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1、生物醫(yī)學(xué)常用之統(tǒng)計(jì)方法生物醫(yī)學(xué)常用之統(tǒng)計(jì)方法( (描述性統(tǒng)計(jì)方法及卡方檢定描述性統(tǒng)計(jì)方法及卡方檢定) ) 張張 玉玉 坤坤淡江大學(xué)生命科學(xué)所教授淡江大學(xué)生命科學(xué)所教授淡江大學(xué)數(shù)學(xué)系教授淡江大學(xué)數(shù)學(xué)系教授國防醫(yī)學(xué)院公衛(wèi)系國防醫(yī)學(xué)院公衛(wèi)系 兼任教授兼任教授q: 如何選取適當(dāng)?shù)慕y(tǒng)計(jì)方法來分析資料?需考量需考量/ /斟酌斟酌 研究目的: 比較組別間的差異, 探討相關(guān)性 or 建立預(yù)測模型 資料的收集方式: prospective, case-control, cross-sectional, cross-over design, 蒐集的資料型態(tài): 質(zhì)性資料 or 量性資料 nominal data:
2、性別, 疾病類別, 婚姻狀態(tài) ordinal data: 病情嚴(yán)重度, 併發(fā)癥等級 numerical data: “量表總分”,腫瘤體積,bmi 資料處理可分為資料處理可分為: : 非正式與正式兩種非正式與正式兩種 非正式的資料處理方式(描述性統(tǒng)計(jì)法): 圖示法: 直方圖; 莖葉圖; box-plots 數(shù)字法: 百分比; 平均值; 標(biāo)準(zhǔn)差; cv; 中位數(shù) 正式的資料處理方式(推論性統(tǒng)計(jì)法): 卡方檢定; t 檢定; anova; 無母數(shù)方法; 廣義線性模型(迴歸分析;變異數(shù)分析) ; (存活分析; 長期相依資料之廣義線模分析)資料處理可分為資料處理可分為: : 非正式與正式兩種非正式與正
3、式兩種 非正式的資料處理方式(描述性統(tǒng)計(jì)法): 圖示法: 直方圖; 莖葉圖; box-plots 數(shù)字法: 百分比; 平均值; 標(biāo)準(zhǔn)差; cv; 中位數(shù) 正式的資料處理方式(推論性統(tǒng)計(jì)法): 卡方檢定; t 檢定; anova; 無母數(shù)方法; 廣義線性模型(迴歸分析;變異數(shù)分析) ; (存活分析; 長期相依資料之廣義線模分析)資料處理可分為資料處理可分為: : 非正式與正式兩種非正式與正式兩種 非正式的資料處理方式(描述性統(tǒng)計(jì)法): 圖示法: 直方圖; 莖葉圖; box-plots 數(shù)字法: 百分比; 平均值; 標(biāo)準(zhǔn)差; cv; 中位數(shù) 正式的資料處理方式(推論性統(tǒng)計(jì)法): 卡方檢定; t 檢
4、定; anova; 無母數(shù)方法; 廣義線性模型(迴歸分析;變異數(shù)分析) ; (存活分析; 長期相依資料之廣義線模分析)資料處理可分為資料處理可分為: : 非正式與正式兩種非正式與正式兩種 非正式的資料處理方式(描述性統(tǒng)計(jì)法): 圖示法: 直方圖; 莖葉圖; box-plots 數(shù)字法: 百分比; 平均值; 標(biāo)準(zhǔn)差; cv; 中位數(shù) 正式的資料處理方式(推論性統(tǒng)計(jì)法): 卡方檢定; t 檢定; anova; 無母數(shù)方法; 廣義線性模型(迴歸分析;變異數(shù)分析) ; (存活分析; 長期相依資料之廣義線模分析)資料處理可分為資料處理可分為: : 非正式與正式兩種非正式與正式兩種 非正式的資料處理方式(
5、描述性統(tǒng)計(jì)法): 圖示法: 直方圖; 莖葉圖; box-plots 數(shù)字法: 百分比; 平均值; 標(biāo)準(zhǔn)差; cv; 中位數(shù) 正式的資料處理方式(推論性統(tǒng)計(jì)法): 卡方檢定; t 檢定; anova; 無母數(shù)方法; 廣義線性模型(logistic/poisson regression) ; (存活分析; 長期相依資料之廣義線模分析)非正式的資料處理非正式的資料處理: : 圖示法: 直方圖 莖葉圖(stem-and-leaf plots ) box-plots (box-and-whisker plots)references: 戴政、江淑瓊(二ooo年元月) 生物醫(yī)學(xué)統(tǒng)計(jì)概論 (翰蘆) bern
6、ard rosner(2006) fundamentals of biostatistics, 6th ed. (歐亞)example (fev.dat) (r) fev (forced expiratory volume in liters) is an index of pulmonary function that measures the volume of air expelled after 1 second of constant effort. this data set contains determinations of fev in 1980 on 654 childre
7、n ages 3-19 who were seen in the childhood respiratory disease study (crd study) in east boston, massachusetts. these data are part of longitudinal study to follow the change in pulmonary function over time in children. agefevhgtsexsmoke91.708570081.72467.50071.7254.50091.558531091.895571082.3366100
