2013_Ch.15_Notes

download 2013_Ch.15_Notes

of 5

Transcript of 2013_Ch.15_Notes

  • 8/17/2019 2013_Ch.15_Notes

    1/5

    QMDS 202 Data Analysis and Modeling

    Chapter 15 Chi-Squared Tests

    The Contingency Table Test

    A Contingency table is one that shows all the classifications of the variables

     being studied, that is, it accounts for all contingencies in a particular situation.

    For a contingency table that has r rows and c columns, the χ2  test can be

    generalized as a test of:

    a !ndependence or 

     b "omogeneity.

    The test statistic #( )

    ∑  −

    =i

    ii

    e

    e f   2

    2 χ 

    v # (r – 1)(c – 1)

     g 

    cr i

     f 

     f  f e   =

     f r  # row total

     f c # column total

     f  g  # grand total

    For a test of independence we have one population and we are testing whether two characteristics in the population are independent. $ith a test of homogeneity we

    have more than one population and we are testing to see if a characteristic is the same

    across populations.

    !ndependence Test

    "%: A and B are independent

     A

    a1   a2   a3

     B b1

    a1∩b1 f 11

    a2∩b1 f 12

    a3∩b1 f 13   fr 1

    b2

    a1∩b2

     f 21

    a2∩b2

     f 22

    a3∩b2

     f 23   fr 2

     fc1   fc2   fc3   f  g 

    !f A and B are independent, then

     g 

     g 

    c

      f  

      f  

      f  

      f  b P a P ba P    &&&&&&   '''   ×=×=∩

    ∴The e(pected fre)uency 'ei of 'a1∩b1

    &

  • 8/17/2019 2013_Ch.15_Notes

    2/5

      # g 

    cr  g 

     g 

     g 

    c g 

      f  

      f    f    f  

      f  

      f  

      f  

      f    f  ba P    &&&&&&   '   =××=×∩

    "omogeneity Test

    "%:  P 'm1| A # P 'm1| B # P 'm1|C 

      and   P 'm2| A # P 'm2| B # P 'm2|C 

    Samples

     A B C 

     M m1

    'm1| A

     f 11

    'm1| B

     f 12

    'm1|C 

     f 13   fr 1

    m2

    'm2| A

     f 21

    'm2| B

     f 22

    'm2|C 

     f 23   fr 2

     fc1   fc2   fc3   f  g 

     fc1 # sample size of first sample

     fc2 # sample size of second sample

     fc3 # sample size of third sample

    !f P 'm1| A# P 'm1| B# P 'm1|C , then all these probabilities e)ual to g 

      f  

      f  m P    && '   =

    .

    *imilarly, P 'm2| A # P 'm2| B # P 'm2|C  # g 

      f  

      f  m P    22 '   = .

    ∴ +(pected fre)uency 'ei of m& in sample A 

    # ei of 'm1| A # P 'm1| A × sample size of A # &&

    c

     g 

    r    f    f  

      f  ×

      +(pected fre)uency 'ei of m2 in sample A 

    # ei of 'm2| A # P 'm2| A × sample size of A # &2

    c

     g 

    r    f    f  

      f  ×

    +(ample & A random sample of % men and -% women were ased whether or not

    they lie a certain brand of product. Their responses are shown in the

    following table:

    /es 0o Total

    1en & 2 %$omen 22 34 -%

    Total 35 -3 &%%

    6se the Chi7s)uare test to test that the liing of both se(es are not

    related, using α # %.%.

    *olution: "%: The liing of both se(es are not related 'independent

    "&: The liing of both se(es are related 'dependentα # %.%

    2

  • 8/17/2019 2013_Ch.15_Notes

    3/5

    χ27distribution will be used as the testing distribution.

    r  # 2   c # 2

    v # 'r  8 &'c 8 & # &

    9eect "% if T* ; 3.4ei:

    4.&,&%%

    35,%=

    ×2.2

    &%%

    -3,%=

    ×

    2.22&%%

    35-%=

    ×4.35

    &%%

    -3-%=

    ×

    ( ) ( ) ( ) ( )%%52.%

    4.35

    4.3534

    2.22

    2.2222

    2.2

    2.22

    4.&,

    4.&,&  2222

    =−

    +−

    +−

    +−

    =TS 

    T* # %.%%52 is not greater than 3.4 ⇒ Cannot reect "%Conclusion: The liing of both se(es are not related.

    +(ample 2 A business firm wants to determine whether the )uality of its product is

    the same at all five of its production units. For this purpose it taes a

    sample of &%% items at each of its production units and after testing the

    items for )uality, obtains the following data. Test whether the

     probability of obtaining a satisfactory item is the same for all five

     production units, using α # %.%.

    6nit & 6nit 2 6nit 3 6nit 6nit Total

    *atisfactory items 42

  • 8/17/2019 2013_Ch.15_Notes

    4/5

    ( ) ( ) ( )&5.&&

    &2

    &2&-...

    44

    44s product have a significant effect in

    losing weight, using α # %.%&.

    *olution: The re)uired contingency table:

    "%:

    "&:

    α # %.%&

    χ27distribution will be used as the testing distribution.

    r  # c #

    v # 'r  8 &'c 8 & #

    9eect "% if T* ;

    ei:

  • 8/17/2019 2013_Ch.15_Notes

    5/5

    T* #

    *tatistical decision:

    Conclusion:

    9eview ?roblems: &.24, &.2