March 30 More examples of casecontrol studies General

Exam Question § • • • p = proportion who smoke cigarettes n =

HRT Use and Colonic Polyps • Case-control study conducted at Digestive Healthcare (DH) in

HRT Use and Polyps Case (Polyps) HRT Use No HRT Use Control (No Polyps)

USING SAS DATA hrt; INFILE DATALINES; INPUT hrt ccstatus count; DATALINES; 1 1 72

hrt ccstatus Frequency| Percent | Row Pct | Col Pct | 1| 2| Total

Interpretation (RO = 0. 46) • Women who use HRT are at a 54%

Analyses of I x J Tables • Two factors, one with I levels, second

Example TOMHS • Rows – 6 treatment groups • Columns – 4 levels of

Cough None Mild Mod/Sev Diuretic 99 28 6 133 ACE 86 36 11 133

Cough- Expected Frequencies None Mild Mod/Sev Diuretic 99 (95. 9) 28 (29. 1) 6

USING SAS DATA cough; INPUT group cough count; DATALINES; 1 1 99 1 2

GROUP cough Frequency| Expected | Percent | Row Pct | Col Pct | 1|

Statistic DF Value Prob ---------------------------Chi-Square 4 5. 4105 0. 2477 Likelihood Ratio Chi-Square 4

1 -Sample Z-Test: Matched Pair Data Control Positive Control Negative Pos a b Neg

Example – Vitamin Use/Disease (440 Pairs) Control Vitamin + Control Vitamin - Vit +

Example of Matched Case-control Study DATA vitamin; INFILE DATALINES; INPUT v_case $ v_cont $

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 |

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

Exam Question § • • • p = proportion who smoke cigarettes n = 10 x = 3 report smoking p = 3/10 = 0. 30 is estimate of p SE(p) = 95% CI for p: = 0. 14 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 (1990 -1995) • 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) HRT Use No HRT Use Control (No Polyps) 72 175 247 102 114 216 289 463 174 RO HRT Use (Case v Control) RO = 72/102 175/114 = 0. 46 c 2 = ( 463 ) (RO)2 ( 174) (289) (247) (216) =16. 04

USING SAS DATA hrt; INFILE DATALINES; INPUT hrt ccstatus count; DATALINES; 1 1 72 ccstatus =1 is case 1 2 175 ccstatus = 2 is control 2 1 102 2 2 114 ; PROC FREQ DATA=hrt; TABLES hrt*ccstatus/CHISQ RELRISK; WEIGHT count; RUN;

hrt ccstatus Frequency| Percent | Row Pct | Col Pct | 1| 2| Total -----+--------+ 1 | 72 | 175 | 247 | 15. 55 | 37. 80 | 53. 35 | 29. 15 | 70. 85 | | 41. 38 | 60. 55 | -----+--------+ 2 | 102 | 114 | 216 | 22. 03 | 24. 62 | 46. 65 | 47. 22 | 52. 78 | | 58. 62 | 39. 45 | -----+--------+ Total 174 289 463 37. 58 62. 42 100. 00 Estimates of the Relative Risk (Row 1/Row 2) Type of Study Value 95% Confidence Limits --------------------------------Case-Control (Odds Ratio) 0. 4598 0. 3135 0. 6744 Cohort (Col 1 Risk) 0. 6173 0. 4855 0. 7849 Cohort (Col 2 Risk) 1. 3424 1. 1562 1. 5587

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: p 1 = p 2 = p 3 (when J=2) • Chi-square (c 2) test with (I-1 x J-1 df): Sum over all cells (Ix. J)

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 None Mild Mod/Sev Diuretic 99 28 6 133 ACE 86 36 11 133 174 45 13 232 Placebo 359 109 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 30 498

Cough- Expected Frequencies None Mild Mod/Sev Diuretic 99 (95. 9) 28 (29. 1) 6 (8. 0) 133 ACE 86 (95. 9) 36 (29. 1) 11 (8. 0) 133 174 (167. 2) 45 (50. 8) 13 (14. 0) 232 Placebo 359 109 30 c 2(4 DF) = (99 -95. 9)2 /95. 9 + (28 -29. 1)2 /29. 1 + … + (13 -14)2/14 = 5. 41 P = 0. 25

