Transformations. Non-parametric testsfolk.uio.no/jonmic/Statkurs/14 - Transformations....
Transcript of Transformations. Non-parametric testsfolk.uio.no/jonmic/Statkurs/14 - Transformations....
Transformations. Non-parametrictests
Jon Michael GranDepartment of Biostatistics, UiO
MF9130 Introductory course in statisticsMonday 06.06.2011
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Overview
Transformations. Non-parametric methods(Aalen chapter 8.8, Kirkwood and Sterne chapter 13 and 30.2)
• Logaritmic transformation• Choice of transformation• Non-parametric methods based on ranks (Wilcoxon signedrank test and Wilcoxon rank sum test)
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Transformations
Motivation• Non-normality means that the standard methods cannot beused
• This includes linear regression, which is often a main tool ofanalysis → major problem
• For data that are skewed to the right, one can often uselog-transformations
• This does not only give normally distributed data; it may alsogive equal variances in different groups
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Logaritmic transformation
• So what do we mean by doing a log transformation?• Take the logaritm of all your data values xi for all i ’s, and doyour analysis on the new dataset of ui ’s, where ui = ln(xi)
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Example: Ln transformation
• Skew distribution. Example of observations: 0.40, 0.96, 11.0
• Ln transformed distribution: ln(0.4) = −0.91,ln(0.96) = −0.04, ln(11) = 2.40
• Do analysis on the ln transformed data• In SPSS: transform → compute
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Example: Ln transformation
• Skew distribution. Example of observations: 0.40, 0.96, 11.0• Ln transformed distribution: ln(0.4) = −0.91,ln(0.96) = −0.04, ln(11) = 2.40
• Do analysis on the ln transformed data• In SPSS: transform → compute
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Example: Ln transformation
• Skew distribution. Example of observations: 0.40, 0.96, 11.0• Ln transformed distribution: ln(0.4) = −0.91,ln(0.96) = −0.04, ln(11) = 2.40
• Do analysis on the ln transformed data• In SPSS: transform → compute
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Choice of transformations• The logarthmic transformation is by far the most frequentlyapplied
• Appropriate for removing positive skewness (data skew to theright)
• There are other types of transformation for data that arestronger or weeker skewed, or data skew to the left
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Other transformations• Skewed to the right:
I Lognormal: Logarithmic (u = ln x)I More skewed than lognormal: Reciprocal (u = 1/x)I Less skewed than lognormal: Square root (u =
√x)
• Skew to the left:I Moderatly skewed: Square (u = x2)I More skewed: Cubic (u = x3)
• Non-linear relationship: Transform only one of the twovariables
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Non-parametric statistics
Motivation• So what if transforming our data doesn’t help and it is stillnot normally distributed?
• Use non-parametric test
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Non-parametric tests• In tests we have done so far, the null hypothesis has alwaysbeen a stochastic model with a few parameters. Models basedon for example:
I Normal distributionI T-distributionI Binomial distribution
• In nonparametric tests, the null hypothesis is not a parametricdistribution, rather a much larger class of possible distributions
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Parametric methods we have seen• Estimation
I Confidence interval for µI Confidence interval for µ1 − µ2
• TestingI One sample T-testI Two sample T-test
• The methods are based on the assumption of normallydistributed data (or normally distributed mean)
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Typical assumptions for parametric methods
1 Independence: All observations are independent. Achieved bytaking random samples of individuals; for paired t-testindependence is achieved by using the difference betweenmeasurements
2 Normally distributed data (Check: histograms, tests fornormal distribution, Q-Q plots)
3 Equal variance or standard deviations in the groups
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How to test for normality
• Visual methods, like histograms and Q-Q plots, are veryuseful
• Also have several statistical tests for normality:I Kolmogorov-Smirnov testI Shapiro-Wilk
• However, with enough data, visual methods are more useful!
