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Nonparametrics and goodness-of-fit

Nonparametrics and goodness-of-fit. Tron Anders Moger 23.10.2006. Nonparametric statistics. In tests we have done so far, the null hypothesis has always been a stochastic model with a few parameters. T tests Tests for regression coefficients Test for autocorrelation …

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Nonparametrics and goodness-of-fit

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  1. Nonparametrics and goodness-of-fit Tron Anders Moger 23.10.2006

  2. Nonparametric statistics • In tests we have done so far, the null hypothesis has always been a stochastic model with a few parameters. • T tests • Tests for regression coefficients • Test for autocorrelation • … • In nonparametric tests, the null hypothesis is not a parametric distribution, rather a much larger class of possible distributions

  3. Parametric methods we’ve seen: • Estimation • Confidence interval for µ • Confidence interval for µ1 -µ2 • Testing • One sample T test • Two sample T test • The methods are based on the assumption of normally distributed data (or normally distributed mean)

  4. Nonparametric statistics • The null hypothesis is for example that the median of the distribution is zero • A test statistic can be formulated, so that • it has a known distribution under this hypothesis • it has more extreme values under alternative hypotheses

  5. Non-parametric methods: • Estimation • Confidence interval for the median • Testing • Paired data: Sign test and Wilcoxon signed rank test • Two independent samples: Mann-Whitney test/Wilcoxon rank-sum test • Make (almost) no assumptions regarding distribution

  6. Confidence interval for the median, example • Betaendorphine concentrations (pmol/l) in 11 individuals who collapsed during a half-marathon (in increasing order): 66.0 71.2 83.0 83.6 101.0 107.6 122.0 143.0 160.0 177.0 414.0 • Find that median is 107.6 • What is the 95% confidence interval?

  7. Confidence interval for the median in SPSS: • Use ratio statistics, which is meant for the ratio between two variables. Make a variable that has value 1 for all data (unit) • Analyze->Descriptive statistics->Ratios Numerator: betae Denominator: unit • Click Statistics and under Central tendency check Median and Confidence Intervals

  8. SPSS-output: • 95% confidence interval for the median is (71.2, 177.0):

  9. The sign test • Assume the null hypothesis is that the median of the distribution is zero. • Given a sample from the distribution, there should be roughly the same number of positive and negative values. • More precisely, number of positive values should follow a binomial distribution with probability 0.5. • When the sample is large, the binomial distribution can be approximated with a normal distribution • Sign test assumes independent observations, no assumptions about the distribution • SPSS: Analyze->nonparametric tests->two related samples. Choose sign under test type

  10. Example from lecture 5: Energy intake kJ Want to test if energy intake is different before and after menstruation. H0: Median difference=0 H1: Median difference≠0 + + + + + + + + + + + SUBJECT PREMENST POSTMENS 1 5260.0 3910.0 2 5470.0 4220.0 3 5640.0 3885.0 4 6180.0 5160.0 5 6390.0 5645.0 6 6515.0 4680.0 7 6805.0 5265.0 8 7515.0 5975.0 9 7515.0 6790.0 10 8230.0 6900.0 11 8770.0 7335.0 Number of cases read: 11 Number of cases listed: 11

  11. Using the sign test: • The number of positive differences (+’s) should follow a binomial distribution with P=0.5 if H0: median difference between premenst and postmenst is 0 • What is the probability of observing 11 +’s? • Let P(x) count the number of + signs • P(x) is Bin(n=11,P=0.5) • The p-value of the test is the probability of observing 11 +’s or something more extreme if H0 is true, hence • However, because of two-sided test, the p-value is 0.0005*2=0.001 • Clear rejection of H0 on significance-level 0.05

  12. In SPSS: • Recall: Found p-value=0.000 using t-test, so p-value from sign-test is a bit larger • This will always be the case P-value

