076059999 7.96 0.0050 | group | .228359987 4 .057089997 5.98 0.0158 treat | .228000009 2 .114000005 11.94 0.0040 | Residual | .076400007 8 .009550001 -----------+---------------------------------------------------- Total | .532760003 14 .038054286 . test _b[treat[1]]=_b[treat[2]] ( 1) treat[1] - treat[2] = 0.0 F( 1, 8) = 8.48 Prob > F = 0.0195 . di min(1,r(p)*3) .05855772 . di min(1,1-(1-r(p))^3) .05742215 . di 1-F(2,8,r(F)/2) .05550787 例4-10 对例4-2资料，试分析各处理组的低密度脂蛋白值是否满足方差齐性, 例题4-10(EX4-2.dta): . one x g Analysis of Variance Source SS df MS F Prob > F ------------------------------------------------------------------------ Between groups 32.1560309 3 10.718677 24.88 0.0000 Within groups 49.967022 116 .43075019 ------------------------------------------------------------------------ Total 82.1230529 119 .690109688 Bartlett's test for equal variances: chi2(3) = 5.2192 Prob>chi2 = 0.156