STA 304H1F Winter 2020 Assignment 2: Analysis of Sampling Techniques

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Running Head: ASSIGNMENT ON SAMPLING TECHNIQUES
ASSIGNMENT ON SAMPLING TECHNIQUES
Name of the Student:
Name of the University:
Author Note:
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1ASSIGNMENT ON SAMPLING TECHNIQUES
Answer 1
Here a population of 6 students is divided into two equal sized strata. It is given that,
N=6 , N1 =3 , N 2=3 , n1=2 , n2=2
Where N=total population size, Ni=size of the ith strata, ni=sample size taken from the ith strata.
a) μ1= 1
3 [66+59+ 70 ] =65
μ2= 1
3 [83+82+ 71 ] =78.67
b) σ 2= 1
N
i=1
N
( Y iY ) 2 , Y i’s are the observations.
σ 1
2=1
3 [ ( 6665 )2 + ( 5965 )2+ (7065 )2 ] =20.67
σ 2
2=1
3 [ ( 8378.67 )2 + ( 8278.67 )2+ ( 7178.67 )2 ]=29.56
c) S2= 1
N1
i=1
N
( Y iY ) 2 S1
2 =31, S2
2=44.33
Var ( ^τ1 ) =N1
2
( N 1n1
N1 n1 )S1
2=46.5,τi is the population total.
Var ( ^τ2 ) =N2
2
( N 2n2
N 2 n2 )S2
2=66.50
d) Suppose a stratified sample is constructed taking scores of 1st and 2nd students (66,59)
from stratum 1 and scores of 4th and 5th students (83,82) from stratum 2.
n1=2, n2 =2 , y1=62.5 , y1=82.5
Var ( ^μstr )= 1
N2 Ni ( N ini ) Si
2
ni
= 1
36 [ 3. 31
2 +3. 44.33
2 ] =3.14
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2ASSIGNMENT ON SAMPLING TECHNIQUES
Var ( ^τ str ) =N2 Var ( ^μstr ) =113.04
e) If SRSWOR(Simple Random Sampling Without Replacement) is adopted to form a
stratified sample of size 4, taking 2 from each strata, then the number of all possible cases
is C2
3 C2
3 =9
The samples are-
((66, 59),(83,82)), ((66,59),(83,71)), ((66,59),(82,71)), ((66,70),(83,82)), ((66,70),
(83,71)), ((66,70),(82,71)), ((59,70),(83,82)), ((59,70),(83,71)), ((59,70),(82,71)).
f)
g) μ= 1
N
i=1
N
Y i=71.83
E ( ^μstr ) = 1
N
i=1
N
Ni Y i =71.83=μ
Therefore ^μ is an unbiased estimator of μ. Similarly, ^τ is an unbiased estimator of τ .
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3ASSIGNMENT ON SAMPLING TECHNIQUES
Answer 4
Here the dataset contains scores of 312 students. The necessary calculations are provided in a
R Markdown File.
a) The histogram shows that the distribution of test 1 scores is skewed to the left. The mean
and standard deviation are 70.79167=71(approx.) and 15.49539 respectively.
b) Here a systematic sample of size 20 is drawn from the population. The estimated mean
score is 65.8. The bound error estimate is found to be 7.155477. Hence, a 95%
confidence interval of the estimate is (58.64452, 72.95548).
c) Here 5 repeated systematic sample are drawn from the population. The estimated average
marks is found to be 67.82. The error bound of the estimate is equal to 9.565965. Hence,
the 95% confidence interval of the estimate is (58.25404, 77.38596).
d) The repeated sampling is better technique since it reduces the bias in sampling and gives
a wider confidence interval. Hence method in c is more appropriate in this case.
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