  13. Wilcoxon signed rank test • Takes into account the strength of the differences, not just if they are positive or negative • Here, the null hypothesis is a symmetric distribution with zero median. Do as follows: • Rank all values by their absolute values, from smallest to largest • Let T+ be the sum of ranks of the positive values, and T-corresponding for negative values • Let T be the minimum of T+ and T- • Under the null hypothesis, T has a known distribution (p.874). • For large samples, the distribution can be approximated with a normal distribution • Assumes independent observations and symmetrical distributions • SPSS: Analyze->nonparametric tests->two related samples. Choose Wilcoxon under test type

  14. The energy intake example: Difference Rank (+) Rank(-) premenst-postmenst 1350,00 6 No 1250,00 4 negative 1755,00 10 differences! 1020,00 3 745,00 2 1835,00 11 1540,00 8.5 1540,00 8.5 725,00 1 1330,00 5 1435,00 7 Rank sum: T+:66 T-:0 • Min(66,0)=0 • Test H0: Median difference =0 vs H1: Median difference≠0 with α=0.05 • From p.874, find T=11 for n=11 and α/2=0.025 • Reject H0 • Also, see that p-value is smaller than 0.01

  15. In SPSS: Sum of ranks P-value based on normal approximation P-value based on Wilcoxon distribution

  16. Sign test and Wilcoxon signed rank test • Often used on paired data. • E.g. we want to compare primary health care costs for the patient in two countries: A number of people having lived in both countries are asked about the difference in costs per year. Use this data in test. • In the previous example, if we assume all patients attach the same meaning to the valuations, we could use Wilcoxon signed rank test on the differences in valuations

  17. Wilcoxon rank sum test (or the Mann-Whitney U test) • Here, we do NOT have paired data, but rather n1 values from group 1 and n2 values from group 2. • Assumptions: Independence within and between groups, same distributions in both groups • We want to test whether the values in the groups are samples from different distributions: • Rank all values as if they were from the same group • Let T be the sum of the ranks of the values from group 1. • Under the assumption that the values come from the same distribution, the distribution of T is known. • The expectation and variance under the null hypothesis are simple functions of n1 and n2.

  18. Wilcoxon rank sum test (or the Mann-Whitney U test) • For large samples, we can use a normal approximation for the distribution of T. • The Mann-Whitney U test gives exactly the same results, but uses slightly different test statistic. • In SPSS: • Analyze->nonparametric tests-> two independent samples tests • Test type: Mann-Whitney U

  19. Example • We have observed values • Group 1: 1.3, 2.1, 1.5, 4.3, 3.2 • Group 2: 3.4, 4.9, 6.3, 7.1 are the groups different? • If we assume that the values in the groups are normally distributed, we can solve this using the T-test. • Otherwise we can try the normal approximation to the Wilcoxon rank sum test • Test H0: Groups come from same distribution vs H1: Groups come from different distributions on 5% significance level

  20. Example (cont.) Rank Group 1 Group 2 Wilcoxon T: 16 E(T)= n1(n1+n2+1)/2 =25 SE(T)= √n1n2(n1+n2+1)/12 =4.08 Z=(T-E(T))/SE(T) =(16-25)/4.08 =-2.20 p-value: 0.032 Reject H0 Rank sum: 16 29

  21. ID GROUP ENERGY 1 0 6.13 2 0 7.05 .... 12 0 10.15 13 0 10.88 14 1 8.79 15 1 9.19 .... 21 1 11.85 22 1 12.79 Number of cases read: 22 Number of cases listed: 22 Example from lecture 4: Energy expenditure in two groups, lean and obese. Want to test if they come from different distributions.

  22. Wilcoxon Rank Sum Test in SPSS: Sum of ranks Z-value P-value based on normal approx. Exact P-value Exactly the same result as we got when using the T distribution!

  23. Spearman rank correlation • This can be applied when you have two observations per item, and you want to test whether the observations are related. • Computing the sample correlation (Pearson’s r) gives an indication. • We can test whether the Pearson correlation could be zero but test needs assumption of normality.

