Business Statistics MEDIAN: NON- PARAMETRIC TESTS

Transcription

1 Business Statistics MEDIAN: NON- PARAMETRIC TESTS

2 CONTENTS Hypotheses on the median The sign test The Wilcoxon signed ranks test Old exam question

3 HYPOTHESES ON THE MEDIAN The median is a central value that may be more suitable for strongly asymmetric distributions and for distributions with fat tails Can we test a population median? e.g., H 0 : M = 400 Note: for a more or less symmetric distribution, M μ, so a t-test of mean is appropriate (if n 15) although perhaps more sensitive to large positive or negative outliers in the sample

4 HYPOTHESES ON THE MEDIAN What is the median of a sample? it is the middle value, i.e. x n/2 So, if H 0 : M = 400 would be true, approximately half of the data in the sample would be lower, and half would be higher Therefore, if we count the number of data points that is lower and compare it to the number of observations, we can develop a test statistic Two varieties of such non-parametric tests today: sign test Wilcoxon signed rank test

5 THE SIGN TEST The sign test involves simply counting the number of positive or negative signs in a sequence of n signs is based on the binomial distribution

6 THE SIGN TEST Computational Steps: for each data point x i compute the difference with the median (M) of the null hypothesis (H 0 ): d i = x i M omit zero differences (d i = 0); effective sample size is n assign +1 to positive differences (d i > 0) and 1 to negative differences (d i < 0) test statistic X is the sum of the positive numbers (= number of positive observations)

7 THE SIGN TEST Example: Context: battery life until failure (in hours) H 0 : M = 400; H 1 : M 400 sample of n = 13 observations (x 1,, x 13 ) reject for large and for small numbers of positive signs

8 THE SIGN TEST Example (H 0 : M = 400): data: x i (i = 1,, 13) difference with M: d i = x i 400 no cases where d i = 0, so n = n s i = 1 if d i > 0 1 if d i < 0 s + i = 1 if d i > 0 0 if d i < 0 n x = + i=1 s i = 8 x i x i -400 s i s (+) i

9 THE SIGN TEST Example (continued): x = 8 under H 0 : X~bbb 13,0.5 P bbb 13,0.5 X 8 = why 8? if we would reject for 8, we would also reject for 9 p-value: = why 2? because it s a two-sided null hypothesis there is no reason to reject H 0

10 THE SIGN TEST In the sign test, we replace the numerical values by signs (+ or ) Advantage: we don t need any assumption on normality, symmetry, etc. that s why we say it s non-parametric: we don t have to assume a certain distribution with parameters Disadvantage: we discard so much information that we also lose power (see later) Are there other non-parametric tests that are more powerful? is there a compromise between value and sign that still needs some assumptions, but not too many assumptions? Yes, replacing data by their rank

11 THE WILCOXON SIGNED RANK TEST Wilcoxon signed rank test involves comparing the sum of ranks of the values larger than the test value with the sum of ranks of the values smaller than the test value Computational Steps: for each data point x i compute the absolute difference with the median (M) of the null hypothesis: d i = x i M omit zero differences (d i = 0); effective sample size is n assign ranks (1,, n ) to the d i reassign + and to the ranks test statistic (W) is the sum of the positive ranks

12 THE WILCOXON SIGNED RANK TEST Example (H 0 : M = 400): data: x i (i = 1,, 13) difference with M: d i = x i 400 no cases where d i = 0, so n = n n w = + i=1 r i = 61 under H 0 : W~? (use table) P H0 W 61 =? x i x i 400 x i 400 r i r i (+)

13 THE WILCOXON SIGNED RANK TEST Testing the median using the Wilcoxon W statistic small samples: using a table of critical values included in tables at exam large samples: using a normal approximation of W valid when n 20

14 THE WILCOXON SIGNED RANK TEST Small samples: critical values of Wilcoxon statistic Table is available at the exam (and on the course website) Lower and Upper Critical Values W of Wilcoxon Signed-Ranks Test one-tail: α = 0.05 α = α = 0.01 α = two-tail: α = 0.10 α = 0.05 α = 0.02 α = 0.01 n 5 0, 15 (lower, upper) ---, , , , 19 0, , , , 25 2, 26 0, , , 31 3, 33 1, 35 0, , 37 5, 40 3, 42 1, , 45 8, 47 5, 50 3, , 53 10, 56 7, 59 5, , 61 13, 65 10, 68 7, , 70 17, 74 12, 79 10, 81 two-sided, α = 0.05, n = 13: w lllll = 17 and w uuuuu = 74 R crit = [0,17] [74,91] w calc = 61, so do not reject H 0 at α = 0.05

15 THE WILCOXON SIGNED RANK TEST Large samples: under H 0 :, it can be shown that E W = var W = n n+1 4 n n+1 2n+1 24 Further, for n 20, approximately: W n n+1 4 n n+1 2n+1 24 ~N 0,1 so you can compute z calc = w calc n n+1 24 and compare it to z crit (e.g., ±1.96) n n+1 4 2n+1

16 THE WILCOXON SIGNED RANK TEST Example, continued: n w = + i=1 r i = 61 under H 0 : W~N E W, var W so, under H 0 : P N W 61 W E W var W = P ~N 0,1 W E W var W P Z 1.08 = p-value: = there is no reason to reject H 0 In fact, not a good idea because n = We do it just to show how it works... =

17 OLD EXAM QUESTION 23 March 2015, Q1l-m