Statistics - Lecture 05

Transcription

1 Statistics - Lecture 05 Nicodème Paul Faculté de médecine, Université de Strasbourg 1/47

2 Descriptive statistics and probability Data description and graphical representation Mean, median, quartiles, standard deviation, interquartile range (IQR) Barplot, histogram and boxplot Decisions are based on probability calculation Notion of random variables and distributions Binomial and normal distributions 2/47 2/47

3 Estimation μ σ 2 Notion of parameters (, ) Di erence between estimate and estimator Xˉ S 2 The sample mean and the sample variance are estimators The Central Limit Theorem and sampling distribution Interval estimate or con dence interval 3/47 3/47

4 Hypothesis testing Parametric tests and test procedure Notion of null and alternative hypotheses Test statistic and its sampling distribution Critical values and critical regions P-values 4/47 4/47

5 Relation between variables and goodness of t Notion of correlation Non parametric tests and goodness of t Notion of association between categorical variables The χ 2 test The Fisher exact test 5/47 5/47

6 Examples You have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the same size and type. Four are allocated at random to an untreated control group, four are treated with the drug Tumostat and four more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour is remeasured. Survival times in ve human cancer (stomach, bronchus, colon, ovary, breast). Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive treatment of cancer: re-evaluation of prolongation of survival times in terminal human cancer. Proceedings of the National Academy of Science USA, 75, Comparison of 5 pretreated patches to reduce mosquito human contact. Bhatnagar, A and Mehta, VK (2007) E cacy of Deltamethrin and Cy uthrin Impregnated Cloth over Uniform against Mosquito Bites. Medical Journal Armed Forces India, 63, /47 6/47

7 Question You have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the same size and type. Four are allocated at random to an untreated control group, four are treated with the drug Tumostat and four more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour is remeasured. What would be the appropriate test here? Parametric test Non parametric test Submit Show Hint Show Answer Clear 7/47 7/47

8 ANOVA: Analysis Of Variance ( μ 1, σ1 2 ) ( X 11, X 12,..., X 1,n1 ) (, ) Population 1: Sample 1: Population 2: Sample 2: Population k: Sample k: Xˉ1 S 2 1 ( μ 2, σ2 2 ) ( X 21, X 22,..., X 2,n2 ) (, ) Xˉ2 S 2 2 ( μ k, σ 2 ) ( X,,..., ) k k1 X k2 X k,nk (, ) Xˉk S 2 k Objective: Comparing the means of multiple populations 8/47 8/47

9 ANOVA assumption and objectives Each of the k population or treatment response distributions is normal σ 1 = σ 2 =... = σ k (The k normal distributions have identical standard deviations) The observations in the sample from any particular one of the k populations or treatments are independent of one another When comparing population means, the k random samples are selected independently of one another μ 1 μ 2 μ k H 0 : = =... = H 1 μ : At least two of the s are di erent Estimation of simultaneous con dence intervals for the mean di erences μ i i, j = 1,..., k and i j μ j for 9/47 9/47

10 Example You have been assigned 12 consenting experimental subjects, each of whom has a brain tumour of the same size and type. Four are allocated at random to an untreated control group, four are treated with the drug Tumostat and four more with the drug Inhibin 4. After two months of treatment, the diameter of each tumour is remeasured. Population parameters: - μ 1 : population mean of the control group - μ 2 : population mean of the Neurohib group - μ 3 : population mean of the Tumostop group. H 0 : There is no di erence in mean tumour diameter among the treatments H 1 μ : There is a di erence in mean tumour diameter among the treatments. At least two of the s are di erent = = μ 1 μ 2 μ /47 10/47

11 Data x ij i j is the th observation resulting from th treatment = T.j n j : total of th treatment. : mean of the th treatment xˉ.j T.j i=1 x ij j = n j j T.. = k j=1 T.j = k j=1 n j i=1 x ij : total of all observations T xˉ.. =.. N : the grand mean where N = k j=1 n j 11/47 11/47

12 Example 1 xˉ.1 = ( ) = xˉ.2 = ( ) = xˉ.3 = ( ) = xˉ.. = ( ) = /47 12/47

13 Within groups sum of squares Variation within group: ssw = 3 j=1 4 i=1( x ij xˉ.j ) 2 ssw = ( ) + ( ) + ( ) = /47 13/47

