Lecture Note
Sampling and Confidence Interval
-
University:
California State University, Northridge -
Course:
MATH 340 | Introduction to Probability and Statistics Academic year:
2024
-
Views:
64
Pages:
27
Author:
regausiu82y
Sampling and Confidence
Interval
How can it be that mathematics, being
after all a product of human thought
independent of experience, is so
admirably adapted to the objects of
reality
Albert Einstein Sampling
An element is the entity on which data are collected.
A population is a collection of all the elements of interest.
Sample is a subset of the population. It contains only a portion
of the population.
Frame is a list of objects
The sample results provide estimates of the values of the
population characteristics
With proper sampling methods, the sample results can provide
“good” estimates of the population characteristics.
2 Sampling from a Finite Population
St. Andrew’s College received 900 applications for admission in
the upcoming year from prospective students. The applicants
were numbered, from 1 to 900, as their applications arrived.
The Director of Admissions would like to select a simple
random sample of 30 applicants.
Generate rand() in column next to the names. Then sort the rand
column. Select the top 30 names.
Sometimes we want to select a sample, but find it is not possible
to obtain a list of all elements in the population.
As a result, we cannot construct a frame for the population. We
cannot use the random number selection procedure.
Most often this situation occurs in infinite population
cases.
3 Sampling from an Infinite Population
Populations are often generated by an ongoing process where
there is no upper limit on the number of units that can be
generated. Examples of on-going processes, with infinite
populations, are:
parts being manufactured on a production line
transactions occurring at a bank
telephone calls arriving at a technical help desk
customers entering a store
These are objects.
A random sample from an infinite population is a sample
selected such that the following conditions are satisfied.
Each element selected comes from the population of interest.
Each element is selected independently.
4 Sampling from an Infinite Population
𝑥𝑥 is the point estimator of the population mean.
s is the point estimator of the population standard deviation.
The target population is the population we want to make
inferences about.
The sampled population is the population from which the
sample is actually taken.
Whenever a sample is used to make inferences about a
population, we should make sure that the targeted population
and the sampled population are in close agreement.
5 10 Samples of size 25
Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6 Sample 7 Sample 8 Sample 9 Sample 10
25
49
0
11
86
78
54
11
80
48
43
98
14
52
14
5
91
53
14
26
75
50
27
84
55
0
35
2
62
85
90
92
31
94
33
30
94
2
48
2
88
39
22
0
47
39
71
48
39
61
7
78
93
81
88
90
58
28
81
49
97
40
35
97
56
13
59
54
50
86
33
27
66
44
77
78
42
10
98
82
18
47
96
23
44
78
17
56
28
72
0
12
3
45
96
29
15
81
94
63
12
39
81
46
3
89
86
84
26
84
22
20
59
29
66
84
29
79
58
80
49
92
88
32
55
57
59
44
45
53
28
21
22
84
60
32
32
14
41
82
82
54
64
53
22
0
14
23
82
72
9
47
57
29
64
85
46
28
37
40
82
93
24
90
66
53
14
7
76
25
21
18
87
88
63
96
81
55
64
64
44
7
38
1
58
87
66
15
48
29
42
5
73
67
83
17
77
65
23
92
3
7
45
21
76
65
5
92
0
20
49
84
25
59
96
19
13
99
67
47
Puplulation Mean
20
65
61
88
5
31
98
7
19
87
50.00
65
42
8
59
72
71
44
4
5
27
Population StdDev
23
77
52
24
3
4
24
78
36
2
28.90
Mean
StdDev
41.1
30.0
48.1
29.9
46.8
29.3
52.0
30.4
55.5
28.3
49.2
33.1
49.0
28.8
41.6
31.8
48.8
27.1
55.1
27.7
Mean StdDev
48.7
4.9
6 Sampling from an Infinite Population
A random variable x1: (µ and σ)
A random variable x2: (µ and σ)
……………………………………..
A random variable xn: (µ and σ)
A random variable y = x1+x2+…..+xn: (?,?)
Mean(y) = Mean(x1)+ Mean(x2)+ ……. + Mean(xn)
µ(y) = µ(x1)+ µ(x2)+ ……. + µ(xn)
µ (y) = n µ(x)
Var(y) = Var(x1)+ Var(x2)+ ……. Var(xn)
σ2(y) = σ2 (x1)+ σ2 (x2)+ ……. σ2(xn)
σ2(y) = n σ2 (x)
σ(y) = 𝑛𝑛 σ(x)
7 Sampling from an Infinite Population
A random Variable y= SUMx: (n µ(x), 𝑛𝑛 σ(x) )
A random Variable SUMx/n = 𝑋𝑋: (?, ?)
A random Variable y= SUMx: (n µ(x), 𝑛𝑛 σ(x) )
A random Variable y/n = SUMx/n = 𝑋𝑋: (?, ?)
