Distribution theory and statistical inference

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Distribution theory and statistical inference Powered By Docstoc
					            Today‟s lecture

• Inferential methods - review
  – Bayesian – frequentist
  – Parametric, non-parametric, semi-parametric
• A more modern approach to non-parametric
  procedures
  – Randomisation tests
  – Bootstraps
• Next week
  – Revision
                                                  1
                 Web site
• Material for last two weeks now on

www.maths.napier.ac.uk/~gillianr

This includes materials for today‟s practical
 workshop on bootstraps


                                                2
 Parametric and non-parametric
            methods
• In both methods we are assuming that the
  data we are observing follow some model
• For parametric methods this is a model
  based on known probability distributions
• What we are saying is
     “IF the model is true – then we can
     conclude …. about the model and its
                 parameters”
                                             3
  Parametric and non-parametric
             methods
• Non-parametric tests also make assumptions
• They imply that the DATA observed are a random
  sample from some unspecified distribution
• What we are saying is
  “IF we have observed these data – then we can
     conclude …. about the distribution(s) they
                     come from”


                                               4
 Modern non-parametric methods
• Parametrics condition (IF) on the model
• Non-parametrics condition(IF) on the data

• Traditional non-parametric tests used ranks
   – This was for practical reasons in pre-computer days
• “Modern” non-parametric methods can use the
  data
   – Randomisation tests for hypotheses (not so modern as
     they go back to RA Fisher
   – Bootstrap methods for confidence intervals
                                                            5
       Randomisation test for a
         difference in means
• Tests the null hypothesis that the two samples
  come from a common distribution
• So in some ways this is more than a difference in
  means or even medians
• It is the same null hypothesis tested by traditional
  rank tests (eg Wilcoxon Mann-Whitney test)
• Rank testsare not just a test of medians
  Mann-Whitney test is not just a test of medians:
  differences in spread can be important
  Anna Hart
  BMJ 2001; 323: 391-393 (get it from BMJ.com)
                                                         6
Obs id sex   weight
   1  1 F        84.0
   2  2 F        98.0   Sample data set
   3  3 F       102.5
   4  4 F        84.5   Data on weights (in pounds)
   5  5 F       112.5   of 19 young people
   6  6 F        50.5
                        9 female
   7  7 F        90.0
                        10 male
   8  8 F        77.0
   9  9 F       112.0   Are males or females
  10 10 M       112.5   heavier?
  11 11 M       102.5   Are the weights of males
  12 12 M        83.0   more variable than those of
  13 13 M        84.0
                        females?
  14 14 M        99.5
  15 15 M       150.0
  16 16 M       128.0
  17 17 M       133.0
  18 18 M        85.0
  19 19 M       112.0                          7
                     T test output from SAS
                 Variable            sex            N    Mean      Std Dev
                 weight               F             9   90.111      19.384
                 weight               M            10   108.95      22.727
                 weight           Diff (1-2)            -18.84       21.22

      T-Tests
      Variable    Method               Variances   DF   t Value      Pr > |t|
      weight      Pooled               Equal       17      -1.93     0.0702
      weight      Satterthwaite        Unequal     17      -1.95     0.0680

      Equality of Variances
      Variable      Method        Num DF       Den DF   F Value     Pr > F
      weight        Folded F           9            8      1.37     0.6645




P value for differences in means is
0.0702 (pooled sd) or
0.0680 (common sd)                                                              8
 Permutation/randomisation test
• Here the difference between the means b
  (girls - boys) was -18.84 pounds
• Is this more than we would expect by
  chance if no difference between M and F?
  –   Consider all 19 people
  –   select 9 of them at random to be female
  –   get weight difference for „females‟ - „males‟
  –   this is the randomisation/permutation
      distribution under H0

                                                      9
 Programming the randomisation
             test
• This can be done easily
• Details of a SAS macro to do this are on the next
  page
• An EXCEL macro to do this is also available. On
  th class web page
• It was downloaded from
   – hhttp://www.bioss.ac.uk/smart/frames.html
   – It incorporates corrections from one of last year‟s
     Napier honours students

                                                           10
         SAS program to do this
• On my web site
• www.maths.napier.ac.uk/~gillianr
  –   macro - randmac.sas (submit this first)
  –   program rand.sas
  –   this reads in data and runs macro
  –   you can alter it for your own data
• Go to SAS now if time (this is V8.1)

                                                11
   Randomisation distribution of difference in means – actual difference was –18.84
   Proportion of the distribution further away from zero than this is 0.0673

      E U N Y
    FR Q E C
          800




         700




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           0

                -   -   -   -   -   -   -   -   -   -   -   -   -   -    -   0   2   4   6   8   1   1   1   1   1   2   2   2   2   2   3   3   3
                3   2   2   2   2   2   1   1   1   1   1   8   6   4    2                       0   2   4   6   8   0   2   4   6   8   0   2   4
                0   8   6   4   2   0   8   6   4   2   0

                                                                        di f f    I P I T
                                                                                 MD O N


This compares with 0.068 or o.07o2 for t-tests                                                                                                       12
  E U N Y
FR Q E C
     1900

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    1400

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    1200

    1100

    1000

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       0

            0   0   0   0   1   1   1   2   2   2   3   3   3   3   4    4   4   5   5   5   6   6   6   6   7   7   7   8   8   8   9   9   9   9   1   1
            .   .   .   .   .   .   .   .   .   .   .   .   .   .   .    .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   0   0
            0   3   6   9   2   5   8   1   4   7   0   3   6   9   2    5   8   1   4   7   0   3   6   9   2   5   8   1   4   7   0   3   6   9   .   .
                                                                                                                                                     2   5

