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Dealing With Statistical Uncertainty Richard Mott Wellcome Trust Centre for Human Genetics

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Synopsis Hypothesis testing P-values Confidence intervals T-test Non-parametric tests Sample size calculations Permutations Bootstrap

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Hypothesis testing Null Hypothesis H 0 Examples of H 0 : – Mean of a population is 3.0 – In a genetic association study, there is no association between disease state and the genotypes of a particular SNP Alternative Hypothesis H 1 Examples of H 1 : – Mean of population > 3.0 – There is an association between disease and genotype

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The Normal Distribution N( , ) Mean Variance 2 Many traits have Normal Distributions – Height – Weight Central Limit Theorem – average of many random quantities often has a Normal Distribution even if the individual observations do not

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P-values The P-value of a statistic Z is the probability of sampling a value more extreme than that observed value z if the null hypothesis is true. For a one-sided test, the p-value would be = Prob( Z > z | H 0 ) For a two-sided test it is = Prob( |Z| > z | H 0 ) Note this is not the probability density at x, but the tails of the cumulative distribution function Upper and lower 2.5% tails of the N(0,1) distribution, corresponding to |z|=1.96

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Power The power of a statistic Z is the probability of sampling a value more extreme than the observed value z if the alternative hypothesis is true. For a one-sided test, the power would be = Prob( Z > z | H 1 ) For a two-sided test it is = Prob( |Z| > z | H 1 ) Note this is not the probability density at x, but the tails of the cumulative distribution function H 0 : N(1,0) H 1 : N(1.5,1) H 2 : N(4,1) 0.05 = Prob( Z> 1.644|H 0 ) (5% upper tail) 0.44 = Prob( Z> 1.644|H 1 ) 0.99 = Prob( Z> 1.644|H 2 )

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The Likelihood Likelihood = “probability of the data given the model” Basis of parametric statistical inference Different hypotheses can often be expressed in terms of different values of the parameters Hypothesis testing equivalent to comparing the likelihoods for different

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The Likelihood Ratio Test General Framework for constructing hypothesis tests: S = Likelihood( data | H 1 ) / Likelihood( data | H 0 ) Reject H 0 if S > s(H 0, H 1 ) Threshold s is chosen such that there is a probability of making a false rejection under H 0. is the size of the test (or false positive rate or Type I error) e.g. =0.05 is the power of the test, the probability of correctly rejecting H 0 when H 1 is true e.g. =0.8 1- is the type II error, the false negative rate Generally, for fixed sample size n, if we fix then we can’t fix If we fix and then we must let n vary. The Neyman Pearson Lemma states that the likelihood ratio test is the most powerful of all tests of a given size Type III error: "correctly rejecting the null hypothesis for the wrong reason".

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Example: The mean of a Normal Distribution H 0 : mean = 0 vs H 1 : mean = 1 data: y 1, … y n independently and identically distributed with a Normal distribution N( , 2 ) Assume variance is the same and is known Therefore we base all inferences on the sample mean compared to the difference in the hypothesised means

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The Normal and T distributions For a sample from a Normal distribution with known mean and variance, the sample mean can be standardised to follow the standard Normal distribution And the 95% confidence interval for the mean is given by But we often wish to make inferences about samples from a Normal Distribution when the variance is unknown: The variance must be estimated from the data as the sample variance Because of this uncertainty the distribution of the sample mean is broader, and follows the T n-1 distribution

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The T distribution The T n and Normal distributions are almost identical for n>20, As a rule of thumb, an approximate 95% confidence interval for n>20 is

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T-tests The T-test compares the means of samples The T-test is optimal (in LR sense) if the data are sampled from a Normal distribution Even if the data are not Normal the test is useful in large samples There are several versions, depending on the details

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T tests for the comparison of sample means One Sample Test One sample y 1 ….. y n – H 0 : the sample mean = 0 – H 1 : the sample mean != 0 – Reject H 0 if – t n-1 ( ) is the quantile of the T n-1 distribution such that the probability of exceeding t n-1 ( ) is – Two sided test – in R, t n-1 (0.025) = qt(0.025,n-1)

