医学统计学6.Chisqua课件

1、医学统计学6.Chisqua医学统计学6.ChisquaMedical Statistics (full English class)Ji-Qian FangSchool of Public Health Sun Yat-Sen University医学统计学6.ChisquaChapter 10 Statistical Analysis of Enumeration Data医学统计学6.ChisquaVocabulary for Chapter 10 (I) s st ta at ti is st ti ic ca al l d de es sc cr ri ip pt ti io on

2、n 统计描述统计描述 e en nu umme er ra at ti io on n d da at ta a 计数资料计数资料 a ab bs so ol lu ut te e mme ea as su ur re e 绝对量绝对量 r re el la at ti iv ve e mme ea as su ur re es s 相对量相对量 c ca at te eg go or ry y 类别类别 f fr re eq qu ue en nc cy y 频数、频率频数、频率 r re el la at ti iv ve e f fr re eq qu ue en nc cy y 相对频

3、数、频率相对频数、频率 p pr ro op po or rt ti io on n 比率比率 i in nt te en ns si it ty y 强度强度 r ra at te e 速率速率 r ra at ti io o 比比 d de en no ommi in na at to or r 分母分母 n nu umme er ra at to or r 分子分子 医学统计学6.Chisquap po oo ol le ed d e es st ti imma at te e 联联合合估估计计 mmy yo op pi ia a 近近视视眼眼 b ba al la an nc ce e

4、 均均衡衡 s st ta an nd da ar rd di iz za at ti io on n 标标准准化化 d di ir re ec ct t s st ta an nd da ar rd di iz za at ti io on n 直直接接标标准准化化 i in nd di ir re ec ct t s st ta an nd da ar rd di iz za at ti io on n 间间接接标标准准化化 s st ta an nd da ar rd d p po op pu ul la at ti io on n 标标准准人人口口 s st ta an nd da a

5、r rd d mmo or rt ta al li it ty y r ra at te es s 标标准准死死亡亡率率 s st ta an nd da ar rd di iz ze ed d mmo or rt ta al li it ty y r ra at te e 标标准准化化死死亡亡率率 s st ta an nd da ar rd d mmo or rt ta al li it ty y r ra at ti io o 标标准准死死亡亡率率比比 w we ei ig gh ht te ed d a av ve er ra ag ge e 加加权权平平均均 i in nc ci i

6、d de en nc ce e r ra at te es s 发发病病率率 医学统计学6.Chisqua医学统计学6.Chisqua Absolute measure:Absolute measure: l The numbers counted for each The numbers counted for each category (frequencies)category (frequencies)l The absolute measure can hardly be The absolute measure can hardly be used for comparison b

7、etween used for comparison between different populations. different populations. 医学统计学6.Chisqua Three kinds of relative measures: Three kinds of relative measures: Frequency (Proportion) Frequency (Proportion) Intensity (Rate) Intensity (Rate) Ratio Ratio医学统计学6.Chisqua condition certain with possibl

8、ely units ofnumber totalTheconditioncertain with units ofnumber Thefrequency Relative ProportionNote: The Chinese text book is wrong!It is not “rate”!It is proportion or frequency!医学统计学6.ChisquaExample 10-1(P.304, revised) %16.15)gradeFirst Myopia(P%89.15)grade SecondMyopia(P%36.18)grade ThirdMyopia

9、(P%08.35)MyopiagradeFirst (P%60.35)Myopiagrade Second(P%32.29)Myopiagrade Third(PQuestion: Which grade has the most serious condition of myopias?Table 10-1 Prevalence rates and constitute of myopia in a junior high school Grade Number of students tested Number of students with myopia Prevalence rate

10、 (%) Constitute Among myopias (%) First grade 442 67 15.16 35.08 Second grade 428 68 15.89 35.60 Third grade 305 56 18.36 29.32 Total 191 100.00 医学统计学6.ChisquaPrevalence rates describe : P(Myopia|First grade) P(Myopia|Second grade) P(Myopia|Third grade)Constitute among myopias describe: P(First grad

11、e | Myopia) P(Second grade | Myopia) P(Third grade | Myopia)Which grade has the most serious condition of myopias?Answer: P(Myopia|Third grade) = Maximum -The third grade has the highest prevalence of myopias P(Second grade | Myopia)= Maximum - Among the myopias, the absolute number of Second grade

12、students is the highest. 医学统计学6.Chisqua year theduring disease ofrisk the toexposing yearsperson year theduring occuring patients ofNumber yearcertain in rate Incidence医学统计学6.Chisqua year theduringdeath ofrisk the toexposing yearsperson year theduring deaths ofNumber yearcertain in rateMortality 医学统

