Statistics - Lecture 04

Transcription

1 Statistics - Lecture 04 Nicodème Paul Faculté de médecine, Université de Strasbourg file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 1/40

2 Correlation In many situations the objective in studying the joint behavior of two variables is to see whether they are related. - X: temperature and Y : body length in drosophila - X: Height and Y : Weight - X: parent Height and Y : child Height - X: eyes Color and Y : hair Color - X: Color and Y : perceived Taste Given n pairs of observations ( x 1, y 1 ), ( x 2, y 2 ),..., ( x n, y n ), it is natural to speak of X and Y having a positive relationship if large x's are paired with large y's and small x's with small y's. Similarly, if large x's are paired with small y's and small x's with large y's, then a negative relationship between the variables is implied. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 2/40 2/40

3 Correlation In a population, we collect a random sample of 38 individuals. For each individual, we measure his or her height (mm) and his or her weight. Height Weight file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 3/40 3/40

4 Independence Is the variable Height is independent of the variable Weight? No Yes I cannot tell Submit Show Hint Show Answer Clear file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 4/40 4/40

5 Correlation A bivariate data set consists of measurements or observations on two variables, X and Y file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 5/40 5/40

6 Sample correlation The Pearson's sample correlation coe cient r of a bivariate data ( x i, y i ) for i = 1,..., n is given by: r = n i=1 x i xˉ y i ȳ ( )( ) ( ( )( ( ) n i=1 x i xˉ) 2 n i=1 y i ȳ ) 2 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 6/40 6/40

7 Sample correlation Some properties of the Pearson correlation coe cent are as following: The value of r does not depend on the unit of measurement for either variable The value of r does not depend on which of the two variables is considered X The value of r is between 1 and -1 A correlation coe cient of r = 1 occurs only when all the points in a scatterplot of the data lie exactly on a straight line that slopes upward. Similarly, r = 1 only when all the points lie exactly on a downward-sloping line. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 7/40 7/40

8 Examples of scatterplots Valeurs possibles: -1, -0.8, -0.4, 0, 0.4, 0.8, 1 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 8/40 8/40

9 Examples of scatterplots with coe cient correlations file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 9/40 9/40

10 Hypothesis testing on ρ X i and Y i be jointly normal random variables R(X, Y ) = n ( X )( ) i=1 i Xˉ Y i Ȳ n i=1 X i Xˉ) 2 n i=1 Y i Ȳ ) 2 ( ( )( ( ) is an estimator for ρ. Hypotheses: H 0 : ρ = 0 H 1 : ρ 0 Test statistic: T = R n 2 1 R 2 follows a t distribution with n 2 degrees of freedoom Decision is based on the critical value t n 2 1 α/2 or p value = P H0 (T > t n 2 ) file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 10/40 10/40

11 Example - linear relationship r = As n = 38, the value of the t statistic is: t n 2 t = r = r 2 For α = 0.05, = As 4.74 > 2.028, we reject the null hypothesis, meaning that there is strong evidence that ρ, the population parameter is signi cantly di erent from 0. Note that you cannot nd this value neither calculate the p-value using the t table provided. Indeed, there is no entry for the t distribution with 36 degrees of freedom in the table. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 11/40 11/40

12 If a signi cant sample correlation coe meaning? cient between two variables X and Y is observed, what could be its X causes Y Y causes X Some third factor, either directly or indirectly, causes both X and Y An unlikely event has occurred and a large sample correlation coe cient has been generated by chance from a population in which X and Y are, in fact, not correlated The correlation is purely nonsensical, a situation that may arise when measurements of X and Y are not taken on a common unit of association Submit Show Hint Show Answer Clear file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 12/40 12/40

13 Goodness of t - checking for normality A normal probability plot is a scatterplot of the (normal score z i, observed value x i ) pairs for i = 1, 2,..., n. The empirical cummulative distribution is de ned as: P(X x i ) = /n 0.5 1/n i n if i = 1 if i = n otherwise And the z i scores are found such that P(Z ) = P(X ). z i x i In software packages, the graphical representation can be obtained using a quantile - quantile plot. A strong linear pattern in a normal probability plot suggests that population normality is plausible. On the other hand, systematic departure from a straight-line pattern (such as curvature in the plot) indicates that it is not reasonable to assume that the population distribution is normal. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 13/40 13/40