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T tests for the comparison of sample means Paired T-test Two samples of paired data (x 1,y 1 ), (x 2,y 2 ) …. (x n,y n ) Example: – two experimental replicates taken from n individuals, we wish to test if the replicates are statistically similar H 0 : mean(x) = mean(y) Take differences d 1 = x 1 - y 1 … d n = x n - y n H 0 : mean(d) = 0 One sample T-test with 0 =0

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T tests for the comparison of sample means Two-Sample T-test Two samples of unpaired data x 1, x 2 ….. x n y 1, y 2 ….. y m H 0 : mean(x) = mean(y) If we assume the variance is the same in each group then we can estimate the variance as the pooled estimator The test statistic is (The case with unequal variances is more complicated)

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T tests in R t.test(x, y = NULL, alternative = c("two.sided", "less", "greater"), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95,...) x a numeric vector of data values. y an optional numeric vector data values. alternative a character string specifying the alternative hypothesis, must be one of "two.sided" (default), "greater" or "less". mu a number indicating the true value of the mean paired a logical indicating whether you want a paired t-test. var.equal a logical variable indicating whether to treat the two variances as being equal. conf.level confidence level of the interval. formula a formula of the form lhs ~ rhs where lhs is a numeric variable giving the data values and rhs a factor with two levels giving the corresponding groups. data an optional data frame containing the variables in the model formula. subset an optional vector specifying a subset of observations to be used.

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T tests in R samp1 <- rnorm( 20, 0, 1) # sample 20 numbers from N(0,1) distribution samp2 <- rnorm( 20, 1.5, 1) # sample from N(1.5,1) t.test( samp1, samp2,var.equal=TRUE ) Two Sample t-test data: samp1 and samp2 t = -3.5892, df = 38, p-value = 0.000935 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -1.9272037 -0.5372251 sample estimates: mean of x mean of y 0.4040969 1.6363113

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Outliers A small number of contaminating observations with different distribution Outliers can destroy statistical genuine significance Reason: the estimate of the variance is inflated, reducing the T statistic And sometimes create false significance : Type III error: "correctly rejecting the null hypothesis for the wrong reason". There are Robust Alternatives to the T-test: Wilcoxon tests > samp3 <- samp2 > samp3[1] = 40 # add one outlier > t.test( samp1, samp3,var.equal=TRUE ) Two Sample t-test data: samp1 and samp3 t = -1.5911, df = 38, p-value = 0.1199 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -7.0519594 0.8450837 sample estimates: mean of x mean of y 0.4040969 3.5075347 > var(samp2) [1] 1.160251 > var(samp3) [1] 74.88977

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Non-parametric tests

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Non-parametric tests: Wilcoxon signed rank test (alternative to Paired T-test) Two samples of paired data (x 1,y 1 ), (x 2 -y 2 ) …. (x n,y n ) Take differences d 1 = x 1 - y 1 … d n = x n – y n H 0 : distribution of d’s is symmetric about 0 Compute the ranks of the absolute differences, rank(|d 1 |) etc Compute the signs of the differences Compute W, the sum of the ranks with positive signs If H 0 true then W should be close to the mean rank n/2 The distribution of W is known and for n>20 a Normal approximation is used. in R wilcox.test(x, y, alternative = c("two.sided", "less", "greater"), mu = 0, paired = TRUE, exact = NULL, correct = TRUE, conf.int = FALSE, conf.level = 0.95,...)

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Wilcoxon rank sum test (alternative to unpaired T-test) Two samples of unpaired data x 1, x 2 ….. x n y 1, y 2 ….. y m Arrange all the observations into a single ranked series of length N = n+m. R 1 = sum of the ranks for the observations which came from sample 1. The sum of ranks in sample 2 follows by calculation, since the sum of all the ranks equals N(N + 1)/2 Compute U = min(R 1,R 2 ) - n(n + 1)/2. z = (U – )/ Normal approximation = nm/2 = nm(n=m+1)/12 The Wilcoxon Rank sum test is only about 5% less efficient than the unpaired T-test (in terms of the sample size required to achieve the same power) in large samples.