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Tests for normality
• Example of SPSS output from a Kolmogorov-Smirnov test:
(measurements of lung function, separated by gender)• According to this test, if the p-value is less than 0.05 the datacannot be considered as Normally distributed
• Note though that these tests are a bit problematic.They’re not very effective at discovering departures fromnormality. The power is low, so even if you don’t get asignificant result it’s a bit of a leap to assume normality
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Tests for normality
• Example of SPSS output from a Kolmogorov-Smirnov test:
(measurements of lung function, separated by gender)• According to this test, if the p-value is less than 0.05 the datacannot be considered as Normally distributed
• Note though that these tests are a bit problematic.They’re not very effective at discovering departures fromnormality. The power is low, so even if you don’t get asignificant result it’s a bit of a leap to assume normality
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Q-Q plots
• Graphical way of comparing two distributions, plotting theirquantiles against each other
• Q for quantile• If the distributions are similar, they Q-Q plot should be closeto a straight line
Q-Q Q-Q
Heavy tailed population Skew population
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Histogram and Q-Q plot for Gender = 1
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Histogram and Q-Q plot for Gender = 2
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Non-parametric statistics
• The null hypothesis is for example that the median of thedistribution is zero
• A test statistic can be formulated, so thatI it has a known distribution under this hypothesisI it has more extreme values under alternative hypotheses
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Non-parametric methods• Estimation
I Confidence interval for the median• Testing
I Paired data: Sign test and Wilcoxon signed rank testI Two independent samples: Mann-Whitney test/Wilcoxon
rank-sum test• Make (almost) no assumptions regarding distribution
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Example: Confidence interval for the median
• Beta-endorphin concentrations (pmol/l) in 11 individuals whocollapsed during a half-marathon (in increasing order): 66.071.2 83.0 83.6 101.0 107.6 122.0 143.0 160.0 177.0 414.0
• We find that the median is 107.6• What is the 95% confidence interval?
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Confidence interval for the median in SPSS• Use ratio statistics, which is meant for the ratio between twovariables. Make a variable that has value 1 for all data (call itunit)
• Analyze → Descriptive statistics → RatiosNumerator: betae Denominator: unit
• Click Statistics and under Central tendency check Median andConfidence Intervals
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SPSS output
• 95% confidence interval for the median is (71.2, 177.0)
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Non-parametric tests
• Most tests are based on rank-sums, and not the observedvalues
• The sums of the rank are assumed approximately normallydistributed, so we can use a normal approximation for thetests
• If two or more values are equal, the tests use a mean rank
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The sign test
• Assume the null hypothesis is that the median is zero• Given a sample from the distribution, there should be roughlythe same number of positive and negative values
• More precisely, number of positive values should follow abinomial distribution with probability 0.5
• When the sample is large, the binomial distribution can beapproximated with a normal distribution
• Sign test assumes independent observations, no assumptionsabout the distribution
• SPSS: Analyze → nonparametric tests → two related samples.Choose sign under test type
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Example: Energy intake kJ
• Want to test if energy intake is different before and aftermenstruation.
• H0: Median difference = 0H1: Median difference 6= 0
• All differences are positive
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Using the sign test
• The number of positive differences should follow a binomialdistribution with P = 0.5 if H0: median difference betweenpremenst and postmenst is 0
• What is the probability of observing 11 positive differences?• Let P(x) count the number of positive signs• P(x) is Bin(n = 11,P = 0.5)• The p-value of the test is the probability of observing 11positives or something more extreme if H0 is true, henceP(x ≥ 11) = P(x = 11) = 11!
0!(11−0)!0.50(1− 0.5)11 = 0.0005
• However, because of two-sided test, the p-value is 0.0005 · 2= 0.001
• Clear rejection of H0 on significance-level 0.05
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Using the sign test
• The number of positive differences should follow a binomialdistribution with P = 0.5 if H0: median difference betweenpremenst and postmenst is 0
• What is the probability of observing 11 positive differences?
• Let P(x) count the number of positive signs• P(x) is Bin(n = 11,P = 0.5)• The p-value of the test is the probability of observing 11positives or something more extreme if H0 is true, henceP(x ≥ 11) = P(x = 11) = 11!
0!(11−0)!0.50(1− 0.5)11 = 0.0005
• However, because of two-sided test, the p-value is 0.0005 · 2= 0.001
• Clear rejection of H0 on significance-level 0.05
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Using the sign test
• The number of positive differences should follow a binomialdistribution with P = 0.5 if H0: median difference betweenpremenst and postmenst is 0
• What is the probability of observing 11 positive differences?• Let P(x) count the number of positive signs• P(x) is Bin(n = 11,P = 0.5)
• The p-value of the test is the probability of observing 11positives or something more extreme if H0 is true, henceP(x ≥ 11) = P(x = 11) = 11!