13、计学6.Chisqua period in the observed years-person Totalperiod in the appearing events ofNumber periodcertain in Intensity In general, Denominator: Sum of the person-years observed in the periodNumerator: Total number of the event appearing in the periodUnit: person/person year, or 1/YearNature: the re

14、lative frequency per unit of time.医学统计学6.ChisquaRatio is a number divided by another related numberExamples Sex ratio of students in this class: No. of males : No. of females = 52% Coefficient of variation: CV=SD/mean Ratio of time spent per clinic visit: Large hospital : Community health station =

15、81.9 min. : 18.6 min. = 4.40医学统计学6.Chisquaa.a.The denominator should be big enough! The denominator should be big enough! Otherwise the absolute measure should be used.Otherwise the absolute measure should be used. Example: Out of 5 cases, 3 were cured 60% ? Example: Out of 5 cases, 3 were cured 60%

16、 ?b.b. Attention to the population where the relative Attention to the population where the relative measure comes from.measure comes from. Mistake in the textbook (P.305) : Mistake in the textbook (P.305) : “Distinguish between constitutes and “Distinguish between constitutes and proportion” !?prop

17、ortion” !? We should sayWe should say “Distinguish between Prevalence “Distinguish between Prevalence rate and Constitute among patients”rate and Constitute among patients” Prevalence rate: Population is the students in thePrevalence rate: Population is the students in thesame grade same grade Const

18、itutes: Population is all the patients Constitutes: Population is all the patients 医学统计学6.ChisquaTable 10-2 Constitute of infectious diseases in a city (Frequency distribution among patients) 1985 1995 Infectious disease No. of cases (Frequency) Relative frequency (%) No. of cases (Frequency) Relati

19、ve frequency (%) Diarrhea 3604 49.39 2032 37.92 Hepatitis 1203 16.49 1143 21.33 Epidemic encephalitis 698 9.56 542 10.11 Measles 890 12.20 767 14.31 Others 902 12.36 875 16.33 Total 7297 100.00 5359 10.00 The above two frequency distributions reflect two populations of all patients; To describe the

20、prevalence rate, one has to look at the general population;医学统计学6.Chisquac. Pooled estimate of the frequencyc. Pooled estimate of the frequency Pooled estimate Pooled estimate = = numerators / numerators / denominators denominators Example: Example: The prevalence of myopia among 3 grades The preval

21、ence of myopia among 3 grades (15.16+15.89+18.37)/3 (15.16+15.89+18.37)/3 The prevalence of myopia among 3 grades The prevalence of myopia among 3 grades = (67+68+56)/(442+428+305) = (67+68+56)/(442+428+305) = 192/1175 = 192/1175 = 16.34 = 16.34d. Comparability between frequencies or between d. Comp

22、arability between frequencies or between frequency distributions Notice the balance of frequency distributions Notice the balance of other conditionsother conditions医学统计学6.Chisquae. If the distributions of other variables are different, to improve the comparability, “Standardization” is needed. f. T

23、o compare two samples, hypothesis test is needed. (See Chi square test)The following will emphasize the above two points: Standardization Hypothesis test医学统计学6.ChisquaCrude incidence rate of city A=28.96; Crude incidence rate of city B=35.03 - Strange!? They are not comparable ! - Because the consti

24、tute are quite different City A City B Age group (Y ear) Population Constitute No.of cases Incidence rate Population Constitute No.of cases Incidence rate 1 - 2542 0.1219 316 124.31 1014 0.2592 117 115.38 5 - 4285 0.2054 168 39.21 1905 0.4870 16 8.40 10 -12 14029 0.6727 120 8.55 992 0.2538 4 4.03 T

25、otal 20856 1.0000 604 28.96 3911 1.0000 137 35.03 Table 10-3 Incidence rates of infectious diseases, children of two cities医学统计学6.ChisquaStandardized incidence rate of city A = 793/24767 = 32.02 Standardized incidence rate of city B = 3523/24767 = 21.12 Two steps:lSelect a standard population taking

26、 as “weight”lWeighted average of the actual incidence ratesdirect standardization rateDirect standardization of the incidence rates of infectious disease for children in city A and B City A City B Age group (Year) Standard population Ni Actual incidence rate () Pa Expected number of cases Ni Pa Actu

27、al incidence rate () Pb Expected number of cases Ni Pb 1 - 3556 124.31 422 115.38 410 5 - 6190 39.21 243 8.40 52 10 -12 15021 8.55 128 4.03 61 Total 24767 (N) 28.96 793 Ni Pa 35.03 3523 Ni Pb 医学统计学6.ChisquaKnown: Age specific populations Ni1, Ni2; Total no.of deaths Di1=432, Di2 =210 Select a set of

