13%20Multiple%20Testing%202_28_08

13%20Multiple%20Testing%202_28_08 - 1 1 Introduction to...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 1 Introduction to Multiple Testing Method for the Analysis of Microarray Data Peng Liu 2/28/2008 2 Possible Errors in Testing ONE Hypothesis ¡ Type I Error : false positives ¡ Type II Error: false negatives (1-power) ¡ Power : true positives correct (Power) Type II Error False Null Type I Error correct True Null Reject Null Accept Null Hypothesis 3 The Multiple Testing Problem ¡ Suppose one test of interest has been conducted for each of the m genes in a microarray experiment. ¡ Often, the test of interest is that for each gene, whether it is differentially expressed between different treatments or not. ¡ Let H 01 , H 02 , ... , H 0m denote the null hypotheses (often interpreted as non-differential expression) corresponding to the m tests (genes). 4 The Multiple Testing Problem (continued) ¡ Suppose m of the null hypotheses are true (corresponding to non-differentially expressed genes) m 1 of the null hypotheses are false (corresponding to differentially expressed genes). 2 5 The Multiple Testing Problem (continued) ¡ If we set level 5% for each test , the number of type 1 error (false positives) is expected to be 5% of m . ¡ Considering the high dimensionality of microarray data, the number of errors is expected to be really big if we only control error at the level of individual test. ¡ We need procedure to control multiple testing error. 6 The Multiple Testing Problem (continued) ¡ Let p 1 , p 2 , ... , p m denote the p-values corresponding to the m tests ( each test corresponding to one gene ). ¡ Let c denote a value between 0 and 1 that will serve as a cutoff for significance :- Reject H 0i if p i ≤ c (declare significant)- Fail to reject (or accept) H 0i if p i > c (declare non-significant) 7 Table of Outcomes Accept Null Reject Null Declare Non-Sig. Declare Sig. No Discovery Declare Discovery Negative Result Positive Result True Nulls U V m False Nulls T S m 1 Total W R m This is the slide that you need for question 4 of homework 3. 8 Accept Null Reject Null Declare Non-Sig. Declare Sig. No Discovery Declare Discovery Negative Result Positive Result True Nulls U V m False Nulls T S m 1 Total W R m U=number of true negatives Table of Outcomes 3 9 Accept Null Reject Null Declare Non-Sig. Declare Sig. No Discovery Declare Discovery Negative Result Positive Result True Nulls U V m False Nulls T S m 1 Total W R m V=number of false positives =number of false discoveries =number of type 1 errors Table of Outcomes 10 Accept Null Reject Null Declare Non-Sig. Declare Sig. No Discovery Declare Discovery Negative Result Positive Result True Nulls U V m False Nulls T S m 1 Total W R m T=number of false negatives =number of type 2 errors Table of Outcomes 11 Accept Null Reject Null Declare Non-Sig. Declare Sig....
View Full Document

This note was uploaded on 08/26/2009 for the course STAT 416 taught by Professor Peng,l during the Spring '08 term at Iowa State.

Page1 / 21

13%20Multiple%20Testing%202_28_08 - 1 1 Introduction to...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online