MAT3377
Assignment 1 on Review section
1. In a certain population, the distribution of number of children in families (younger than
18 years old) is tabulated in the table below.
k
0
2
3c 2c
P (x = k)
1
c
3
4
c/2 c/2
(i) Find c.
(ii) Find the mean and the
MAT3377
Solution to Assignment
2
1. (i) In here n = 20 and Observations are
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2.
Therefore an estimate for average number of cracked eggs/box is
y=
In this case
s2 =
1
n1
8+6
= 0.7.
20
n
2
yi n2 =
MAT3377
Assignment 3
1. [Bonus question 15 extra marks]. With the same notations as in lectures show that if
1/Ni are not negligible then we get
Varran () = Varprop (st ) +
y
y
L
1f
n(N 1)
1
Ni (Yi Y )2
N
i=1
L
(N Ni )Si2
i=1
Now, discuss when stratied s
Mid-term test/
MATH3377 Surveys and
Sampling
Time : 80 Minutes
Professor : M. Zarepour
Name :
Student Number :
Calculators are permitted. This is an open book exam.
1. In a small city with the population size of N=8500 families a simple random
sample of n
MAT3377
Assignment 5
[28 Points] 1. In some cases, it is reasonable to assume that our population follows the
following model, for given constants and
Yi = + i + i ,
where E( i ) = 0, E( 2 ) = 2 and
i
for i = 1, 2, . . . , N = nk.
1, . . . , N
are uncorr
MAT3377
Assignment 4
1. We know from lectures that r = y / is not an unbiased estimate for R. Here is one way
x
to nd an unbiased estimate. Dene ri = yi /xi and Ri = Yi /Xi and r =
1
n
n
i=1 ri
Show that (i)
1
E( R) =
r
XN
N
Ri (Xi X).
i=1
Show that (ii
MAT3377
Solution to Assignment 5
Problem 1. First of all notice that
N
N + 1 2 N (N 2 1)
) =
.
2
12
(i
i=1
This can be proved by expanding the sum and using the fact that
N
=
i=1
N (N + 1)(2N + 1)
.
6
It is also easy to see that
1
Y =
N
N
( + i + i ) = +
MAT3377
Assignment 2
1. Out of consignment of 100 boxes of eggs, each containing 40 eggs, 20 boxes werele selected
without replacement. Among the boxes in the sample, nine were found to contain no cracked
eggs, eight contain 1, and three contain two crack
MAT3377
Solution to Assignment 3
1. As it was discussed in the lectures we can write
L
Nh
(N 1)S 2 =
Nh
L
(Yhi Y )2 =
h=1 i=1
L
(Yhi Yh + Yh Y )2 =
h=1 i=1
L
Nh (Yh Y )2
2
(Nh 1)Sh +
h=1
h=1
Therefore
V arran () =
y
L
L
1f
S2
(1 f ) =
n
n(N 1)
Nh (Yh
MAT3377
Solution to Assignment 4
1. Part (i),
Dene ri = yi /xi , i = 1, 2, . . . , n and Ri = Yi /Xi , i = 1, 2, . . . , N and R = Y /X. In
SRS we have, E() = R. Therefore
r
1
E( R) =
r
N
N
i=1
Yi
Y
.
Xi X
On the other hand
1
NX
N
i=1
1
Ri (Xi X) =
NX
MAT3377
Fall 2013
Simple Random Sampling
N = Population size
n = Sample size
N
Number of sample of size n without replacement =
.
n
Population: Y1 , . . . , YN .
Sample: y1 , . . . , yn .
Population mean (average)
Y1 + + YN
=
Y =
N
N
i=1
y1 + + yn
=
n
MAT3377
Fall 2013
Simple Random Sampling
N = Population size
n = Sample size
Number of sample of size n without replacement =
N
.
n
Population: Y1 , . . . , YN .
Sample: y1 , . . . , yn .
Population mean (average)
Y1 + + YN
=
Y =
N
N
i=1
y1 + + yn
y=
=
Source: Frerichs, R.R. Rapid Surveys (unpublished), 2008. NOT FOR COMMERCIAL DISTRIBUTION
3
Simple Random Sampling
3.1
INTRODUCTION
Everyone mentions simple random sampling, but few use this method for population-based surveys.
Rapid surveys are no except
MAT3377
Fall 2011
Assignment 1
Due in class on October 17, 2011
1. Assume a population of size N = N1 + N2 is partitioned to two disjoint parts 1 and 2 with
population sizes N1 and N2 . In first part we arrange a census and Y1 is given. In the second
part
MAT3377
Fall 2011
Assignment 2
Due in class on October 31, 2011.
1. [Bonus question 15 extra marks]. With the same notations as in lectures show that if
1/Ni are not negligible then we get
L
L
X
1f
1 X
Varran (
y ) = Varprop (
yst ) +
Ni (Yi Y )2
(N Ni )
MAT3377
Fall 2015
Assignment 5
Due on December 10, 2015 in my oce
[40 Points] 1. In some cases, it is reasonable to assume that our population follows the
following model, for given constants and
Yi = + i + i ,
where E( i ) = 0, E( 2 ) = 2 and
i
for i =