Statistics in medicine

Transcription

1 Statistics in medicine Lecture 3: Bivariate association : Categorical variables Proportion in one group One group is measured one time: z test Use the z distribution as an approximation to the binomial distribution Fatma Shebl, MD, MS, MPH, PhD Assistant Professor Chronic Disease Epidemiology Department Yale School of Public Health Fatma.shebl@yale.edu S L I D E 0 S L I D E 1 Steps of the statistical testing, z test 1- Calculate the test statistic 2- Determine the of 3-Compare the test statistic with the 4- Calculate the p value 5- Draw a conclusion 1- Calculate the Critical ratio () Z = p π π(1 π)/n S L I D E 2 S L I D E 3 2-Determine the of Critical value: E.g., the critical value for.05 alpha level, one-tailed test, is Compare the test statistic with the critical value The test statistic is < the (in absolute sense) acceptance area The test statistic is > the (in absolute sense) rejection area Source of table: S L I D E 4 S L I D E 5 1

2 4- Calculate the p value Using the z-table Find the z value that is =test statistic by scrolling down the z column then scrolling to the right in the row The entries of the tables are used to calculate the p values For negative z values One-tailed, the p value is the entry of the z table Two-tailed, the p value is the entry of the z table multiplied by 2 For positive z values One-tailed, the p value = 1- the entry of the z table Two-tailed,: p value = (1- the entry of the z table )multiplied by 2 4- Calculate the p value E.g., for a z=3.1 one-tailed test, p=(1-.999) =.001 two-tailed test, p=(1-.999) *2=.002 Source of table: S L I D E 6 S L I D E 7 5- Calculate the confidence interval CI of proportion: p ± z p(1 p) n E.g. 95% CI=p ± 1.96 p(1 p) n The value that defines the central 95% of the area 6- Draw a conclusion Reject or fail to reject the null hypothesis Fail to reject the null If the p value > alpha level If the confidence interval crosses the 1 Reject the null If the p value < alpha level If the confidence interval does not cross the 1 S L I D E 8 S L I D E 9 example A study of success of vaccination revealed that 97.1% responded to the vaccine. The investigators would like to examine whether this proportion is significantly higher than 95%. If the calculated z value was Using a.05 alpha level, what is the and the corresponding p value? Answer: Critical value=1.645 (one tail) p=( )=.0401 Conclusion: we reject the null hypothesis Source: paired/matched One group is measured twice (or matched design): McNemar Chi-square Is a chi-square test for comparing proportions from two DEPENDENT or paired groups. Use the chi-square distribution S L I D E 10 S L I D E 11 2

3 paired/matched One group is measured twice (or matched design): McNemar Chi-square E.g., E.g., Is there a change in bowel functions after cholecystectomy? Steps of the statistical testing, McNemar Chi-square 1- Calculate the test statistic (critical ratio) 2- Calculate degrees of freedom 3- Determine the of 4-Compare the test statistic with the 5- Calculate the p value 6- Calculate the confidence interval 7- Draw a conclusion S L I D E 12 S L I D E Calculate the Critical ratio () McNemar X 2 = b c 2 b+c 2- Calculate degrees of freedom = (R-1)(C-1)=1 Matched design Cases Control Total Paired design Positive risk Before intervention After intervention Total Positive Negative risk Positive risk a b a+b Negative risk c d c+d Negative Positive a b a+b Negative c d c+d 3-Determine the of Critical value: Depends on: Alpha level Degrees of freedom E.g., for an alpha of.05, 1, the value of the critical x 2 is S L I D E 14 S L I D E 15 4-Compare the test statistic with the critical value The test statistic is < the (in absolute sense) acceptance area The test statistic is > the (in absolute sense) rejection area 5- Calculate the p value From the x 2 table Locate the row of the relevant Find the x 2 value that is < test statistic Find the p value at the top of the column E.g., for a, 1, x 2 =4.05 > p value >.025 S L I D E 16 S L I D E 17 3

4 6- Draw a conclusion Reject or fail to reject the null hypothesis Fail to reject the null If the p value > alpha level Reject the null If the p value < alpha level paired/matched, McNemar test example Researcher examined whether bowel movement changed postoperative. If the calculated McNemar statistic is 15, what is: the and the p value? = > p value S L I D E 18 S L I D E 19 Readings and resources Chapter 5, p93-133: Dawson, B. and Trapp, R. G. (2004). Basic and Clinical Biostatistics (4th edition). New York: McGraw-Hill. Chapter 11, p : Jekel's epidemiology, biostatistics, preventive medicine, and public health by David L. Katz et al (4th edition). Proportions in two groups, z approximation test Z test: Definition: Test of the equality of two independent proportions Is an approximation of the binomial distribution when p*n>5 Requires at least 5 observed frequencies in each group Parametric test S L I D E 20 S L I D E 21 Steps of the statistical testing, z test 1- Calculate the test statistic 2- Determine the of 3-Compare the test statistic with the 4- Calculate the p value 5- Draw a conclusion Proportions in two groups, z test 1- Calculate the Z = Where, p 1 p 2 p 1 p [ 1/n 1 + 1/n 2 ] p 1, and p 2 are proportions in each group P is the pooled proportion p = n 1p 1 +n 2 p 2 n 1 +n 2 S L I D E 22 S L I D E 23 4