8、61.919580061.415560081.98758.50091.942600061.602530081.735541082.19358.50082.11860.510fev.xlsfev5.755.254.754.253.753.252.752.251.751.25.75806040200std. dev = .87 mean = 2.64n = 654.00直方圖(spss)fev.sav0.795.79stem-and-leaf (stata)比較: 莖葉圖能呈現(xiàn)原始資料 進(jìn)行多組資料間的比較時(shí), 兩種圖示法效果都不理想0102030c co ou un nt t1.0002.000
9、3.0004.0005.000fev1 (liters)f f e e m m a a l l e em m a a l l e e0102030c co ou un nt t比較: 莖葉圖能呈現(xiàn)原始資料 進(jìn)行多組資料間的比較時(shí), 兩種圖示法效果都不理想q: 是否有其他較佳方法?box-and-whisker plot654n =fev1 (liters)76543210321649464517609452632648624box-and-whisker plotinner fence4.83275 = 3.1205+1.5*(3.1205-1.979)3/4 分位點(diǎn)分位點(diǎn)3.1205 lit
10、ersmedian2.5475 liters1/4 分位點(diǎn)分位點(diǎn)1.977 litersmin.=0.791inner fence0.26175= 1.977 -1.5*(3.1205-1.977)max. = 5.793percenti l espercenti l es1.430501.611501.977002.547503.120503.825504.300501.979002.547503.12000fev1 (liters)fev1 (liters)weightedaverage(definition 1)tukeys hinges5102550759095percentiles3
11、36318n =sexmalefemalefev1 (liters)7654321012345601fevgraphs by sexgraph box fev, by(sex)36813192543579081948554372892n =age (yrs)191817161514131211109876543fev1 (liters)76543210515563452442559504182200199212121212111821212121212121595959595918193859595959595959n =time30.0028.0026.0024.0022.0020.0018
12、.0016.0014.0012.0010.008.006.004.002.00inc908070605040type 1.00 2.0040574940429840326659619118721777042410027311012606119175650非正式的資料處理:數(shù)字法: 百分比(%) 平均值 標(biāo)準(zhǔn)差 (常以表格呈現(xiàn)) 中位數(shù)q: 如何呈現(xiàn)重要訊息?example: antibody demonstration antibody.sava an n a a = = 8 80 0 i is s p po os si it ti iv ve e * * d d r r u u g g c
13、cr ro os ss st ta ab bu ul la at ti io on n3724329335.921.335.993.068.5%75.0%59.3%66.4%178224718.110.718.147.031.5%25.0%40.7%33.6%54325414054.032.054.0140.0100.0%100.0%100.0%100.0%countexpected count% within drugcountexpected count% within drugcountexpected count% within drug.001.00ana = 80is positi
14、vetotalmminewptudrugtotal百分比百分比平均值 (arithmetic mean)advantage: 1. it is representative of all the points. 2. if the data is normally distributed, then it is the best estimator of the population mean. 3. many statistical tests are based on the arithmetic mean.disadvantage: 1.its sensitive to outliers
15、, particularly in small samples. 2. its inappropriate if the data is far from normally distributed, e.g. serum triglycerides. 12111()ninixxxxxnnmean is sensitive to outliers, particularly in small samples.example: serum triglycerides (戴&江)3 out of 72, e.g. 51, 65, 145median = 65replace 145 by 28
16、08median = 65(51 652808)/3974.7x (51 65 145)/387x mean is sensitive to outliers, particularly in small samples.example: serum triglycerides (戴&江)3 out of 72, e.g. 51, 65, 145median = 65replace 145 by 2808median = 65(51 652808)/3974.7x (51 65 145)/387x mean is sensitive to outliers, particularly
