Access over 20 million homework & study documents
search

# All Soln

User Generated

Statistics

Homework

### Rating

Showing Page:
1/3
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).

Showing Page:
2/3
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:
     




  

Showing Page:
3/3

### Unformatted Attachment Preview

Question 2.3 Given: 𝑁 = 1000 𝑁1 = 600, 𝜋𝑘 = 0.1, 𝑘 ∈ 𝑈1 𝑁2 = 300 𝜋𝑘 = 0.1, 𝑘 ∈ 𝑈2 𝑁3 = 100 𝜋𝑘 = 0.8, 𝑘 ∈ 𝑈3 a) The expected value of the sample size is: 𝐸(𝑛𝑠 ) = 𝐸(𝑛1 ) + 𝐸(𝑛2 ) + 𝐸(𝑛3 ) 𝐸(𝑛𝑠 ) = 𝑁1 𝜋1 + 𝑁2 𝜋2 + 𝑁3 𝜋3 𝐸(𝑛𝑠 ) = 600(0.1) + 300(0.1) + 100(0.8) 𝐸(𝑛𝑠 ) = 170 b) The variance of the sample size is: 𝑉(𝑛𝑠 ) = 𝑉(𝑛1 ) + 𝑉(𝑛2 ) + 𝑉(𝑛3 ) 𝑉(𝑛𝑠 ) = 𝑁1 𝜋1 (1 − 𝜋1 ) + 𝑁2 𝜋2 (1 − 𝜋2 ) + 𝑁3 𝜋3 (1 − 𝜋3 ) 𝑉(𝑛𝑠 ) = 600(0.1)(1 − 0.1) + 300(0.1)(1 − 0.1) + 100(0.8)(1 − 0.8) 𝑉(𝑛𝑠 ) = 54 + 27 + 16 𝑉(𝑛𝑠 ) = 97 c) Assuming 𝜋𝑘 is constant for all 𝑘 ∈ 𝑈, that is 𝜋𝑘 = 𝜋, 𝐸(𝑛𝑠 ) = 𝑁𝜋 The value of 𝜋 so that 𝐸(𝑛𝑠 ) is same as in ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Review
Review

Anonymous