# X Treatment population Control population 0 Examples: Drug vs. Placebo, Drugs vs. Surgery, New Tx vs. Standard Tx Let X = decrease (–) in cholesterol.

## Presentation on theme: "X Treatment population Control population 0 Examples: Drug vs. Placebo, Drugs vs. Surgery, New Tx vs. Standard Tx Let X = decrease (–) in cholesterol."— Presentation transcript:

X Treatment population Control population 0 Examples: Drug vs. Placebo, Drugs vs. Surgery, New Tx vs. Standard Tx Let X = decrease (–) in cholesterol level (mg/dL); Patients satisfying inclusion criteria RANDOMIZERANDOMIZE Treatment Arm Control Arm RANDOM SAMPLES End of Study T-test F-test (ANOVA ) Experiment significant? possible expected distributions:

S(t) = P(T > t) 0 1 T Examples: Drug vs. Placebo, Drugs vs. Surgery, New Tx vs. Standard Tx Let T = Survival time (months); End of Study Log-Rank Test, Cox Proportional Hazards Model Kaplan-Meier estimates population survival curves: significant? S 2 (t) Control S 1 (t) Treatment AUC difference survival probability

Case-Control studies Cohort studies

E+ vs. E– statistically significant Observational study designs that test for a statistically significant association between a disease D and exposure E to a potential risk (or protective) factor, measured via odds ratio, relative risk, etc. Lung cancer / Smoking PRESENT E+ vs. E– ?D+ vs. D– ? Case-Control studies Cohort studies Both types of study yield a 2 2 contingency table of data: D+D+D–D– E+E+ aba + b E–E– cdc + d a + cb + dn relatively easy and inexpensive relatively easy and inexpensive subject to faulty records, recall bias subject to faulty records, recall bias D+ vs. D– FUTUREPAST measures direct effect of E on D expensive, extremely lengthy expensive, extremely lengthy… Example: Framingham, MA study where a, b, c, d are the numbers of individuals in each cell. cases controlsreference group End of Study Chi-squared Test McNemar Test H 0 : No association between D and E.

–1 0 +1 As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. But what if the two variables – say, X and Y – are numerical measurements? Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? JAMA. 2003;290:1486-1493 Correlation Coefficient measures the strength of linear association between X and Y X Y Scatterplot r positive linear correlation negative linear correlation

–1 0 +1 As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? JAMA. 2003;290:1486-1493 Correlation Coefficient measures the strength of linear association between X and Y X Y Scatterplot r positive linear correlation negative linear correlation But what if the two variables – say, X and Y – are numerical measurements?

–1 0 +1 As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? JAMA. 2003;290:1486-1493 Correlation Coefficient linear measures the strength of linear association between X and Y X Y Scatterplot r positive linear correlation negative linear correlation But what if the two variables – say, X and Y – are numerical measurements?

As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? Correlation Coefficient linear measures the strength of linear association between X and Y But what if the two variables – say, X and Y – are numerical measurements? For this example, r = –0.387 (weak, negative linear correl) For this example, r = –0.387 (weak, negative linear correl)

For this example, r = –0.387 (weak, negative linear correl) For this example, r = –0.387 (weak, negative linear correl) residuals As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? But what if the two variables – say, X and Y – are numerical measurements? Want the unique line that minimizes the sum of the squared residuals. Simple Linear Regression Simple Linear Regression gives the best line that fits the data. Regression Methods

For this example, r = –0.387 (weak, negative linear correl) For this example, r = –0.387 (weak, negative linear correl) For this example, r = –0.387 (weak, negative linear correl) Y = 8.790 – 4.733 X (p =.0055) For this example, r = –0.387 (weak, negative linear correl) Y = 8.790 – 4.733 X (p =.0055) residuals As seen, testing for association between categorical variables – such as disease D and exposure E – can generally be done via a Chi-squared Test. Furthermore, if sample data does suggest that one exists, what is the nature of that association, and how can it be quantified, or modeled via Y = f (X)? Regression Methods But what if the two variables – say, X and Y – are numerical measurements? Want the unique line that minimizes the sum of the squared residuals. Simple Linear Regression Simple Linear Regression gives the least squares regression line. Furthermore however, the proportion of total variability in the data that is accounted for by the line is only r 2 = (–0.387) 2 = 0.1497 (15%).

Sincere thanks to Judith Payne Judith Payne Heidi Miller Heidi Miller Rebecca Mataya Rebecca Mataya Troy Lawrence Troy Lawrence YOU! YOU!

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