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MH4511 Sampling & Survey Final Solution

AY2019/20 Semester 1

Problem F.1 (Solution) a) Sampling units: the 100 rabbits involved in the study. Variable of interest: the weight of each rabbit at 2 months. Auxiliary variable: the weight of each rabbit at pre-study. b) The scatterplot seems to suggest a linear relationship between the current weight and prestudy weight for the rabbits c) First, we need to find the value of the regression coefficients: () =  +  ∙   =

∑( − )( − )  0.0211 = 2 = ≈ 0.950 ∑( −  )  0.0222

 =  −  ∙  = 4.0 − 0.950 × 3.0 = 1.150

 =  +   = 1.150 + 0.950 × 3.1 = 4.095 We also see that  =  =  (1 −   ) =



. √.×. 0.0289 × (1 − 0.833

   = 󰇡1 − 



=

= 0.833, and

) = 0. 094

    󰇢 = 󰇡1 − 󰇢 (1 −   )    

= 1 −

  = 0.0282

10 0.0289 (1 − 0.833 ) ≈ 0.000795 = 0.0282  10 100

Hence, a 95% confidence interval for the average current weight of the 100 rabbits is:  ± ,. ×    = 4.095 ± 2.306 × 0.0282 = 4.095 ± 0.065 = (4.030, 4.160)

d) Using regression estimator with margin of error = 0.025 marks (for mean), . = 1.96 ,  = 0. 094 / ×   1.96 × 0.094   =   = 54.31  =   0.025

=

 54.31 = ≈ 35.2 ≈ 36 1 +  / 1 + 54.31/100

We know that there are 2 strata, one with phone ( = 0.85 ) the other without ( = 0.15 ). We also know that for optimal allocation,

Problem F.2 (Solution)

 ∝

a) With the budget constrain, we have

  

$400 + $10 ×  + $10 ×  = $2000 $400 + $10 ×  = $2000  = 160

If all households are interviewed in person, then  is the same for both strata.

But,  =  (1 −  ) = 0.5(1 − 0.5) = 0.5 , and  = 0.2(1 − 0.2) = 0.4

With  = 160, then  = and similarly  =

/ 0.85 0.15 1

Stratum 1 2 Total





 

 0.5 0.2

∙ =

∙ =

.×.

.×..×.

.×.

.×..×.

 =  (1 −  ) 0.5 0.4

× 160 ≈ 140,

× 160 ≈ 20. ( /) 0.425 0.060 0.485

≈  140 20 160

b) In this case,  ’s are the same as above, while  = 4 and  = 25.     

Hence,  = Similarly,





      





∙ =



Or,  =  −  .

∙ =

1 2 Total

/ 0.85 0.15 1

.×. √ .×. .×.  √ √

∙ =

. ∙ .

 = 0.94655 , and

∙  = . ∙  = 0.05345 . .

$400 + $4 ×  + $25 ×  = $2000

Bringing in the cost,

Stratum

.×. √ .×. .×.  √ √

$400 + $4 × 0.94655 + $25 × 0.05345 = $2000 

0.5 0.2

 ≈ 312,  ≈ 295,  ≈ 17. 

4 25

 = (1 − ) 0.5  0.4 

( /)/  0.2125 0.0120 0.2245

≈  295 17 312

To determine the variance for Part (a), note that 󰇡1 − Stratum

/



1 2 Total

0.85 0.15 1

0.5 0.2

Hence,

 

= (1 −  ) 0.5  0.4 





≈  140 20 160

󰇢 ≈ 1 with large population size.  1−

          0.001290 0.000180 0.001470

  = 0.001470 = 0.0383

To determine the variance for Part (b), note that 󰇡1 −  󰇢 ≈ 1 with large population size. Stratum

/



1 2 Total

0.85 0.15 1

0.5 0.2

Hence,

  = (1 −  ) 0.5  0.4 



≈  295 17 312



 1−

  = 0.000824 = 0.0287

          0.000612 0.000212 0.000824

Problem F.3 (Solution) a) Primary sampling units (): the 40 blocks of household in the community. Secondary sampling units (): the 4000 household in the community. b) The probability of a household being selected is the same as the probability that the block in which the household is in is being selected. Since the manager uses SRS to select 8 blocks out of a total of 40 blocks, the probability is   󰇡 󰇢󰇡 󰇢    󰇡 󰇢 

=





= 0.20.

Therefore, Pr(household  is in the sample) = 0.20.

c) The probability that 2 households from different blocks are included in the sample is the same probability that the 2 different blocks are selected in the sample. Under the SRS, The probability is    󰇡 󰇢󰇡 󰇢󰇡 󰇢     󰇡 󰇢 

×

= × = 0.0359.

Therefore, Pr(households  and  from different blocks are in the sample) = 0.20.

d) The two events are not independent as Pr(households  and  from different blocks are in the sample) ≠ Pr(household  is in the sample)× Pr(household  is in the sample), when households  and  are from different blocks.

e) We know  = 40,  = 8. (See table on next page.) Using ratio estimator,  = 



∑  ∑   

=

∑    ∙ ∑   

=



 

= 1.40

1 1 1 ( −  ∙  ) = × 6218 = 888.286 ( ∙  −  ∙  ) = = −1 −1 7 



1 1  8 󰇢 ∙  = 1 −     ∙ 888.286 = 0.0942,  40 8 ∙   ∙ 4000 = where  = 100 40

 ( ) = 󰇡1 −

A 95% CI for the average number of newspapers purchased is given by:

 ± , ×  ( ) = 1.40 ± ., × 0.0942 = (1.177, 1.623) 

where ., = 2.365

1 2 3 4 5 6 7 8

Number of households ( ) 80 85 100 125 120 125 130 95

total

860

Block

Average number of newspapers purchased () 1.4 1.6 1.5 1.8 1.4 1.2 1.0 1.4

 =   ∙  112 136 150 225 168 150 130 133 1204

 ∙ 

112 119 140 175 168 175 182 133

( −  ∙  )  0 289 100 2500 0 625 2704 0 6218

Problem F.4 (Solution) a)

1 st R.N. (psu ) 6 8 3 5 1 10

 20 3 16 4 8 8

2 nd R.N. 12 4 15 2 16 7

 > 2nd R.N. Yes No Yes Yes No Yes

Action Select Plant 6 Select Plant 3 Select Plant 5 Select Plant 10

According to set of random numbers given, the Lahiri’s method gives the sample index: {6, 3, 5, 10}. b) (i) Primary sampling units (): the 10 noodle manufacturing plants. Secondary sampling units (): the 100 machines in the 10 plants.

(ii) This Sampled Plant () 3 5 6 10 Total Average Std Dev





 

 

16 4 20 8

16/100 4/100 20/100 8/100

2, 6 8, 4 11, 5 3, 9

4 6 8 6

 =

 =  ×  64 24 160 48 296

  400 600 800 600 2400 6 00 163.30



1 1   = [400 + 600 + 800 + 600] = 600  4  

 1  163.3 1  󰇩Stdev 󰇧 󰇨󰇪 =    =   󰇩 −  󰇪 = = 81.65   ( − 1)  √ √4  ...