USING SAS DATA cough; INPUT group cough count; DATALINES; 1 1 99 1 2 28 cough =1 (none) 1 3 6 cough = 2 (mild) 2 1 86 2 2 36 cough = 3 (mod/severe) 2 3 11 3 1 174 3 2 45 3 3 13 ; PROC FREQ DATA=hrt; TABLES group*cough/CHISQ CMH; WEIGHT count; RUN;

GROUP cough Frequency| Expected | Percent | Row Pct | Col Pct | 1| 2| 3| Total -----+--------+----+ 3 | 99 | 28 | 6 | 133 | 95. 878 | 29. 11 | 8. 012 | | 19. 88 | 5. 62 | 1. 20 | 26. 71 | 74. 44 | 21. 05 | 4. 51 | | 27. 58 | 25. 69 | 20. 00 | -----+--------+----+ 5 | 86 | 36 | 11 | 133 | 95. 878 | 29. 11 | 8. 012 | | 17. 27 | 7. 23 | 2. 21 | 26. 71 | 64. 66 | 27. 07 | 8. 27 | | 23. 96 | 33. 03 | 36. 67 | -----+--------+----+ 6 | 174 | 45 | 13 | 232 | 167. 24 | 50. 779 | 13. 976 | | 34. 94 | 9. 04 | 2. 61 | 46. 59 | 75. 00 | 19. 40 | 5. 60 | | 48. 47 | 41. 28 | 43. 33 | -----+--------+----+ Total 359 109 30 498 72. 09 21. 89 6. 02 100. 00

Statistic DF Value Prob ---------------------------Chi-Square 4 5. 4105 0. 2477 Likelihood Ratio Chi-Square 4 5. 2873 0. 2591 Mantel-Haenszel Chi-Square 1 0. 0245 0. 8756 Phi Coefficient 0. 1042 Contingency Coefficient 0. 1037 Cramer's V 0. 0737 Effective Sample Size = 498 Frequency Missing = 7 Cochran-Mantel-Haenszel Statistics (Based on Table Scores) Statistic Alternative Hypothesis DF Value Prob -------------------------------1 Nonzero Correlation 1 0. 0245 0. 8756 2 Row Mean Scores Differ 2 4. 9233 0. 0853 3 General Association 4 5. 3996 0. 2487 Takes into account the ordering of the row categories

1 -Sample Z-Test: Matched Pair Data Control Positive Control Negative Pos a b Neg c d Case Ho: p = 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) c 2 = (b-c)2/(b+c) Analyses is done on discordant pairs b and c Called Mc. Nemar’s chi-square

Example – Vitamin Use/Disease (440 Pairs) Control Vitamin + Control Vitamin - Vit + 100 50 Vit - 90 200 Case Ho: p = 0. 5 where n=140 and b = 50 c 2 = (50 -90)2/(50+90) = 11. 43 (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 100 pairs where both case and control took vitamins 1 -YES 2 -NO 50 2 -NO 1 -YES 90 2 -NO 200 ;

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 -----+--------+ 1 -YES | 100 | 50 | 150 | 22. 73 | 11. 36 | 34. 09 | 66. 67 | 33. 33 | Only the off diagonals | 52. 63 | 20. 00 | give information -----+--------+ 2 -NO | 90 | 200 | 290 Table gives evidence that vitamins were | 20. 45 | 45. 45 | 65. 91 protective. | 31. 03 | 68. 97 | | 47. 37 | 80. 00 | -----+--------+ Total 190 250 440 43. 18 56. 82 100. 00

v_case(Case Use of Vitamins) v_cont(Control Use of Vitamins) Frequency| Percent | Row Pct | Col Pct |1 -YES |2 -NO | Total -----+--------+ 1 -YES | 100 | 50 | 150 | 22. 73 | 11. 36 | 34. 09 | 66. 67 | 33. 33 | | 52. 63 | 20. 00 | c 2 = -----+--------+ 2 -NO | 90 | 200 | 290 | 20. 45 | 45. 45 | 65. 91 | 31. 03 | 68. 97 | | 47. 37 | 80. 00 | -----+--------+ Total 190 250 440 43. 18 56. 82 100. 00 Statistics for Table of v_case by v_cont Mc. Nemar's Test ------------Statistic (S) 11. 4286 DF 1 Pr > S 0. 0007 (50 -90)2/(50+90) = 11. 43 Obtained with AGREE Option