14 Between groups sum of squares Variation between groups: ssb = 3 j=1 4( xˉ.j xˉ.. ) 2 ssb = = /47 14/47

15 Total sum of squares Variation between groups: sst = 3 i=1 4 i=1( x ij xˉ.. ) 2 sst = ( ) + ( ) + ( ) = 102 sst = ssw + ssb = = /47 15/47

16 Check yourself Under the null hypothesis, the ratio should be close to 1. True False Submit Show Hint Show Answer Clear ssb ssw 16/47 16/47

17 Test for equal means The hypotheses are: H 0 : H 1 : μ 1 = μ 2 =... = μ k some means are different If the null hypothesis is true, we combine k samples to estimate overall mean and the sample mean for the group as: i Xˉ.. k i=1 n j X ij Xˉ.j 1 = = N j=1 n 1 j X i,j n j i=1 SST = k j=1 n j ( sum of squares within groups and groups i=1 X ij Xˉ..) 2 Total sum of squares, SSW = ( We can show that: SST = SSW + SSB k j=1 n j i=1 X ij Xˉ.j) 2 SSB = k j=1 n j( Xˉ.j Xˉ..) 2 sum of squares between 17/47 17/47

18 Check yourself X 1, X 2,..., X n N (μ, σ 2 ) μ σ S 2 = 1 ( n 1 n i=1 X i Xˉ) 2 Let with and unknown. Let. What is the distribution of? σ N (μ; 2 ) n t n 1 χ 2 1 n 1 S 2 σ 2 χ 2 n 1 χ 2 n Submit Show Hint Show Answer Clear 18/47 18/47

19 Check yourself In the anova framework, we de ned SSW Let n 1 n 2 n k, what is the distribution of? N = SSW = k j=1 n j i=1( X ij Xˉ.j) 2 σ 2 as the sum of squares within group. χ 2 1 χ 2 k χ 2 k 1 χ 2 N χ 2 N 1 χ 2 N k χ 2 N k 1 Submit Show Hint Show Answer Clear 19/47 19/47

20 Parameter estimation Comparing means from multiple populations assuming the variances are the same and equal to σ 2 Pooled variance estimator: S 2 pool k j=1( 1) = n j S 2 j = k j=1( n j 1) k j=1 n j i=1( X ij Xˉ.j) 2 N k N = n 1 + n n k = where, and Xˉ.j 1 n j n j i=1 X i,j S 2 1 j = n j i=1( X ij Xˉ.j) 2 n j 1 Notice that: S 2 pool ( n 1 1) S1 2 ( n 2 1) S2 2 ( n k 1) S 2 k = σ 2 σ 2 σ 2 σ 2 χ 2 N k j = 1, 2,..., k As for, ( n j 1) S 2 j σ 2 χ 2 1 n j 20/47 20/47

21 Parameter estimation = Xˉ.j 1 n j n j is an estimator of i=1 X ij μ j As,,..., N (, ), then X 1j X 2j X nj j μ j σ 2 N (, ) Xˉ.j μ j σ 2 n j Xˉ.j μ j 1 S pool nj t N k (1 α) con dence intervals for population means are: = [, + ] 1 α xˉ.j t N k 2 n 1 α j 2 n j I α xˉ.j t N k s pool s pool 21/47 21/47

22 Example A 95% con dence interval for respectively μ control, μ neurohib and μ mitostop is: - [ /9 1/2; /9 1/2] = [6.935; ] - [ /9 1/2; /9 1/2] = [3.935; 8.065] - [ /9 1/2; /9 1/2] = [0.935; 5.065] 22/47 22/47

23 Check yourself In the anova framework, we de ned SST distribution of? σ 2 SST = k j=1 n j ( i=1 X ij Xˉ..) 2 as the sum of total squares. What is the χ 2 1 χ 2 k χ 2 N 1 χ 2 N χ 2 N k Submit Show Hint Show Answer Clear 23/47 23/47

24 Check yourself In the anova framework, we de ned SSB What is the distribution of? σ 2 SSB = k j=1 n j( Xˉ.j Xˉ..) 2 as the sum of squares between groups. χ 2 1 χ 2 k χ 2 N 1 χ 2 N k χ 2 k 1 Submit Show Hint Show Answer Clear 24/47 24/47