  24. Spearman rank correlation • The Spearman rank correlation tests for association without any assumption on the association: • Rank the X-values, and rank the Y-values. • Compute ordinary sample correlation of the ranks: This is called the Spearman rank correlation. • Under the null hypothesis that X values and Y values are independent, it has a fixed, tabulated distribution (depending on number of observations)

  25. Concluding remarks onnonparametric statistics • Tests with much more general null hypotheses, and so fewer assumptions • Often a good choice when normality of the data cannot be assumed • If you reject the null hypothesis with a nonparametric test, it is a robust conclusion • However, with small amounts of data, you can often not get significant conclusions • In SPSS, you only get p-values from non-parametric tests, more interesting to know the mean difference and confidence intervals

  26. Some useful comments: • Always start by checking whether it is appropriate to use T tests (normally distributed data) • If in doubt, a useful tip is to compare p-values from t-tests with p-values from non-parametric tests • If very different results, stay with the p-values from the non-parametric tests

  27. Contingency tables • The following data type is frequent: Each object (person, case,…) can be in one of two or more categories. The data is the count of number of objects in each category. • Often, you measure several categories for each object. The resulting counts can then be put in a contingency table.

  28. Testing if probabilities are as specified • Example: Have n objects been placed in K groups, each with probability 1/K? • Expected count in group i: • Observed count in group i: • Test statistic: • Test statistic has approx. distribution with K-1 degrees of freedom under null hypothesis.

  29. The Chi-square distribution • The Chi-square distribution with n degrees of freedom is denoted • It is equal to the sum of the squares of n independent random variables with standard normal distributions.

  30. Testing association in a contingency table Treatment G R O U P n values in total

  31. Testing association in a contingency table • If the assignment of observations to the two categories is independent, the expected count in a cell can be computed from the marginal counts: • Actual observed count: • Test statistic: • Under null hypothesis, it has distribution with (r-1)(c-1) degrees of freedom • In SPSS: Analyze->Descriptives->Crosstabs

  32. Example from Lecture 4: • Spontanous abortions among nurses helping with operations and other nurses • Want to test if there is any association between abortions and type of nurse

  33. Example: • Expected cell values: Abortion/op.nurses: 13*36/70=6.7 Abortion/other nurses: 13*34/70=6.3 No abortion/op.nurses: 57*36/70=29.3 No abortion/other nurses: 57*34/70=27.7

  34. We get: • This has a chi-square distribution with (2-1)*(2-1)=1 d.f. • Want to test H0: No association between abortions and type of nurse at 5%-level • Find from table 7, p. 869, that the 95%-percentile is 3.84 • This gives you a two-sided test! • Reject H0: No association • Same result as the test for different proportions in Lecture 4!

  35. In SPSS: Check Expected under Cells, Chi-square under statistics, and Display clustered bar charts!

  36. Goodness-of-fit tests • Sometimes, the null hypothesis is that the data comes from some parametric distribution, values are then categorized into K groups, and we have the counts from the categories. • To test this hypothesis: • Estimate the m parameters in the distribution from data. • Compute expected counts. • Compute the test statistic used for contingency tables. • This will now have a chi-squared distribution with K-m-1 d.f. under the null hypothesis.

  37. Poisson example from lecture 3: No. of No. Expected no. observed cases observed (from Poisson dist.) 0 20 18.4 1 16 17.2 2 8 8.1 3 2 2.5 4 0 0.6 5 0 0.11 6 1 0.017 • Calculation: P(X=2)=0.9362*e-0.936/2!=0.17 • Multiply by the number of weeks: 0.17*47=8.1 • Poisson distribution fits data fairly well!

  38. Test if the data can come from a Poisson distribution: • Test statistic: approx. with 6-1-1=4 d.f. • Calculate: • From table 7, p. 869, see that 95%-percentile of chi-square ditribution with 4 d.f. is 9.49 • Reject H0: The data come from a Poisson distribution

  39. Tests for normality • Visual methods, like histograms and normality plots, are very useful. • In addition, several statistical tests for normality exist: • Kolmogorov-Smirnov test (which I have showed you before) • Bowman-Shelton test (tests skewness and kurtosis) • However, with enough data, visual methods are more useful

  40. Next time: • Seen tests for comparing means in two independent groups • What if you have more than two groups? • Analysis of variance

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