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					Comparing more than two population means
We can use two sample test to test the equality of more
 than two population means, but this procedure
          – Require large number of two sample tests
          – Performing many two sample tests at α tends to inflate
            the overall α risk.
   For example, To test the equality of 10-population
      means, we have to perform 45 test
   If the tests are independent and each test use α =0.05,
      then overall α=45(0.05)=2.25
we require a procedure for carrying out test of hypothesis
about the equality of several population means
simultaneously
      –we can use F-distribution in ANOVA that yields a single test
      statistic for comparing all means.
                Analysis of Variance
                     (ANOVA)
      Analysis of Variance is a procedure that partitions the
       total variability present in the data set into meaningful
       and distinct components. Each component represents
       the variation due to a recognized source of variation,
       in addition, one component represents the variation
       due to uncontrolled factors and random errors
       associated with the response measurements


NORMALITY:-The K-populations from which sample are drawn should be normal
INDEPENDENCE:-The k-samples should be independent
Randomness: The k-Samples should be random
 Four groups of students were subjected to different teaching
 techniques and tested at the end of a specified period of
 time. Due to drop outs in the experimental groups (sickness,
 transfers etc) the number of students varied from group to
 group
Method 1   Method 2   Method 3   Method 4
   65         75         59         94
   87         69         78         89
   73         83         67         80
   79         81         62         88
   81         72         83
   69         79         76
              90
  454        549        425        351      1779



Do the data provide sufficient evidence to indicate a
difference in the mean achievements for the 4
teaching techniques?
  Construction of hypotheses
  Ho : 1=2=3=4(i.e Mean achievements from 4 methods are same)
  H1: At least two ’s are different
    Level of significance
                                = 5%
        Test Statistic             2

                         F    S   b
                                   2
                               S   w


                                       A N O V A TABLE
 Source Of Variation      Degree of Freedom Sum of Squares    Mean Sum of Squares   Fcal
       (S.O.V)                   DF               SS               MSS=SS/df
Between Methods           4-1 =3            712.6            237.5    S2 b          3.77*
Within Methods            22-3=19           1196.6           63.0   S2w(MSE)
(Error)
TOTAL                     23-1=22           1909.2




      CALCULATION FOR ANOVA TABLE ?
 Method 1   Method 2   Method 3   Method 4
    65         75         59         94      Correction Factor=(G.T)2/Obs= (1779)2/23 = 137601.8
    87         69         78         89
    73         83         67         80      Total SS=(65)2+(87)2 …(88)2 – CF= 139511 – 137601.8
    79         81         62         88
    81         72         83                                 = 1909.2
    69         79         76
               90
   454        549        425        351      1779

                    (454 ) 2 (549 ) 2 (425 ) 2 (351 ) 2
BetweenMet hodsSS                                    CF  138314 .4  137601 .8  712 .6
                       6        7        6        4
Within Methods SS                   =Total SS – Between Methods SS=1909.2 – 712.6 = 1196.6

       SOV                                DF        SS        MSS=SS/d f F cal
       Between Methods                    4-1=3     712.6     237.5              237.5/63=3.77
       Within Methods                     19        1196.6    63
       TOTAL                              23-1=22   1909.2

  Decision Rule:- Reject Ho if F cal  F(3,19)
Result:-As F cal =3.77 > F.05(3,19) =3.10 So reject Ho and conclude that
there is difference in the mean achievements for the four teaching methods.
• Let 6 different wheat varieties with three plots
each was observed and measured the larval
population (%).The data is given below.
• Test the Hypothesis that all Varieties have the
same Larval Population?
       H 0 ; U1  U 2  U3  U 4  U5  U 6
       H1; At least one mean is not equall
Now, the layout is as follows:
      1-V3 2-V6 3-V4 4-V2 5-V2 6-V2
     74.00 48.57 49.24 59.75 45.75 60.22