                                                                                    I P I T
                                                                        f r at i o M D O N



                                                                                                                                                         13
              Conclusions
• For this problem, all methods give very
  similar answers for both means and
  variances
• This is usually true
• Exceptions are for odd distributions with
  possible outliers
• For these a randomisation test is a good
  choice
• To go with it, use a bootstrap C.I.
                                              14
   Comparing parametric and bootstrap confidence
                    intervals
                        Statistics
                              Lower CL             Upper CL
Variable        sex       N        Mean     Mean      Mean
 weight           F      9     75.211     90.111    105.01
 weight           M     10     92.692     108.95    125.21
 weight    Diff (1-2)          -39.41     -18.84    1.7313

     Bootstrap 95% confidence interval for
          difference (1000 bootstraps)
                (-37.87 to 0.16)
      So again, very similar to parametric
                                                              15
                Bootstraps
• Methods developed in the 1970s
  – Brad Efron
  – Rob Tibshirani
• Text book in library by these authors
  – also describes randomisation tests




                                          16
          What is a bootstrap?
• The data consist of n items
• Take a sample of size n from the data WITH
  replacement
  – data 1 2 3 4
  – possible bootstraps 1 1 2 3 , 1 1 1 1, 1 2 3 4,
• Take lots of bootstrap samples and calculate
  the quantity of interest (e.g. the mean) from
  each one.
  – The variability between the quantities calculated
    from the bootstraps is like the variation you
    expect in your estimate.                         17
Example of 3 bootstrap samples
 DATA           Number     Bootstrap samples
           12            8         23           2    12
          114                       2          12   114
           16                      16          23    16
           15                      15          15     2
           23                     114          12     2
           12                      16          15    16
            2                     114          15    12
            9                      16          16     9

 Mean             25.375     39.5    13.75   22.875
 Median             13.5       16       15       12
 s.d.           36.30796 46.34652 5.849298 37.22302
                                                      18
Go to EXCEL demo
     Found on web at
http://www.mste.uiuc.edu/

 At University of Illonois
     M2T2 project
 Mathematics material for
  tommorow‟s teachers
 Copied on my web page
                             19
     Bootstrap macro for SAS
First submit macro file (randmac.sas)
Then run macro(example in file bootstrap.sas)
%boots(in=new,var=x,seed=233667,fun=mean,
  nboots=100,out=result);
Explanation of each is in the sample program
bootstrap.sas
Go to SAS now

                                               20
       Other bootstrap macros
Does bootstrap CI for differences in means
Example in program rand.sas
%bootdiff(in=class,var=weight,group=sex,n
  boots=1000,seed=45345,out=bootdiff);
Does bootstraps for correlation coefficients
Example in rand.sas
%bootscor(in=new2,var1=score1,var2=score2
  ,seed=5465,nboots=50     ,out=corr);



                                               21
Pearson‟s correlation coefficient
• Calculated for sample data


    r
          ( x  x )( y  y )
          (x  x) ( y  y)
                    2           2



• It has values between -1 and +1
• 0 represents no association
• We can think of our sample value of r
  estimating a population quantity r
• So we can calculate a bootstrap CI for r
                                             22
      Bootstrap for correlation
Data consist of pairs of values
(x1, y1) (x2 ,y2) (x3, y3) (x4,y4) (x5,y5)
Bootstraps are samples of pairs – with
  replacement eg
(x1,y1) (x1,y1) (x3,y3) (x5,y5) (x5,y5)
The correlations from bootstrap samples will
  always be between –1 and +1.
Sample data next
                                               23
     Confidence interval for a correlation
                 coefficient
• Data on agreement in scoring between two
  observers from a sample of 30 items scored
• Sample value is 0.966
• How well might this represent the true population
  value ?
• Bootstrap confidence interval gives us the interval
  and also the bootstrap distribution (see next slide)
• Next we will look at the classical approach to this

                                                     24
scor e1
     80




    70




    60




    50




    40




    30

          30   40   50     60      70   80   90

                         scor e2



                                             25
26
            Distribution of r
• If we took many small samples, the
  distribution of r would not be normal.
• Fisher showed that, for bivariate normal
  distributions,
  – z = 0.5 ln[(1+r)/(1-r)]
  – is approximately normal with a standard error
    of 1 / sqrt(n-3)
  This can be used to get a confidence interval for r
  – r = [exp(2z)-1]/[exp(2z)+1]
                                                   27
    Example used for bootstraps
R = 0.966 n=30
z = 0.5 ln[ (1+0.966)/(1-0.966)]
  = 2.0287
standard error of z = 1/(30-3) = 1/ 27
  =0.192
95% C Int for z (2.0287 +/- 1.96 x 0.1925)
                  (0.93 - 0.98)

Bootstraps gave (0.94-0.99) - near enough
                                             28
                 Conclusions
• Randomisation tests and bootstraps are a useful
  alternative to parametric tests
• BUT they are more bother to do
• But should get easier in future
• In general they vindicate the robustness of
  classical tests, even when their assumptions are
  not true
• An exception may be data with outliers
• We will investigate this in the practical

                                                     29
                Summary
• F tests for ratios of variances
• How F ratios are used in regression and
  analysis of variance (sketch)
• Randomisation / permutation tests
• Bootstraps
• Correlation coefficient (introduction)


                                            30

				
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posted:9/19/2011
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