28、 standard mortality rates Standard mortality ratio: SMR1 = Di1/ Ni1Pi = 432/100.67 = 4.2912 (smoker) SMR2 = Di2/ Ni2Pi = 210/100.67 = 0.8620 (non-smoker) Standardized mortality rate P1=34.60 SMR1=148.48 (1/105), P2=34.60 SMR2=29.83 (1/105)Table 10-6 Indirect standardization of death rates for lung c

29、ancer, smokers and non-smokers Smoker Non-smoker Age group (Year) Standard mortality rate of lung cancer (1/105) Pi Observed person-years Ni1 Expected number of deaths Ni1 Pi Observed person-years Ni2 Expected number of deaths Ni2 Pi 35 - 7.04 49705 3.50 189370 13.33 45 - 25.70 42633 10.96 104762 26

30、.92 55- 108.25 28117 30.44 60043 65.00 65- 263.94 10624 28.04 27540 72.69 75- 451.87 6137 27.73 14532 65.67 Total 34.60 137216 100.67 396247 243.61 医学统计学6.Chisqua10.2 Statistical Inferencefor Enumeration Data医学统计学6.ChisquaVocabulary of Chapter 10 (II) chi-square test 卡卡方方检检验验 2 test 卡卡方方检检验验 u test

31、u检检验验 contingency table 列列联联表表 observed frequency 观观察察频频数数 theoretical frequency 理理论论频频数数 row 行行 column 列列 adjustment 校校正正 positive rate 阳阳性性率率 equivalent to 等等价价于于 large sample 大大样样本本 significant difference 有有统统计计学学意意义义的的差差异异 医学统计学6.ChisquaExample Suppose the death rate is 0.2, if the ratsare fed w

32、ith a kind of poison. What will happen when we do the experiment on n=1, 2, 3 or 4 rat(s)?医学统计学6.Chisquan d Frequency distribution Sample rate 1 0 1 0.8 0.2 0/1=0 1/1=1 2 0 1 2 0.80.8=0.64 0.80.2+0.20.8=0.32 0.20.2 0/2=0 1/2=0.5 2/2=1 3 0 1 2 3 0.80.80.8=0.512 3(0.80.80.2)=0.384 3(0.80.20.2)=0.096 0

33、.20.20.2=0.008 0/3=0 1/3=0.3 2/3=0.7 3/3=1 4 0 1 2 3 4 0.80.80.80.8=0.4096 4(0.80.80.80.2)=0.4096 6(0.80.80.20.2)=0.1536 4(0.80.20.20.2)=0.0256 0.20.20.20.2=0.0016 0/4=0 1/4=0.25 2/4=0.5 3/4=0.75 4/4=1 医学统计学6.Chisqua In general, Supposed the population proportion is , sample size =n l The frequency

34、is a random variablelWhen is unknown and n is big enough, is approximately equal tonPP)1 (nXP nppsP)1 ( P医学统计学6.Chisqua%5 . 32007nXP%30. 10130. 0200)035. 01 (035. 0)1 (nppsP医学统计学6.Chisqua If the sample size n is big enough, and observed frequency is p , then we have approximately )1 (,(nppNP医学统计学6.C

35、hisqua If the sample size n is big enough, and observed frequency is p , then95% Confidence interval 99% Confidence intervalnppp)1 (96. 1:nppp)1 (58. 2:医学统计学6.Chisqua%05. 6%95. 0%30. 196. 1%5 . 3)1 (96. 1:nppp%85. 6%15. 0%30. 158. 2%5 . 3)1 (58. 2:nppp医学统计学6.ChisquaXnxpnXnp)1 (nssXnppsp)1 ( Xstx2/:p

36、sup2/:医学统计学6.Chisqua1. Comparison of sample proportion and population proportion Example 10.6 Cerebral infarction Cases Cure rateNew Method 98 50% Routine 30% 3 . 0:3 . 0:10HH医学统计学6.ChisqualStatistic ulDecision rule If , then reject Otherwise, no reason to reject (accept ) Since , reject uu 0H0H96.

37、1u0Hnpu)1(00032.498)3.01(3.03.05.0)1(000npu0H医学统计学6.Chisqua2. Comparison of two sample proportionsExample 10.7 Carrier rate of Hepatitis B City: 522people were tested, 24 carriers, 4.06% (population carrier rate: 1) Countryside: 478people were tested, 33 carriers, 6.90% (population carrier rate: 2)2

38、11210:HH医学统计学6.ChisqualPooled estimatelStandard error of P1-P2 2121nnXXpc057. 04785223324cp)11)(1 (2121nnppsccpp0147. 0)47815221)(057. 01 (057. 021pps医学统计学6.ChisqualStatistic ulDecision rule If , then reject Otherwise, no reason to reject (accept ) Since , not reject 2121ppsppuuu 0H0H565. 10147. 006