     7-V5 8-V3 9-V5 10-V4 11-V5 12-V6
     50.30 75.28 36.70 60.84 36.06 45.33

     13-V1 14-V1 15-V6 16-V3 17-V1 18-V4
     42.93 31.20 39.15 61.53 33.28 60.58
           V1      V2       V3       V4      V5       V6

       42.93    59.75    74.00    49.24 50.30       48.57

       31.20    45.75    75.28    60.84 36.70       45.33
       33.28    60.22    61.53    60.58 36.06       39.15
TOTAL 107.41 165.72 210.81 170.6 123.06 133.05 910.71



  1. Group the data by treatment and calculate the treatment
     totals (T) and grand total (G).

  2. Construct an outline of the analysis of variance as follows:
SOURCE OF      DEGREE OF   SUM OF  MEAN   COMPUTED TABULAR F
VARIATION      FREEDOM     SQUARES SQUARE    F      5% 1%


TREATMENT         t-1
EXPERIMENTAL
ERROR            t(r-1)

 TOTAL          (r)(t)-1

3. Calculation of Degree of Freedom
 Total d.f = (r) (t) – 1 = (3) (6) -1 = 18 – 1 = 17
 Treatments d.f. = t – 1      =6–1=5
 Error d.f. = t(r-1) = 6(3 – 1)   = 6(2) = 12
   The error d.f. can also can also be obtained through
   subtraction as:

   Error d.f. = total d.f. – treatment d.f.   = 17 – 5 = 12
4. Using X i   to represent the measurement of the

  ith plot, Ti as the total of the ith treatment, and n as

  the total number of experimental plots [i.e., n =(r)(t)],

  Calculate the correction factor and the various sums

  of squares (SS) as:
  STEP-1
                                    2
Correction factor (C.F.) =
                                 G
                                  n
Where G indicates        X
         (42.93  31.20  33.28  59.75  45.75  ........  39.15)2
C.F.   
                                   18
                     2
         (910.71)
       
            18
         = 46077.37245
STEP-2
                                      n

Total SS (sum of squares) =           X i2  C.F
                                     i 1


 (42.93)2  (31.20)2  .......... ..  (45.33)2  (39.51)2  C.F


= 49062.475-46077.37


= 2985.105
STEP-3              t    2
                       Ti
SS treatment =      r  C.F.
                   i 1 i


  (107.41)2  (165.72)2  .......... .  (133.05)2
                                                   46077.37
                      3
=48470.59-46077.37


=2393.22

Note: Treatment = Variety
STEP- 4

 SS error   = Total SS – Treatment SS


 SS error   = 2985.10 – 2393.22


 SS error   = 591.88
We can calculate the men square (MS) for each source
of variation by dividing each SS by its corresponding d.f:

                      TreatmentSS
Treatment MS =
                          t 1
                     2393.22 2393.22
                           
                      6 1      5

                   = 478.64
             ErrorSS
Error MS =
               t(r  1)
           591.88 591.88 591.88
                       
           6(3  1) 6(2)   12
       = 49.32
ANOVA Table

 S.O.V.   D.F.     SS      MSS F-value

  Treat    5     2393.22 478.64

  Error   12     591.88    49.32   9.70

  total   17     2985.10
Calculate the F value for testing significance of the treatment
difference as:
         TreatmentMS
      F=
           ErrorMS
           478.64
         
           49.32

         = 9.70

The F value should be computed only when the error d.f.
is large enough for a reliable estimate of the error variance.
We can calculate the tabular F value from F-distribution table
where

   f1   = treatment d.f. = (t-1)

        = 6-1 = 5

   f 2 = error d.f. = t(r-1)
        = 6(3-1) = 6(2) =12

For our example, tabular F values are 5.06 for the 1% level of
significance.
Since Fcal(9.70) is greater than Ftab = F(5,12;0.99) = 5.06, so
 result is significant and we Conclude their is evidence that
varieties have different larval Populations.




    V1      V2       V3       V4       V5      V6
    35.81 55.24 70.27 56.88 41.02 44.35

				
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