25 Test for equal means SST σ 2 SSW σ 2 X ij = k j=1 n j Xˉ.. i=1( ) 2 χ 2 σ N 1 k j=1 n j X ij Xˉ.j = i=1( ) 2 χ 2 σ N k = ( σ ) 2 χ 2 k 1 SSB k σ 2 j=1 Xˉ.j Xˉ.. n j 25/47 25/47

26 Test for equal means De nition: Let Y and W be independent random variables such that Y has the χ 2 m distribution with m degrees of freedom and has the distribution with n degrees of freedom, where m and n are given positive integers. The random variable T de ned as follows: T F m,n Then the distribution of is, the F distribution with m and n degrees of freedom. W T = Y /m W/n χ 2 n Under the null hypothesis, the random variable: T = SSB/(k 1) SSW/(N k) F k 1,N k k 1 N k has a distribution with and degrees of freedom. 26/47 26/47

27 Test for equal means α f To test with a signi cant level, we calculate the value of the test statistic from the samples H 0 f > f 1 α Reject the null distribution if where is the critical value. The p value for the test is: k 1,N k f 1 α k 1,N k p value = P(T > f) where T F k 1,N k If the null hypothesis is rejected, what next? - - Tests for contrasts Pairwise comparison 27/47 27/47

28 Example 72/2 k 1 = 2 N k = 9 f = = 10.8 We have:,,, and. 30/9 2,9 = p value = f 0.95 We reject the null hypothesis of equal mean of tumour diameters 28/47 28/47

29 Example: Comparison of 5 pretreated patches to reduce mosquito human contact Reference: Bhatnagar, A and Mehta, VK (2007) E cacy of Deltamethrin and Cy uthrin Impregnated Cloth over Uniform against Mosquito Bites. Medical Journal Armed Forces India, 63, /47 29/47

30 Example: parameter estimation model01 = aov(measure~treatment, data=bites) model.tables(model01, type="means") Tables of means Grand mean Treatment Treatment C+O Cyfluthrin Deltamethrin D+O Odomos xˉ = xˉc = xˉd = xˉo = xˉc+o = xˉd+o = /47 30/47

31 Example: test summary(model01) Df Sum Sq Mean Sq F value Pr(>F) Treatment ** Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We have: If then, we reject the null hypothesis of equal mean mosquito bite rates. k 1 = 4 SSB = N k = 145 SSW = f = 4.48 p value = f4,145 1 α α = 0.05 = /47 31/47

32 Example: Survival times in terminal human cancer Reference: Cameron, E. and Pauling, L. (1978) Supplemental ascorbate in the supportive treatment of cancer: reevaluation of prolongation of survival times in terminal human cancer. Proceedings of the National Academy of Science USA, 75, STOMACH BRONCHUS COLON OVARY BREAST /47 32/47

33 Example: Survival times in terminal human cancer 33/47 33/47

34 Example: Survival times in terminal human cancer 34/47 34/47

35 Example: parameter estimation model02 = aov(survival~type, data=cdat) model.tables(model02, type="means") Tables of means Grand mean Type stomach bronchus colon ovary breast rep xˉ = xˉs = xˉb = xˉc = xˉo = xˉb = /47 35/47

36 Example: test summary(model02) Df Sum Sq Mean Sq F value Pr(>F) Type ** Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We have: f4,59 1 α α = 0.05 = k 1 = 4 SSB = N k = 59 SSW = f = p value = If then, we reject the null hypothesis of equal mean survival days. 36/47 36/47

37 Contrasts A contrast is any linear combination of the population means k i=1 c i = 0 such that and integer. μ 1, μ 2,..., μ 5 k = 5 c i C = c 1 μ 1 + c 2 μ c k μ k If are means of populations, some examples of contrasts are: - μ 1 μ 2-2μ 1 μ 3 μ 4 - μ 1 + μ 2 + μ 3 μ 4 2μ /47 37/47