39、9. 0046. 02121ppsppu96. 1u0H0H医学统计学6.ChisqualThe parameter estimation and hypothesis testing of proportion are based on the normal approximation (when sample size is big enough)lHow big is enough? By experience, n 5 and n(1- ) 5l If the sample size is not big, u test cant be used and there is no t-t

40、est for proportion. (see more detailed text book)医学统计学6.Chisqua医学统计学6.Chisqua TheThe u u test test can only be used for can only be used for comparing comparing with a given with a given 0 (one sample) (one sample)or comparing or comparing 1 with 2 (two samples). (two samples). If we need to compare

41、 more than If we need to compare more thantwo samples,two samples, Chi-square testChi-square test is widelyis widelyused.used. 医学统计学6.ChisqualGiven a set of observed frequency distribution Given a set of observed frequency distribution A A1 1, , A A2 2, , A A3 3 to test whether the data follow certa

42、in theory.to test whether the data follow certain theory.lIf the theory is true, then we will have a set If the theory is true, then we will have a set of theoretical frequency distribution:of theoretical frequency distribution: T T1 1, , T T2 2, , T T3 3 lComparing Comparing A A1 1, , A A2 2, , A A

43、3 3 and and T T1 1, , T T2 2, , T T3 3 If they are quite different, then the theory If they are quite different, then the theory might not be true; might not be true; Otherwise, the theory is acceptable. Otherwise, the theory is acceptable.医学统计学6.ChisquaExample10-8 Example10-8 AcuteAcute lower respi

44、ratory infectionlower respiratory infection TreatmentEffectNon-effectTotalEffect rateDrug A68(64.82) a6(9.18) b74 (a+b)91.89 %Drug B52(55.18) c11(7.82) d63(c+d)82.54 %Total120 (a+c)17 (b+d)13753.59 %0 01 1n To calculate the theoretical frequencies To calculate the theoretical frequencies0 0 is true,

45、 is true, 医学统计学6.ChisqualTo compare A and T by a statistic To compare A and T by a statistic 2 2 If If H H0 0 is true, is true, 2 2 follows a chi-square follows a chi-square distribution. distribution. =(row-1)(column-1)=(row-1)(column-1) If the If the 2 2 value is big enough, we doubt value is big

46、enough, we doubt about about H H0 0 , then reject , then reject H H0 0 ! !.)()(122121211211112TTATTAnnnTCRRC totalRow :RnalColumn tot :Cn医学统计学6.ChisquaTo Example10-8To Example10-8 , , =(row-1)(column-1)=(2-1)(2-1)=1, =(row-1)(column-1)=(2-1)(2-1)=1, 2 20.05(1)0.05(1)=3.84, =3.84, Now, Now, 2 2=2.734

47、3.84, =2.7340.05, 0.05, H H0 0 is not rejected. is not rejected. We have no reason to say the effects We have no reason to say the effects of two treatments are different.of two treatments are different.734. 282. 7)82. 711(18. 9)18. 96 (18.55)18.5552(82.64)82.6468(22222医学统计学6.Chisqua For For 2 2 2 2

48、 table table, there is a specific , there is a specific formula of chi-square calculation:formula of chi-square calculation:)()()()(22dbcadcbanbcad 734. 2171206374137)5261168(22To Example10-8To Example10-8 , , 医学统计学6.Chisqua Large sample is requiredLarge sample is required (1) (1) N N 40,40, T Ti i

49、5, 5, N N 4040(2)If (2)If n n 40 or 40 or T Ti i 1, 1, 2 2 test is not test is not applicableapplicable(3)If (3)If N N 40,40, 1 1 T Ti i 5 , needs adjustment: 40) For large sample (b+c40) Otherwise, needs adjustmentOtherwise, needs adjustmentl If If the the 2 2 value is value is too big, then reject

50、 too big, then reject H H0 0 bA 1cA 2221cbTTcbcbcbcbccbcbb2222)(2)2(2)2(cbcb22) 1(医学统计学6.Chisqua92. 4112) 1112(22医学统计学6.ChisquaTable 6.6 Blood types of patient with different diseases Blood type Total Disease status A B O Digestive ulcer 679 134 983 1796 Stomach cancer 416 84 383 883 Control 2625 57

51、0 2892 6087 Total 3720 788 4258 8766 Remark: There is no order among the categories! 0H: The distributions of blood types in three populations are all same 1H: The distributions are not all same 医学统计学6.ChisquannnTCRRC.)()(122121211211112TTATTA122CRRCnnAn543.401425860872892372017966798766222 = (31) (31) =4, 205. 0=9.488 , p0.