HSS4303B – Intro to Epidemiology Mar 8, 2010 – Matched Studies

Summary from Last Time Case control study design Sources of cases and controls Problems in selection of controls Practical and conceptual problems Matching Recall problems Limitation in recall Recall bias Multiple controls Type of case control studies Nested case control study Prevalence study

Summary of related studies Table 10-10. Finding Your Way in the Terminology Jungle Case-control study   = Retrospective study Cohort study Longitudinal study Prospective study Prospective cohort study Concurrent cohort study Concurrent prospective study Retrospective cohort study Historical cohort study Nonconcurrent prospective study Randomized trial Experimental study Cross-sectional study Prevalence survey

Review of Cohort studies

Design of cohort study (1)

Design of cohort study (2)

Prospective vs Retrospective Cohort Prospective study Identify a population and follow them prospectively until events develop Concurrent cohort Longitudinal study

Cohort study: prospective design Pitfalls of the study Loss of subjects Loss of investigators Lifestyle changes in the subjects

Prospective vs Retrospective Cohort Retrospective study Identify a population and observe the events as they occur and retrospectively determine their exposure status from historical records Non-current prospective study Historical cohort study

Cohort study: retrospective study Pitfalls of the study: Availability of records Quality of records Recall bias

Prospective and retrospective studies The designs of both prospective and retrospective study are similar Exposed and unexposed population are compared for the events Difference in time frame: Prospective study – forward time frame Retrospective study – historical records for similar period of time as prospective study

Prospective and retrospective studies

Potential biases in cohort studies Bias in assessment of the outcome Information on exposure status biases outcome status Information bias Difference in available information for the exposed and unexposed Biases from non-response and losses to follow-up Attrition rate creates study power problems Analytic biases Subjectivity at the time of analyses

Table 8–2. Comparison of the Attributes of Retrospective and Prospective Cohort Studies. Retrospective Approach Prospective Approach Information Less complete and accurate More complete and accurate Discontinued exposures Useful Not useful Emerging new exposures Expense Less costly More costly Completion time Shorter Longer

Advantages and Disadvantages of Cohort Studies. Direct calculation of risk ratio (relative risk) Time consuming May yield information on the incidence of disease Often requires a large sample size Clear temporal relationship between exposure and disease Expensive Particularly efficient for study of rare exposures Not efficient for the study of rare diseases Can yield information on multiple exposures Losses to follow-up may diminish validity Can yield information on multiple outcomes of a particular exposure Changes over time in diagnostic methods may lead to biased results Minimizes bias   Strongest observational design for establishing cause and effect relationship

Review of Odds Ratios (Case-Control Study)   Cases Controls Exposed 6 3 Nonexposed 4 7 10 Odds ratio = 3.5 Compute odds ratio of this dataset

Case control study of 10 unmatched subjects: summary Figure 11-8 A case-control study of 10 cases and 10 unmatched controls.   Cases Controls Exposed 6 3 Nonexposed 4 7 10

But What if Data is Matched? Why do we match again?

Matched case control study Cases are matched with the controls on specific variables Cases and controls are analyzed in pairs rather than individual subjects Concordant pairs Discordant pairs 1. Pairs in which both the case and the controls were exposed 2. Pairs in which neither the case nor the control was exposed 3. Pairs in which the case was exposed but not the control 4. Pairs in which the control was exposed and not the case

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Eg, outcome = getting the runs (the cases) vs not getting the runs (controls) exposure = did you attend the picnic and eat the egg salad? confounder = lactose intolerance (?)

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Assume this is an unmatched study. How does the contingency table look?

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Outcome (cases) No outcome (controls) Totals Exposed (picnic) 2 3 5 Not exposed (no picnic) 1 4 8 Odds ratio? (2x1)/(3x2) =0.33

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Now let’s organize the data considering that it’s a matched study

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed   Controls Exposed Not Exposed Cases

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed   Controls Exposed Not Exposed Cases 1 2

Concordant and discordant pairs   Control Exposed Not Exposed Case a b c d a pairs – both case and the control were exposed b pairs – case was exposed but not the control c pairs – case was not exposed but the control is exposed d pairs – neither case nor control was exposed a and d pairs are concordant pairs b and c pairs are discordant pairs

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed   Controls Exposed Not Exposed Cases 1 2 concordant

Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed   Controls Exposed Not Exposed Cases 1 2 concordant discordant

Individual matching (1:1) Echovirus meningitis outbreak, Germany, 2001 Was swimming in pond “A” risk factor? Case control study with each case matched to one control Concordant pairs Discordant pairs Source: A Hauri, RKI Berlin

Odds ratio for matched pairs Odds ratio for matched pairs is: The ratio of the ratio of the discordant pairs The ratio of the number of pairs in which the case was exposed and the control was not, to the number of pairs in which the control was exposed and the case was not exposed b / c The ratio of the number of pairs that support the hypothesis of an association to the number of pairs that negate the hypothesis of an association ?