0!(11−0)!0.50(1− 0.5)11 = 0.0005
• However, because of two-sided test, the p-value is 0.0005 · 2= 0.001
• Clear rejection of H0 on significance-level 0.05
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Using the sign test
• The number of positive differences should follow a binomialdistribution with P = 0.5 if H0: median difference betweenpremenst and postmenst is 0
• What is the probability of observing 11 positive differences?• Let P(x) count the number of positive signs• P(x) is Bin(n = 11,P = 0.5)• The p-value of the test is the probability of observing 11positives or something more extreme if H0 is true, henceP(x ≥ 11) = P(x = 11) = 11!
0!(11−0)!0.50(1− 0.5)11 = 0.0005
• However, because of two-sided test, the p-value is 0.0005 · 2= 0.001
• Clear rejection of H0 on significance-level 0.05
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Using the sign test
• The number of positive differences should follow a binomialdistribution with P = 0.5 if H0: median difference betweenpremenst and postmenst is 0
• What is the probability of observing 11 positive differences?• Let P(x) count the number of positive signs• P(x) is Bin(n = 11,P = 0.5)• The p-value of the test is the probability of observing 11positives or something more extreme if H0 is true, henceP(x ≥ 11) = P(x = 11) = 11!
0!(11−0)!0.50(1− 0.5)11 = 0.0005
• However, because of two-sided test, the p-value is 0.0005 · 2= 0.001
• Clear rejection of H0 on significance-level 0.05
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In SPSS
• Find p-value = 0.000 using t-test, so p-value from sign-test isa bit larger
• This will often be the case
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Wilcoxon signed rank test
• Takes into account the strength of the differences, not just ifthey are positive or negative
• Here, the null hypothesis is a symmetric distribution withzero median. Do as follows:
I Rank all values by their absolute values, from smallest tolargest Let T+ be the sum of ranks of the positive values, andT-corresponding for negative values
I Let T be the minimum of T+ and T-I Under the null hypothesis, T has a known distribution
• For large samples, the distribution can be approximated with anormal distribution
• Assumes independent observations and symmetricaldistributions
• SPSS: Analyze → nonparametric tests → two related samples.Choose Wilcoxon under test type
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Energy intake example
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In SPSS
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Sign test and Wilcoxon signed rank test
• Main difference: The sign test assume nothing about thedistribution, while the Wilcoxon signed rank test assumessymmetry.
• In other words: It is a little bit harder to get significant resultsusing the Sign test, because of lesser assumptions
• Often used on paired data
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Wilcoxon rank sum test (or the Mann-Whitney U test)
• Not for paired data, but rather n1 values from group 1 andn2 values from group 2
• Assumptions: Independence within and between groups, samedistributions in both groups
• We want to test whether the values in the groups are samplesfrom different distributions:
I Rank all values as if they were from the same groupI Let T be the sum of the ranks of the values from group 1I Under the assumption that the values come from the same
distribution, the distribution of T is knownI The expectation and variance under the null hypothesis are
simple functions of n1 and n2
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• For large samples, we can use a normal approximation for thedistribution of T
• The Mann-Whitney U test gives exactly the same results, butuses slightly different test statistic
• In SPSS:I Analyze → nonparametric tests → two independent samples
testsI Test type: Mann-Whitney U
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Example 1• We have observed values
I Group 1: 1.3, 2.1, 1.5, 4.3, 3.2I Group 2: 3.4, 4.9, 6.3, 7.1
• Are the groups different?• If we assume that the values in the groups are normallydistributed, we can solve this using the T-test
• Otherwise we can try the normal approximation to theWilcoxon rank sum test
• Test H0: Groups have equal medians vs. H1: Groups havedifferent medians on 5% significance level
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Example 1 (cont.)
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Example 1 (cont.)
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Example 2
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Example 2 (cont.)
• Wilcoxon Rank Sum Test in SPSS:
• We would have gotten exactly the same result by using the Tdistribution!
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Some useful general comments
• Always start by checking whether it is appropriate to uset-tests (normally distributed data)
• If in doubt, a useful tip is to compare p-values from t-testswith p-values from non-parametric tests
• If similar results, stay with the parametric tests
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Exercises
• Data files can be found on www.med.uio.no/imb/stat/kursfiler• They are called Children, oppg-9-2 and oppg-9-4
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Summary
Key words
• Skew data• Logaritmic transformations (and other alternatives)• Testing for normality• Non-parametric tests:
I Sign test and Wilcoxon (paired data)I Mann-Whitney/Wilcoxon rank-sum test (two independent
samples)
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