 # From the homework: Distribution of DNA fragments generated by Micrococcal nuclease digestion mean(nucs) = 113.5 bp median(nucs) = 110 bp sd(nucs+ = 17.3.

## Presentation on theme: "From the homework: Distribution of DNA fragments generated by Micrococcal nuclease digestion mean(nucs) = 113.5 bp median(nucs) = 110 bp sd(nucs+ = 17.3."— Presentation transcript:

From the homework: Distribution of DNA fragments generated by Micrococcal nuclease digestion mean(nucs) = 113.5 bp median(nucs) = 110 bp sd(nucs+ = 17.3 bp

http://www.stat.wisc.edu/courses/language Link to Jean-Yves Sgro’s R & Bioconductor Manual http://akka.genetics.wisc.edu/groups/genetics875/wiki/7a793/M_Oct_1_.html

R 2 = 0.9897 R 2 = 0.686 R 2 = 0.804 Why did the correlation go up?

4 Expression difference Gene X expression under condition 1 Gene X expression under condition 2 Select differentially expressed genes to focus on Methods of gene selection: -- arbitrary fold-expression-change cutoff example: genes that change >3X in expression between samples -- statistically significant change in expression requires replicates

5 Test if the means of 2 (or more) groups are the same or statistically different The ‘null hypothesis’ H 0 says that the two groups are statistically the same -- you will either accept or reject the null hypothesis Choosing the right test: parametric test if your data are normally distributed with equal variance nonparametric test if neither of the above are true Why do the data need to be normally distributed?

6 Test if the means of 2 groups are the same or statistically different The ‘null hypothesis’ H 0 says that the two groups are statistically the same -- you will either accept or reject the null hypothesis T = X 1 – X 2 difference in the means standard error of the difference in the means SED If your two samples are normally distributed with equal variance, use the t-test If T > T c where T c is the critical value for the degrees of freedom & confidence level, then reject H 0 Notice that if the data aren’t normally distributed mean and standard deviation are not meaningful.

7 T = D Average difference in expression Standard error of the mean difference SEM If your two samples are normally distributed with equal variance AND your data were paired before collection, use the paired t-test The paired t-test for gene expression ratios Example: Tumor sample before and after treatment Gene expression differences expressed as ratios eg) mutant vs. wt log 2 [ratio]: 5.0 4.3 6.7 If T > T c where T c is the critical value for the degrees of freedom (n-1) & confidence level, then reject H 0

8 Test if the means of 2 (or more) groups are the same or statistically different The ‘null hypothesis’ H 0 says that the two groups are statistically the same -- you will either accept or reject the null hypothesis ANOVA (ANalysis Of Variance): for comparing 2 or more means variation between samples variation within samples F = ANOVA only tells you that at least one of your samples is different … may need to identify which is different for >2 sample comparisons If F > F c where F c is the critical value for the degrees of freedom (n-1) & confidence level, then reject H 0

9 Assessing & minimizing error in calls Type I error = false positives FDR = False Discovery Rate Type II error = false negatives Balance between minimizing false positives vs. false negatives Assessing false positives vs. false negatives: sensitivity vs. specificity Sensitivity (how well did you find what you want): # of true positives # of total significant calls ( = #true positives + # false negatives) Specificity (how well did you discriminate): # of true negatives # of total negative calls (= #true negatives + #false positives)

Assessing accuracy based on known truth: Receiver-Operator Curves (ROC) Plot the fraction of True Positives (TPR) vs False Positives (FPR) called at each significance threshold Known Truth is either a set of positive controls … or can come from simulated data

11 When working with many genes must correct for multiple testing … p < 0.01 means that there is a 1 in 100 chance that the observation is H 0 But if you have 30,000 genes, with 0.01 change that each conclusion is wrong then you will get 300 false positives! Adjust the p-value cutoff such that there is a 1 in 100 chance of false identification for each gene: p = 0.01 / 30,000 ‘trials’ p < 3 x 10 -7 is significant (this is also known as Bonferroni correction)

Newer, better way of dealing with this is FDR correction FDR: false discovery rate How many of the called positives are false? 5% FDR means 5% of calls are false positive This is different from the false positive rate: The rate at which true negatives are called significant 5% false positives means 5% of true negatives are incorrectly called significant “The p-value cutoff [and false positive rate] says little about the content of the features actually called significant” (Storey and Tibshirani 2003) Storey and Tibshirani 2003: q-value to represent FDR

FDR = expected ratio of false positives vs all positives (Expected [F/S]) q value: for a given region of data space, what fraction of genes in that region are false? eg) Gene X has a q = 0.04 … this means that if you call Gene X significant, then all the genes with better statistics must also be called significant -> 4% of all of these genes are false positives “The q-value for a particular feature is the expected proportion of false positives incurred when calling that feature significant.”

FDR = expected ratio of false positives vs all positives: Expected [F/S] ~ Expected[F] / Expected [S] -- can initially estimate S based on a simple p-value cutoff We need to estimate  0 = m 0 / m = fraction of all features that are truly negative Genes with p > 0.5 show a relatively flat density … because we expect that p-values of null genes are randomly distributed, we assume that most of these genes are true nulls … (The tuning parameter is the p cutoff above which nulls are assumed) The density for genes with p>0.5 allows us to estimate the # of true negatives and thus  0  in this case

Download ppt "From the homework: Distribution of DNA fragments generated by Micrococcal nuclease digestion mean(nucs) = 113.5 bp median(nucs) = 110 bp sd(nucs+ = 17.3."

Similar presentations