March 30 More examples of case-control studies General I x J table Chi-square tests

Exam Question   proportion who smoke cigarettes n = 10 x = 3 report smoking p = 3/10 = 0.30 is estimate of  SE(p) == % CI for  0.30 ± 1.96 (0.14) 0.30 ± 0.28 or (0.02 – 0.58)

HRT Use and Colonic Polyps Case-control study conducted at Digestive Healthcare (DH) in Minneapolis ( ) Patients underwent colonoscopy –Cases were patients with adenomatous polyps –Controls were patients without polyps Questionnaires completed –Use of hormone replacement therapy ascertained Published CEBP 1996 (Potter et al) Since patients undergoing colonoscopy were not a random sample RR of polyps cannot be estimated

HRT Use and Polyps Case (Polyps)Control (No Polyps) HRT Use RO = 72/ /114 = 0.46 No HRT Use 247 RO HRT Use (Case v Control)    463 ) (RO) 2  174) (289) (247) (216) =

USING SAS DATA hrt; INFILE DATALINES; INPUT hrt ccstatus count; DATALINES; ; PROC FREQ DATA=hrt; TABLES hrt*ccstatus/CHISQ RELRISK; WEIGHT count; RUN; ccstatus =1 is case ccstatus = 2 is control

hrt ccstatus Frequency| Percent | Row Pct | Col Pct | 1| 2| Total | 72 | 175 | 247 | | | | | | | | | | 102 | 114 | 216 | | | | | | | | | Total Estimates of the Relative Risk (Row1/Row2) Type of Study Value 95% Confidence Limits Case-Control (Odds Ratio) Cohort (Col1 Risk) Cohort (Col2 Risk)

Interpretation (RO = 0.46) Women who use HRT are at a 54% decreased risk of developing polyps compared to women who do not use HRT. The odds of developing polyps is 54% lower in women who use HRT compared to women who do not use HRT.

Analyses of I x J Tables Two factors, one with I levels, second with J levels I rows, J columns General hypothesis: –Are rows independent of columns? –Ho:    2  3 (when J=2) Chi-square  2 ) test with (I-1 x J-1 df)  Sum over all cells (IxJ)

Example TOMHS Rows – 6 treatment groups Columns – 4 levels of side-effect –1 = none; 2=mild, 3=moderate, 4=severe Do side effect distribution depend on group? –Group 1 = diuretic –Group 2 = ACE –Group 3 = placebo –Side effects: 1=none, 2=mild, 3 = moderate/severe Look at cough

Cough Diuretic ACE Placebo NoneMild Mod/Sev Expected Frequencies if Ho is true: Cell (1,1) = 359*133/498 = 95.9 Cell (1,2) = 109*133/498 = 29.1 Compute for all 9 cells 498

Cough- Expected Frequencies 99 (95.9)28 (29.1)6 (8.0) 86 (95.9)36 (29.1)11 (8.0) 174 (167.2)45 (50.8)13 (14.0) Diuretic ACE Placebo NoneMild Mod/Sev    4DF)   / ( )2 / … + (13-14) 2 /14 = 5.41 P = 0.25

USING SAS DATA cough; INPUT group cough count; DATALINES; ; PROC FREQ DATA=hrt; TABLES group*cough/CHISQ CMH; WEIGHT count; RUN; cough =1 (none) cough = 2 (mild) cough = 3 (mod/severe)

GROUP cough Frequency| Expected | Percent | Row Pct | Col Pct | 1| 2| 3| Total | 99 | 28 | 6 | 133 | | | | | | 5.62 | 1.20 | | | | 4.51 | | | | | | 86 | 36 | 11 | 133 | | | | | | 7.23 | 2.21 | | | | 8.27 | | | | | | 174 | 45 | 13 | 232 | | | | | | 9.04 | 2.61 | | | | 5.60 | | | | | Total

Statistic DF Value Prob Chi-Square Likelihood Ratio Chi-Square Mantel-Haenszel Chi-Square Phi Coefficient Contingency Coefficient Cramer's V Effective Sample Size = 498 Frequency Missing = 7 Cochran-Mantel-Haenszel Statistics (Based on Table Scores) Statistic Alternative Hypothesis DF Value Prob Nonzero Correlation Row Mean Scores Differ General Association Takes into account the ordering of the row categories

1-Sample Z-Test: Matched Pair Data ab cd Control PositiveControl Negative Case Pos Neg Ho:  0.5 where n=b+c and x=b Z = (b/(b+c) – 0.5)/sqrt(.5*.5/(b+c)) Z = (b-c)/sqrt(b+c)  2 = (b-c) 2 /(b+c) Analyses is done on discordant pairs b and c Called McNemar’s chi-square

Example – Vitamin Use/Disease (440 Pairs) Control Vitamin +Control Vitamin - Case Vit + Vit - Ho:  0.5 where n=140 and b = 50  2 = (50-90) 2 /(50+90) = (p=.0007)

Example of Matched Case-control Study DATA vitamin; INFILE DATALINES; INPUT v_case $ v_cont $ count; LABEL v_case = 'Case Use of Vitamins'; LABEL v_cont = 'Control Use of Vitamins'; DATALINES; 1-YES 1-YES YES 2-NO 50 2-NO 1-YES 90 2-NO 2-NO 200 ; 100 pairs where both case and control took vitamins

PROC FREQ DATA=vitamin; TABLES v_case*v_cont/AGREE; WEIGHT COUNT; TITLE 'Matched Case-control Study'; RUN;

v_case(Case Use of Vitamins) v_cont(Control Use of Vitamins) Frequency| Percent | Row Pct | Col Pct |1-YES |2-NO | Total YES | 100 | 50 | 150 | | | | | | | | | NO | 90 | 200 | 290 | | | | | | | | | Total Only the off diagonals give information Table gives evidence that vitamins were protective.

v_case(Case Use of Vitamins) v_cont(Control Use of Vitamins) Frequency| Percent | Row Pct | Col Pct |1-YES |2-NO | Total YES | 100 | 50 | 150 | | | | | | | | | NO | 90 | 200 | 290 | | | | | | | | | Total Statistics for Table of v_case by v_cont McNemar's Test Statistic (S) DF 1 Pr > S Obtained with AGREE Option  2 = (50-90) 2 /(50+90) = 11.43