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Question 2.3
Given:

,





a) The expected value of the sample size is:
 
 
 
 
    

b) The variance of the sample size is:
 
 
  
 
  
 
  



  



  
 
  

c) Assuming
is constant for all , that is
,

The value of so that 
is same as in case (a) is computed as follows:
 



Therefore the constant

The new variance is:

  



  

The new variance, 141.1 is greater than the variance of 97 in case (a).

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Solution to Question 2.4
a
 
 
1
250



0.3125
0.3125
2
350



0.8750
0.0000
3
150



0.5625
0.5625
4
50



0.2500
1.0000
Total
800
1.0000
 

 

Average cluster size  . Therefore, expected number of individuals to be interviewed is:
     




  

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Solution to Question 2.5
0.4
0.3
0.2
0.1
Solution 3(a)
Computation of
:
 
 
     
 
 
     
 
 
     
Computation of

:



 
   

 
   
Computation of

:






 
   
Computation of

:








Solution to 3(b)
i. Direct calculation of 
from the definition





    
ii. Calculation of 
from




 
 
    


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Unformatted Attachment Preview

Question 2.3Given: = 10001 = 600, = 0.1, 12 = 300 = 0.1, 23 = 100 = 0.8, 3a) The expected value of the sample size is:( ) = (1 ) + (2 ) + (3 )( ) = 1 1 + 2 2 + 3 3( ) = 600(0.1) + 300(0.1) + 100(0.8)( ) = 170b) The variance of the sample size is:( ) = (1 ) + (2 ) + (3 )( ) = 1 1 (1 1 ) + 2 2 (1 2 ) + 3 3 (1 3 )( ) = 600(0.1)(1 0.1) + 300(0.1)(1 0.1) + 100(0.8)(1 0.8)( ) = 54 + 27 + 16( ) = 97c) Assuming is constant for all , that is = ,( ) = The value of so that ( ) is same as in ...
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