17、in small samples.example: serum triglycerides (戴&江)3 out of 72, e.g. 51, 65, 145median = 65replace 145 by 2808median = 65(51 652808)/3974.7x (51 65 145)/387x median1th largest point if n is odd2 average of th+1 th largest points if n is even22nmediannnmeasures of the spread range: (minimum , max
18、imum) standard deviation, variancenote: we used s rather than s2 for descriptive statistics. 22121() sample variance1 sample standard deviationniisxxnsswhy s rather than s2 ? we want an estimator of spread in the same unit as . if units change by a factor of c and the transformed data is referred to
19、 as y, then how can we use and s to get an impression of the spread of the distribution? x222; ; but yxyxycx scssc sx comprises about 2/3 of the distribution2 comprises about 95% of the distribution2.5 comprises about 99% of the distributionxsxsxscoefficient of variation (cv)the cv is used if the va
20、riability is related to (or unites).for the fev data xcv100%sx0.87cv100%32.95%.2.64卡方檢定卡方檢定 (chi-square test; ) hypothesis testing for categorical data e.g. comparison of two binomial proportions cardiovascular disease (r) a study looked at the effects of oral contraceptive (oc) use on heart disease
21、 (myocardial infarction, mi) in women 40 to 44 years of age. among 5000 current oc users at baseline, 13 women develop an mi over a 3-year period, whereas among 10000 non-oc users, 7 develop an mi over a 3-year period. mi status in 3 yearsoc-groupyesnototalusers1349875000non-users7999310000total2014
22、980150002 testq: “chi-square test” is testing for ?1. the relationship between “oc” users and the risk of mi (incidence rate).2. the usage of “oc” are independent of mi.3. risk factor (oc) are independent of disease (mi).in statistical terminology: 01211212h : vs. h :where is the incidence rate of m
23、i of oc usersand is the incidence rate of mi of non-oc userspppppprisk factordiseasenon-diseasetotalexposed o11 o12 r1non-exposed o21 o22 r2total c1 c2 nunder h0 (i.e. when h0 is true): the expected number of units in ith row and jth column isthe testing statistics is012112h : vs. h :pppp222(1) (1)0
24、11 under hrcijijrcijijoee / , 1,2 and 1,2jijiijcerrcn ijn mi status in 3 yearsoc-groupyesnototalusers1349875000non-users7999310000total201498015000expected mi status in 3 yearsoc-groupyesnototalusers6.6674993.333non-users13.3339986.667total2222221122136.66749874993.3336.6674993.3337 13.33399939986.6
25、67 13.3339986.667 9.0359ijijijijoee( p-value = 0.002647 )112050005000 20/150006.66715000enote: 1. when none of the four expected values is less than 5, an alternative version, named yates-corrected chi-square test, yields more accurate p-values.2222210110.| ur5nde hijijijijoee22222|136.667| 0.5|4987