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Wilcoxon rank sum test (alternative to unpaired T-test) >wilcox.test(samp1,samp2) Wilcoxon rank sum test data: samp1 and samp2 W = 82, p-value = 0.001041 alternative hypothesis: true mu is not equal to 0 > wilcox.test(samp1,samp3) # with outliers Wilcoxon rank sum test data: samp1 and samp3 W = 82, p-value = 0.001041 alternative hypothesis: true mu is not equal to 0

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Binomial Sign Test (T-test on one sample) One sample y 1 ….. y n H 0 : the sample mean = 0 Count M, the number of y’s that are > 0 If H 0 is true, this should be about n/2, and should be distributed as a Binomial random variable B(n/2,n)

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Contingency Tables SmokerNon-smoker Male10 20 Female51217 152237 Do Males and Females have different rates of smoking?

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Contingency Tables Data are a table of counts, with the row and column margins fixed. H o : the counts in each cell are consistent with the rows and columns acting independently n 11 r 1 c 1 /N n 12 r 1 c 2 /N r 1 = n 11 + n 12 n 21 r 2 c 1 /N n 22 r 2 c 2 /N r 2 = n 21 + n 22 c 1 = n 11 + n 21 c 2 = n 12 + n 22 N

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The Chi-Squared statistics

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Fisher’s Exact Test for 2 x 2 Contingency tables Similar idea to a permutation test (see below) Given the marginal totals in a 2x2 table, we can compute the probability P of getting the observed cell counts assuming the null hypothesis of no associations between classifying factors. – hypergeometric distribution. Find all tables with same margins and with probabilities no bigger than P P-value of the table is the sum of these probabilities

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Fisher’s Exact Test SmokerNon-smoker Maleaba+b Femalecdc+d a+cb+dn R implementation: m <- matrix(c(10,10,5,12),nrow=2) fisher.test(m) Fisher's Exact Test for Count Data data: m p-value = 0.3152 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.513082 11.942154 sample estimates: odds ratio 2.342796 > SmokerNon-smoker Male10 20 Female51217 152237

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Permutation Tests Another non-parametric Alternative Permutation is a general principle with many applications to hypothesis testing. But it is not useful for parameter estimation The method is best understood by an example: – Suppose a data set is classified into N groups. We are interested in whether the differences between group means (eg if N=2 then the two sample T test or Wilcoxon Rank Sum test is appropriate) – Compute a statistic S (eg the difference between the maximum and minimum group mean) – If there are no differences between groups then any permutation of the group labellings between individuals should produce much the same result for S – Repeat i = 1 to M times: permute the data compute the statistic on the permuted data, S i – The Permutation p-value is the fraction of permutations that exceed S It’s that simple!

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Bootstrapping Given a set of n things S = {x 1 …. x 2 }, a bootstrap sample is formed by sampling with replacement from S, to create a new set S* of size n Original Resample Some elements of S will be missing in S* and some will be present multiply often (about 1/e = 63% of the elements will be present in S* on average) S* can be used as a replacement for S in any algorithm that works on S, and we can repeatedly draw further bootstrap samples. Any statistic Z that can be calculated on S can also be calculated on S*, and so we can construct the bootstrap sampling distribution of Z from repeated samples. From this we can make inferences about the variation of Z, eg construct a 95% confidence interval (if Z is numeric) Can be used for parameter estimation

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Bootstrapping Bootstrapping is a general way to evaluate the uncertainty in estimates of statistics whilst making few assumptions about the data It is particularly useful for complex statistics whose distribution cannot be evaluated analytically Example: – In a gene expression study, we wish to find a minimal set of SNPs that in combination will predict the variation in a phenotype. – Suppose we have a deterministic algorithm A that will find us a plausible set of SNPs (ie a model). This is likely to be a complex programme and contain some arbitrary steps. It is also unlikely to give any measure of uncertainty to the model. – There may be many possible sets of SNPs with similar claims, so we would like a way to average across different models, and evaluate the evidence for a SNP as the fraction of models in which the SNP occurs To do this, we need a way to perturb the data, to generate new data sets on which we can apply the procedure A Bootstrapping is one way to do this

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