14 Checking for normality file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 14/40 14/40

15 Assess normality with Shapiro test Hypotheses: H 0 : L(X) = N H 1 : L(X) N Shapiro-Wilk normality test data: rnorm (50, mean = 1, sd = 3) W = , p-value = Shapiro-Wilk normality test data: rf (50, df1 = 3, df2 = 2) W = , p-value = 2.687e-09 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 15/40 15/40

16 Assess normality with Shapiro test Hypotheses: H 0 : L(X) = N H 1 : L(X) N Shapiro-Wilk normality test data: scatdat$height W = , p-value = Shapiro-Wilk normality test data: scatdat$weight W = , p-value = file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 16/40 16/40

17 Dealing with non normal data Transform the data with the log or the square root function. Wilcoxon signed rank test can be used in place of a 1-sample or paired t-test Wilcoxon rank sum test substitutes for the 2-sample t-test Bootstrappping (testing for a population mean) 1. Extract a new sample of n observations from the original set of n. 2. Calculate the mean of this new sample of n. 3. Repeat steps (1) and (2) an arbitrarily large number of times, say 5000 times. 4. Use the estimated distribution of the sample mean for inference file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 17/40 17/40

18 Goodness of t Suppose data is generated from an experiment, given the following frequencies for the values of X. X k Frequencies n 1 n 2... n k We want to test the hypotheses: : L(X) = (,,..., ) : L(X) (,,..., ) H 0 p 1 p 2 p k H 1 p 1 p 2 p k with a signi cant level α. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 18/40 18/40

19 Goodness of t Genetics Consider two di erent characteristics of tomatoes: leaf shape and plant size. The leaf shape may be potato-leafed or cut-leafed, and the plant may be tall or dwarf. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 19/40 19/40

20 Goodness of t - Genetics Tall cut-leaf (TTCC, TTCc, TtCc, TtCC): 9 16, Tall potato-leaf (TTcc, Ttcc): 3 16 Dwarf cut-leaf (ttcc, ttcc): 3 16, Dwarf potato-leaf (ttcc): 1 16 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 20/40 20/40

21 Notation Given X or cells): a categorical random variable with k possible values (k di erent levels or categories L(X) = P 0 = ( p 1, p 2,..., p k ) p i is the true proportion of category i and we have p 1 + p p k = 1. X = {1, 2, 3, 4} with tall cut-leaf: 1, tall potato-leaf: 2, dwarf cut-leaf: 3, dwarf potato-leaf: 4 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 21/40 21/40

22 Goodness of t Suppose data is generated from an experiment, given the following frequencies for the values of X. X k Frequencies n 1 n 2... n k We want to test the hypotheses: : L(X) = (,,..., ) : L(X) (,,..., ) H 0 p 1 p 2 p k H 1 p 1 p 2 p k with a signi cant level α. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 22/40 22/40

23 Goodness of t Under the null hypothesis, given a sample size n, the expected count for category i is: n p i. We can de ne the goodness-of- t statistic, KH as: k KH = i=1 ( n i n p i ) 2 n p i Expressed di erently and referring to category as cell, we write: KH = all cells (observed cell count expected cell count) 2 expected cell count Under H 0, with additionnal conditions such as: n 50, n 5, for i = 1,..., k we have: p i L(KH) χ 2 k 1 d d is the number of estimated parameters in the null hypothesis. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 23/40 23/40

24 Goodness of t - Example Suppose we perform an experiment with tomato plants, and we observed the following data : We want to test : H 0 : L(X) = (,,, ) H 1 : L(X) (,,, ) with a level of signi cance α = file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 24/40 24/40

25 Goodness of t The value of the test statistic: ( ) 2 ( ) 2 kh = ( ) ( ) We verify: 25/40 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 25/40

26 Goodness of t For a signi cance level α, we calculate the critical value c α such that P( > ) = α. We would fail to reject H 0 si kh < c α. χ 2 k 1 d In the example, kh 1.47, d = 0, α = 0.01, number of degrees of freedom is 3. c α The critical value is Therefore, we fail to reject the null hypothesis. The p-value is approximately file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 26/40 26/40

27 Contingency table A rectangular table used to summarize a categorical data set; two-way tables are used to compare several populations on the basis of a categorical variable or to determine if an association exists between two categorical variables. Example of a contingency table : file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 27/40 27/40

28 Review Given two discrete random variables X and Y, the joint distribution is de ned by: = P (X =, Y = ) p ij x i y j Y X x 1 x 2 y 1 p 11 p 21 y 2 p 12 p y m p 1m p 2m x l p l1 p l2... p lm The marginal distribution of X : P(X = x i ) = p = m p i. j=1 ij The marginal distribution of Y : P(X = y j ) = p = n p.j i=1 ij If X and Y are independent : p ij = p i. p.j file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 28/40 28/40