38 Check yourself You want to test: 1 H 0 : μ 1 ( ) = 0 : + ( ) 0 4 μ 1 2 μ 3 μ 4 μ 5 H 1 μ 1 4 μ 2 μ 3 μ 4 μ 5 What would be the contrast: 1 μ 1 ( ) 4 μ 2 μ 3 μ 4 μ 5 4 μ 1 μ 2 μ 3 μ 4 μ 5 5 μ 1 μ 2 μ 3 μ 4 μ 5 Submit Show Hint Show Answer Clear 38/47 38/47

39 Test for a contrast The hypotheses: H 0 k k : c j μ j = 0 versus H 1 : c j μ j 0 j=1 j=1 S w Note = SSW/(N k), the test statistic: T = k j=1 c jxˉ.j S w k j=1 c 2 j n j has a t-distribution with N k degrees of freedom. The (1 α) 100% con dence interval for the contrast is: k c 2 k c 2 [ t N k j ; + t N k j ] k j=1 c j xˉ.j 1 α 2 s w j=1 n j k j=1 c j xˉ.j 1 α 2 s w j=1 n j 39/47 39/47

40 Check yourself In the anova framework with k conditions or treatments, when you reject the null hypothesis how many comparisons would you do to compare the means? k 1 k k 2 k(k+1) 2 k(k 1) 2 Submit Show Hint Show Answer Clear 40/47 40/47

41 Pairwise comparisons There are k(k 1) 2 pairwise comparisons or tests Recall that when testing a single hypothesis H 0, a type I error is made if it is rejected, even if it is actually true. The probability of making a type I error in a test is usually controlled to be smaller than a α certain level of, typically equal to 0.05 H 01 H 02 H 0m α When there are several null hypotheses,,,...,, and all of them are tested simultaneously, one may want to control the type I error at some level. A type I error is then made if at least one true hypothesis in the family of hypotheses being tested is rejected. This signi cance level is called the familywise error rate (FWER). If the hypotheses in the family are independent, then: α i i = 1, 2,..., m FWER = 1 (1 α i ) m where for are individual signi cance levels. 41/47 41/47

42 Pairwise comparisons FWER α H 0i H 01, H 02,..., p-value is less than α/m. H 0m Bonferroni: To control, reject all among for which the Studentized range distribution (Tukey) procedure: - Rank the k sample means - Two population means μ i and μ j are declared signi cantly di erent if the (1 α)100 μ i μ j con dence interval of : xˉi xˉj ± q N k,k,1 α s w ( + ) 2 n i n j q N k,k,1 α populations. is the upper-tail critical value of the Studentized range for comparing k di erent 42/ /47

43 Comparison of 5 pretreated patches Tukey multiple comparisons of means 95% family-wise confidence level factor levels have been ordered Fit: aov(formula = Measure ~ Treatment, data = bites) $Treatment diff lwr upr p adj D+O-C+O Odomos-C+O Cyfluthrin-C+O Deltamethrin-C+O Odomos-D+O Cyfluthrin-D+O Deltamethrin-D+O Cyfluthrin-Odomos Deltamethrin-Odomos Deltamethrin-Cyfluthrin /47 43/47

44 Comparison of 5 pretreated patches Cy uthrin patches when applied in presence of odomos were found to have much more repellent action as compared to only odomos. The di erence in the repellent action was very highly signi cant (p < 0.01). Thus it can be inferred that signi cant bene t is achieved in reducing man-mosquito contact when cy uthrin patches are applied over the uniform by the troops in addition to using odomos as compared to those using odomos only 44/47 44/47

45 Survival times in terminal human cancer Tukey multiple comparisons of means 95% family-wise confidence level factor levels have been ordered Fit: aov(formula = Survival ~ Type, data = cdat) $Type diff lwr upr p adj stomach-bronchus colon-bronchus ovary-bronchus breast-bronchus colon-stomach ovary-stomach breast-stomach ovary-colon breast-colon breast-ovary /47 45/47

46 Survival times in terminal human cancer is signi cantly di erent to 0. In fact, ascorbate, when used with the treatment, seems to improve survival times better in breast cancer than in bronchus cancer. μ breast log( ) log( ) μ breast μ bronchus μ stomach log( ) log( ) is also signi cantly di erent to 0, showing a signi cant improvement of survival in breast cancer compared to stomach cancer when ascorbate supplement is used in the treatment. 46/47 46/47

47 See you next time 47/47 47/47