Matched cases and controls 2 x 2 table   Control Exposed Not Exposed Case 2 4 1 3 Concordant pairs: 2 pairs (exposed and exposed) and 3 pairs (not exposed and not exposed) Discordant pairs: 4 pairs (exposed and not exposed and 1 pair (not exposed and exposed) Odds ratio = b / c = 4 / 1 = 4

Picnic Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Matched odds ratio =b/c =1/2 =0.5   Controls Exposed Not Exposed Cases 1 2

Remember this example? Concordant pairs Discordant pairs Source: A Hauri, RKI Berlin

Individual matching (1:1): Analysis Echovirus meningitis outbreak, Germany, 2001 Was swimming in pond “A” risk factor? Case control study with each case matched to one control

What Else Can We Do With These Data? Remember the Chi-Square test?

Chi-square Chi square is a non-parametric test of statistical significance for bivariate tabular analysis It lets you know the degree of confidence you can have in accepting or rejecting an hypothesis It provides information on whether or not two different samples are different enough in some characteristic or aspect of their behaviour

Chi Square There are actually all sorts of chi-square tests out there Pearson’s Yate’s Mantel-Haenszel Portmanteau Fisher’s Exact <- We’ll be using this one

Also need to compute something called “degrees of freedom”

Chi-square calculation Variable 1  Variable 2  Data type 1  Data type 2  Totals  Category 1  a b a + b  Category 2  c d c + d  Total a + c b + d a + b + c + d = N Chi square = N(ad-bc)2 / (a+b) (c+d) (b+d) (a+c) The degrees of freedom =(number of columns minus one) x (number of rows minus one) not counting the totals for rows or columns. For our data this gives (2-1) x (2-1) = 1.

Chi-square calculations Number of animals that survived the treatment    Dead  Alive Total  Treated  36  14  50  Not treated  30  25  55  Total  66  39  105 (36x25)/(14x30) = 2.14 Odds ratio = Chi square = 105[(36)(25) - (14)(30)]2 / (50)(55)(39)(66) = 3.418 (2-1)x(2-1) = 1 DOF= Now what do we do with this?

Degrees of freedom and chi square table Df 0.5 0.10 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.210 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.070 13.388 15.086 20.517 Using the Chi square table The corresponding probability is 0.10<P<0.05. This is below the conventionally accepted significance level of 0.05 or 5%, so the null hypothesis that the two distributions are the same is verified. In other words, when the computed x2 statistic exceeds the critical value in the table for a 0.05 probability level, then we can reject the null hypothesis of equal distributions. Since our x2 statistic (3.418) did not exceed the critical value for 0.05 probability level (3.841) we can accept the null hypothesis that the survival of the animals is independent of drug treatment

p-value The p-value is the probability that your sample could have been drawn from the population being tested given the assumption that the null hypothesis is true. A p-value of .05, for example, indicates that you would have only a 5% chance of drawing the sample being tested if the null hypothesis was actually true. A p-value close to zero signals that your null hypothesis is false, and typically that a difference is very likely to exist. Large p-values closer to 1 imply that there is no detectable difference for the sample size used. A p-value of 0.05 is a typical threshold used to evaluate the null hypothesis.

p-value So what does a p-value of 0.10 mean? We “fail to reject the null hypothesis”

What is the null hyp that we are testing? In cohort studies, the chi-square test tells us whether to accept or reject the null hypothesis that RR=1 In case-control studies, the chi-square test tells us whether or accept or reject the null hypothesis that OR=1 Pierson’s chi-square is NOT appropriate to test the null hypothesis of whether the matched study pairs are related For that we use something called McNemar’s test, which we will not cover in this class

Remember the Picnic Data Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Matched odds ratio =b/c =1/2 =0.5   Controls Exposed Not Exposed Cases 1 2 Pretend it’s unmatched and construct the contingency table…

Data Can you compute a chi-square for this? Pair Outcome – Yes (cases) Outcome –No (controls) 1 2 3 4 Exposed Not exposed Outcome (cases) No outcome (controls) Totals Exposed (picnic) 2 3 5 Not exposed (no picnic) 1 4 8 Odds ratio? (2x1)/(3x2) =0.33

Caveat to Pierson’s Chi Square Typically, does not work well if any cell has a count of <5 If it does, better off using Fisher’s Exact Test or some other similar test We will not be doing that in this class

Summary Chi square = (ad-bc)2 (a+b+c+d) / (a+b) (c+d) (b+d) (a+c) The degrees of freedom equal (number of columns minus one) x (number of rows minus one) not counting the totals for rows or columns.

If you’re lazy… Lots of online OR, RR and chi-square calculators Eg, http://faculty.vassar.edu/lowry/odds2x2.html

Homework 12 women with uterine cancer and 12 without were asked if they’d ever used supplemental estrogen. Each woman with cancer was matched by race, weight and parity to a woman without cancer: pair Women with cancer Women without cancel 1 2 3 4 5 6 7 8 9 10 11 12 Estrogen user Estrogen nonuser

Homework What is the estimated relative risk of cancer when analyzing this study as a matched-pairs study? What is the estimated relative risk of cancer when analyzing this study as an unmatched study? What is the chi square statistic of the (unmatched) relationship between cancer and estrogen intake? What is the null hypothesis being tested by the chi-square test? What does the p-value of the statistic tell you about whether to reject or accept the null hypothesis? Estimated RR is the same as OR 3.00 4.00 2.67 Null hypothesis = “the OR =1” 0.5<p<1.0 therefore we fail to reject the null hypothesis. However, cell values are <5, therefore Pierson’s chi-square cannot be accurately computed for this sample