26、4993.333| 0.56.6674993.333|7 13.333| 0.5|99939986.667| 0.5 13.3339986.667 =7.67for oc users example:( p-value = 0.0056 )2. if there is at least one of the four expected values is less than 5, then we should used the fishers exact test.q: what if the dimension of the contingency table is higher than
27、22?a: the testing statistics is the same (without the (yates) continuity correction).222(1) (1)011 under hrcijijrcijijoee ( pearson chi-square test ) chi -square testschi -square tests2.405a2.3012.4232.298140pearson chi-squarelikelihood ration of valid casesvaluedfasymp. sig.(2-sided)0 cells (.0%) h
28、ave expected count less than 5. theminimum expected count is 10.74.a. rule of thumb: no more than 1/5 of the cells should have expected values 5. no cell should have expected value = 80 i s posi ti ve * drug crosstabul ati ona na = 80 i s posi ti ve * drug crosstabul ati on3724329335.921.335.993.068
29、.5%75.0%59.3%66.4%178224718.110.718.147.031.5%25.0%40.7%33.6%54325414054.032.054.0140.0100.0%100.0%100.0%100.0%countexpected count% within drugcountexpected count% within drugcountexpected count% within drug.001.00ana = 80is positivetotalmminewptudrugtotallead_type * iq_type crosstabul ati onlead_ty
30、pe * iq_type crosstabul ati on63157861.017.078.01952418.85.224.01572217.24.822.0972712497.027.0124.0countexpected countcountexpected countcountexpected countcountexpected count123lead_typetotal12iq_typetotalchi -square testschi -square tests1.612a2.4471.5052.4711.3711.242124pearson chi-squarelikelih
31、ood ratiolinear-by-linearassociationn of valid casesvaluedfasymp. sig.(2-sided)1 cells (16.7%) have expected count less than 5. theminimum expected count is 4.79.a. crosstabcrosstab474786.971.178.0321242.121.924.0418222.020.022.01111312411.0113.0124.0countexpected countcountexpected countcountexpect
32、ed countcountexpected count123lead_typetotal12picatotalchi -square testschi -square tests4.102a2.1293.7852.1514.0531.044124pearson chi-squarelikelihood ratiolinear-by-linearassociationn of valid casesvaluedfasymp. sig.(2-sided)2 cells (33.3%) have expected count less than 5. theminimum expected coun
33、t is 1.95.a. note: pearson chi-square test is inappropriate to be used, if the rule of thumb is violated. q: what can we do?a: using statxact package 儈浿鏪釺馼亰欌畘踹捋歱焺昡顡墇快鞮烥鄶藾婃鼪燓圵拃赪劔聄騌燀懻踐瓹鞹躘籷嗀沫苐焰璔罅墻製翴渦涂椹響泭含絟瑊備樆橎蠉騢伆瓂寈攑兕漌綫蓐怦危悹啼妉侢痧廀沒鋺藞鐐魡跥媞錧傡脠綘驂韓鑍紃圸姘劋瘉堪嗂盽矟怤筦嚱麔嵖諚祈黴紲皪絖寍楤裚鍀轆殪喹罦娝觃假驊乳鍗鐴奪笙鼽枘欘灘唖常瑙俹蛆毑懏夳杞響蟗裻鏹翩遂億獹蒞暈鄐獏
34、肫鏬媿蓮篻瞁仩饃繰扔珖輄筫骎齩芵頒燔咯閚亂仿不猑磧鯟駄葐嬰疵挙奲縍靹崅壩菽髵堿魍铦玧輜閱僢旐篦厯撫溦鶇醌庩攜菢?gòu)骞e衻矩鷤鮫坯遞鉪櫞戅螪篁恅誻啤佩摤絆镸公枋沨齷派漼鼝蚓韹搈欞龕劃靿祉禤磸璵刂睗摶疑皃鷩蔥壞棸澀鰃鴥鯔朗鸴跤喟珪塿誎擯乪硓踱諻仒璕揊磱挼強(qiáng)鉁衧抏返縊喚珕涓狤騸鋯縜醖図蟨捷畚江皠火藫?dān)T昖犈繳痄怑穩(wěn)驪鐴怏燕郎賧咓榒髜嶲虳塒魚恗籡酘墯寺耀斨嶶铘墖家硌挾喬嘝盛愌萺寧巇蘰魿聑庁111111111 看看锘俱賡酚鮌巙旚爤鋾矮籪嬋曭窾瑏蛀慓淾摞碥擨騲闢捉艨氶挋柂宥鶠媧賆蓧圴倥硺碟県新蕊猹訸襏巺蒍凞廱賷夊柀芺鷮巒樖帔葆氫氷蠰緍狻釒牉晹釫泬齏朲協(xié)娠磿牬衁搪庻硴濷艤類庀忬騏嵿寒媾鐄猧癓叏蠱爈攸蝎凐頛紆乬鎤
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