5 Proportions in two groups, z test 2-Determine the of 3-Compare the test statistic with the 4-Calculate the p value Calculate the confidence interval CI of proportion difference = (p 1 p 2 ) ± z SE p1 p 2 where., SE p1 p 2 = p(1 p) (1/n 1 ) + (1/n 2 ) 5- Draw a conclusion Proportions in two groups, z test example A study of the physicians screening practices of domestic violence (DV), revealed that 86.6% of 202 physicians who were trained for DV detection versus 58.3% of 266 physicians who were not trained for DV detection screened their patients for DV. The investigators would like to examine whether the proportion among trained physicians is the same as proportion among non trained physicians. If the calculated z value was 6.6, and the SE(p1-p2)=.043. Using a.05 alpha level, what are: the, the corresponding p value, and the 95% CI? S L I D E 24 S L I D E 25 Proportions in two groups, z test example Answer: Critical value=1.96 (two tail) P<(1-.999)*2 < % CI= ( )+1.96*.043= (.201,.367) Conclusion: we reject the null hypothesis Proportions in two groups, Chi-square test : Definition: the test of the null hypothesis that proportions are equal, or equivalently, that factors or characteristics are independent or not associated Nonparametric test Could be used for two or more groups Use the X 2 distribution Could be used if n*p>5 Should not be used if expected frequencies <2 S L I D E 26 S L I D E 27 Chi-square distribution (X 2 ) The distribution used to analyze counts in frequency tables. A nonsymmetrical distribution with mean (µ) and variance (σ 2 ) Used for categorical (nominal) data Properties: Degrees of freedom = υ µ = υ σ 2 = υ*2 Approaches normal distribution with the increase in Steps of the statistical testing, for independence 1- Calculate the test statistic 2- Calculate degrees of freedom 3- Determine the of 4-Compare the test statistic with the 5- Calculate the p value 6- Draw a conclusion Graph generated by R S L I D E 28 S L I D E 29 5

6 Proportions in two groups, chisquare test of independence 1- Calculate the X 2 = O E 2 E Short cut formula for 2x2 table Exposure outcome Total Disease X 2 n ad bc = 2 (a+c)(b+d)(a+b)(c+d) Where, O=observed frequencies row total x column total E=expected frequencies= grand total 2- Calculate degrees of freedom = (R-1)(C-1) Undiseased Exposed a b a+b Unexposed c d c+d Proportions in two groups, chisquare test of independence 3-Determine the critical value of from the x 2 table 4-Compare the test statistic with the 5-Calculate the p value 6-Draw a conclusion S L I D E 30 S L I D E 31 Proportions in two groups, chi-square test of independence, example Researcher examined whether there is an association between DV training and screening for DM by physicians. If the calculated x 2 statistic is 44.4, what are: the and the p value, and your conclusion? =(2-1)(2-1)=1 p value < Conclusion: there is a significant association between training and screening. Proportions in two groups, Fisher s exact test Fisher s exact test: Definition: An exact test for 2x2 contingency table. It is used when the sample size is too small to use a chi-square test Should be used if expected frequencies <2 Fisher p= (a+b)!(c+d)!(a+c)!(b+d)! N!a!b!c!d! S L I D E 32 S L I D E 33 Proportions in two groups, Risk ratios Allows examining associations between two nominal measures Allows estimating confidence intervals as a test Proportions in two groups, Relative risk Relative risk: The ratio of the incidence of a given disease in exposed to the incidence of the disease in the unexposed S L I D E 34 S L I D E 35 6

7 Proportions in two groups, Relative risk 1- Calculate the relative risk (RR) RR= a/(a+b) c/(c+d) 2- Calculate the confidence interval 1 [ a a+b exp ln (RR) ± z ] c 1 [ c+d α/2 + ] a c Exposure outcome Total Disease Exposed a b a+b Unexposed c d c+d Proportions in two groups, Odds ratio Odds ratio An estimate of relative risk in case-control studies. It is the odds that a patient was exposed to a given risk factor divided by the odds that a control was exposed to a given risk factor S L I D E 36 S L I D E 37 Proportions in two groups, Odds ratio 1- Calculate the odds ratio (OR) OR= a c /[ a+c a+c ] ad b d /[ ]= bc b+d b+d 2- Calculate the confidence interval exp ln (OR) ± z α/2 1 a + 1 b + 1 c + 1 d Exposure outcome Total Disease Undiseased Undiseased Exposed a b a+b Unexposed c d c+d Readings and resources Chapter 6, p : Dawson, B. and Trapp, R. G. (2004). Basic and Clinical Biostatistics (4th edition). New York: McGraw-Hill. Chapter 11, p : Jekel's epidemiology, biostatistics, preventive medicine, and public health by David L. Katz et al (4th edition). S L I D E 38 S L I D E 39 Proportions in three or more groups, : Definition: the test of the null hypothesis that proportions are equal, or equivalently, that factors or characteristics are independent or not associated Nonparametric test Could be used for two or more groups Use the X 2 distribution Could be used if n*p>5 Should not be used if expected frequencies <2 Steps of the statistical testing, for independence 1- Calculate the test statistic 2- Calculate degrees of freedom 3- Determine the of 4-Compare the test statistic with the 5- Calculate the p value 6- Draw a conclusion S L I D E 40 S L I D E 41 7