29 Testing independence The sample data observed from (X, Y ) is a contingency table: Y X x 1 x 2 y 1 n 11 n 21 y 2 n 12 n y m n 1m n 2m x l n l1 n l2... n lm n i. The marginal frequencies for X : = n.j The marginal frequencies for Y : = m j=1 n ij n i=1 n ij If X and Y are independent : = n i. n.j n ij n file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 29/40 29/40

30 Testing independence H 0 H 1 : X and Y are independent : X and Y are not independent p ij = p i. p.j avec i = 1, 2,..., let j = 1, 2,..., m The number of parameters to estimate is l 1 + m 1. Test statistic: KH = n i. n.j l m ( n i. n.j n ij n )2 n i. n.j i=1 j=1 n Under H 0, when n 50 and 5 for all i, j: n L(KH) χ 2 (l 1)(m 1) file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 30/40 30/40

31 Example Let X a random variables with values F (Family) and S (alone) and Y another random variable with values P (Present) et A (Absent). From the sample, we can estimate: 40 P(F, P ) P(F, A) P(S, P ) P(S, A) P(F ) P(S) P(P ) P(A) file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 31/40 31/40

32 Example H 0 : X and Y are independent H 1 : X and Y are not independent X and Y are independent if: P(X = x, Y = y) = P(X = x) P(Y = y) file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 32/40 32/40

33 Example 100 P(F ) P(P ) P(F ) P(A) P(S) P(P ) P(S) P(A) file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 33/40 33/40

34 Example expected cell count = (row marginal total) (column marginal total) grand total df = (number of rows 1) (number of columns 1) (40 54) 2 (60 46) 2 ( ) 2 ( ) 2 kh = The χ 2 distribution has (2 1) (2 1) = 1 degrees of freedom. c α = for α = We reject then the null hypothesis. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 34/40 34/40

35 Can she taste the di erence? Tea-tasting experiment: a woman claimed to be able to judge whether tea or milk was poured in a cup rst. Fisher designed an experiment to test her ability What should be done about chance variations in the temperature, sweetness, and so on? How many cups should be used in the test? Should they be paired? In what order should the cups be presented? Fisher suggests: if discrimination of the kind under test is absent, the result of the experiment will be wholly governed by the laws of chance. What conclusion could be drawn from a perfect score or from one with one or more errors? file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 35/40 35/40

36 Fisher's exact test Tea-tasting experiment: a woman claimed to be able to judge whether tea or milk was poured in a cup rst. The woman was given eight cups of tea, in four of which milk was poured rst, and was told to guess which four had milk poured rst. The contingency for this design is: p 1 p 2 : probability to select a cup milk rst : probability to select a cup tea rst The hypotheses to test: H 0 : p 1 = p 2 H 1 : p 1 > p 2 file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 36/40 36/40

37 Fisher's exact test Tea-tasting experiment: a woman claimed to be able to judge whether tea or milk was poured in a cup rst. The woman was given eight cups of tea, in four of which tea was poured rst, and was told to guess which four had tea poured rst. X random variable with avlues x = 0, 1, 2, 3, 4 The Hypergeometric Distribution: 4 4 ( )( ) x 4 x P H0 ( ) (X = x) = x = 0, 1, 2, 3, ( ) = 70 There are 4 possible ways to classify 4 of the 8 cups as milk rst. file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 37/40 37/40

38 Check yourself Tea-tasting experiment: a woman claimed to be able to judge whether tea or milk was poured in a cup rst. The woman was given eight cups of tea, in four of which milk was poured rst, and was told to guess which four had milk poured rst. Suppose she identi es 3 cups milk rst correctly. The p-value is: 1/ /70 17/70 Submit Show Hint Show Answer Clear file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 38/40 38/40

39 Fisher's exact test The marginal counts a + b, c + d, a + c, and b + d are xed The hypotheses to test: H 0 : p 1 = p 2 H 1 : p 1 > p 2 or p 1 < p 2 or p 1 p 2 The test statistic is T the number of observations cell (1, 1) and a+c b+d ( )( ) x a+b x P H0 n ( ) (T = x) = x = 0, 1, 2,..., min{a + b, a + c} a+b file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 39/40 39/40

40 See you next time file:///users/home/npaul/enseignement/esbs/ /cours/04/index.html#40 40/40 40/40