8 Proportions in three or more groups, 1- Calculate the test statistic X 2 = O E 2 E Where, O=observed frequencies row total x column total E=expected frequencies= grand total 2- Calculate degrees of freedom = (R-1)(C-1) Exposure outcome Total Sever disease Mild disease Control Exposed y11 y12 y13 y11+y12+y13 Unexposed y21 y22 y23 y21+y22+y23 Total y11+y21 y12+y22 y13+y32 y11+y12+y13 +y21+y22+y23 Proportions in three or more groups, 3-Determine the critical value of from the x 2 table 4-Compare the test statistic with the 5-Calculate the p value 6-Draw a conclusion S L I D E 42 S L I D E 43 Proportions in three or more groups, Chisquare test, example Researchers examined the response to three vaccine s dilutions. Subjects were classified as responders or nonresponders. Researchers would like to determine whether there is an association between vaccine s dilutions and the response to the vaccine. If the calculated x 2 statistic is 2.589, and for an alpha of.05, what are: the, of and the p value, and your conclusion? =(3-1)(2-1)=2 Critical value=5.991 p value >.1 Conclusion: there is no significant association between vaccine s dilutions and the response S L I D E 44 Proportions in three or more groups, Chisquare test, example Researchers examined the association between eyes color and hair color. Eyes color was classified into three groups, and hair s color was classified into 5 groups. If the calculated x 2 statistic is 20.92, and for an alpha of.05, what are: the, of and the p value, and your conclusion? =(3-1)(5-1)=2x4=8 Critical < p value <.01 Conclusion: there is significant association between eyes' color and hair s color S L I D E 45 Proportions in three or more groups, odds ratio 1. Pick exposure reference category 2. Calculate OR for each exposure level Proportions in three or more groups, risk ratio 1. Pick exposure reference category 2. Calculate RR for each exposure level Outcome Exposure level Total High Middle Low Outcome Exposure level Total High Middle Low Diseased y11 y12 y13 y11+y12+y13 Not diseased y21 y22 y23 y21+y22+y23 Total y11+y21 y12+y22 y13+y32 y11+y12+y13 +y21+y22+y23 Diseased y11 y12 y13 y11+y12+y13 Not diseased y21 y22 y23 y21+y22+y23 Total y11+y21 y12+y22 y13+y23 y11+y12+y13 +y21+y22+y23 P (y11) (y11 + y21) (y12) (y12 + y22) (y13) (y13 + y23) P (y11) (y11 + y21) (y12) (y12 + y22) (y13) (y13 + y23) OR (y11)(y23) (y13)(y21) (y12)(y23) (y13)(y22) 1 RR (y11)/(y11 + y21) (y13)/(y13+y23 ) (y12)/(y12 + y22) (y13)/(y13 + y23) 1 S L I D E 46 S L I D E 47 8

9 Proportions in three or more groups, Cochran-Armitage Trend Test Is a test of linear trend in proportions Used if one variable is dichotomous and the other variable is ordinal A test of trend of ordinal categories has higher power than the chi-square test of the nominal categories Ordinal levels values will depend on the type of the variable, ex.: Cancer stage: levels 1, 2, 3, 4 are reasonable Number of cigarettes smoked: median values of the intervals such as 5, 15, 25, 35 could be used Proportions in three or more groups, Cochran-Armitage Trend Test A clinical trial investigated whether larger drug doses increase adverse response. Patients were randomized to receive placebo, or one of the four drug doses. Source: SAS examples Adverse Table of Adverse by Dose Dose Total No Yes Total Cochran-Armitage Trend Test Statistic (Z) One-sided Pr < Z <.0001 Two-sided Pr > Z <.0001 S L I D E 48 S L I D E 49 Readings and resources Chapter 7, p : Dawson, B. and Trapp, R. G. (2004). Basic and Clinical Biostatistics (4th edition). New York: McGraw-Hill. Chapter 11, p : Jekel's epidemiology, biostatistics, preventive medicine, and public health by David L. Katz et al (4th edition). S L I D E 50 9