Mean (𝑋𝑋) = n µ(x)/n = µ(x)
StdDev (𝑋𝑋) = (1/n)* 𝑛𝑛 σ(x) ) = σ(x)/ 𝑛𝑛
x: (µ and σ)
𝑋𝑋: (µ, σ/ 𝑛𝑛)
8 Standard Deviation of 𝑋𝑋 for N=25 and N=50
Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6 Sample 7 Sample 8 Sample 9 Sample 10
25
49
0
11
86
78
54
11
80
48
43
98
14
52
14
5
91
53
14
26
75
50
27
84
55
0
35
2
62
85
90
92
31
94
33
30
94
2
48
2
88
39
22
0
47
39
71
48
39
61
7
78
93
81
88
90
58
28
81
49
97
40
35
97
56
13
59
54
50
86
33
27
66
44
77
78
42
10
98
82
18
47
96
23
44
78
17
56
28
72
0
12
3
45
96
29
15
81
94
63
12
39
81
46
3
89
86
84
26
84
22
20
59
29
66
84
29
79
58
80
49
92
88
32
55
57
59
44
45
53
28
21
22
84
60
32
32
14
41
82
82
54
64
53
22
0
14
23
82
72
9
47
57
29
64
85
46
28
37
40
82
93
24
90
66
53
14
7
76
25
21
18
87
88
63
96
81
55
64
64
44
7
38
1
58
87
66
15
48
29
42
5
73
67
83
17
77
65
23
92
3
7
45
21
76
65
5
92
0
20
49
84
25
59
96
19
13
99
67
47
Puplulation Mean
20
65
61
88
5
31
98
7
19
87
50.00
65
42
8
59
72
71
44
4
5
27
Population StdDev
23
77
52
24
3
4
24
78
36
2
28.90
Mean
StdDev
Mean
StdDev
41.1
30.0
48.1
29.9
44.6
29.9
46.8
29.3
52.0
30.4
49.4
29.7
55.5
28.3
49.2
33.1
52.4
30.6
49.0
28.8
41.6
31.8
45.3
30.2
48.8
27.1
55.1
27.7
52.0
27.3
Mean StdDev
48.7
4.9
48.7
3.7
TRUE
9 Standard Deviation of 𝑋𝑋 for N=25 and N=50
Puplulation Mean
50.00
Population StdDev
28.90
Mean
StdDev
Mean
StdDev
41.1
30.0
48.1
29.9
44.6
29.9
46.8
29.3
52.0
30.4
49.4
29.7
55.5
28.3
49.2
33.1
52.4
30.6
49.0
28.8
41.6
31.8
45.3
30.2
48.8
27.1
55.1
27.7
52.0
27.3
Mean StdDev
48.7
4.9
48.7
3.7
TRUE
10 Sampling from an Infinite Population
When the population from which we are selecting a random
sample does not have a normal distribution, the central limit
theorem is helpful in identifying thes hape of the sampling
distribution of 𝑋𝑋.
CENTRAL LIMIT THEOREM: In selecting random samples of
size n from a population, the sampling distribution of the
sample mean 𝑋𝑋 can be approximated by a normal distribution
as the sample size becomes large.
11 Sampling from an Infinite Population
𝑋𝑋: (44.8,5.56)
What is the probability that a simple random sample of 25 will
provide an estimate of the population mean that is within +/−
10 of the actual population mean µ ?
Probability that 𝑋𝑋 is between 44.8-4.48 and 44.8+4.48
Probability that 𝑋𝑋 is between about 40.3 and 49.3
That is 4.48/5.56 = 0.81 standard deviation to left or right
=NORM.S.DIST(0.81,1) = 0.79
From negative infinity to +0.82 is 0.79
Therefore, from +0.82 to infinity is 0.21
Consequently from – infinity to -0.82 is also 0.21
Thus between -0.81 and +0.81 is 0.58
12 Interval Estimation vs Point Estimation
𝑋𝑋provides a point estimate for µ
Point estimate is based on just one sample, we can not expect it to
be equal to the corresponding population parameter.
Indeed, each sample will have a different non of them is equal to
µ. But they are all unbiased estimates of µ. ( Unbiased means
their expected value is equal to µ.)
Point estimate ;
µ=x
x
Interval Estimate ; with a certain confidence (probability)
x −a ≤μ ≤ x + a
13 Relationship between μ and x
.95
1-α
X ≤ µ + z(1−0.05 / 2 )σ X
X ≤ µ + z(1−α / 2 )σ X
X ≤ µ − z(1−0.05 / 2 )σ X
X ≤ µ − z(1−α / 2 )σ X
14 Relationship between μ and x
X ≤ µ + z(1−α / 2 )σ X
X ≥ µ − z(1−α / 2 )σ X
X − z(1−α / 2 )σ X ≤ µ X + z(1−α / 2 )σ X ≥ µ
X + z(1−α / 2 )σ X ≥ µ ≥ X − z(1−α / 2 )σ X
15 Relationship between μ and x
X + z(1−α / 2 )σ X ≥ µ ≥ X − z(1−α / 2 )σ X
X + z(1−0.05 / 2 )σ X ≥ µ ≥ X − z(1−0.05 / 2 )σ X
X + z0.975σ X ≥ µ ≥ X − z0.975σ X
X + 1.96σ X ≥ µ ≥ X − 1.96σ X
16 Interval Estimation of a Population Mean: σ is known
A mail ordering company want to know the quality of its product from the point of view of its customers.
Each month a sample of 100 customers are interviewed. Sample size : n=
25
They measure quality on a scale of 0 to 100.
Data from previous months indicates that the standard deviation of the measure of the quality is s =
We want to find an interval estimate for µ
This month interview scors are given below
Find the interval estimate for µ with
95% confidence interval.
90
115
77
70
67
83
119
61
91
101
96
72
66
95
76
105
70
72
105
48
60
103
77
87
105
81
63
70
64
77
Xbar
StdDevXbar
Z0.975
UCL
LCL
81.69
2.5
1.96
86.58991
76.79009
TRUE
79
62
55
126
111
101
99
78
54
68
132
62
46
51
53
57
85
41
81
70
88
116
97
74
76
78
60
59
121
87
82
110
78
56
64
94
51
103
84
33
82
119
77
88
68
44
92
87
75
85
77
71
67
78
119
75
113
56
63
54
25
101
139
82
119
115
71
90
49
151
74
17 3-Ways to Compute CI in Excel: σ is known
Using Z
Xbar=
SigmaX=
CI=
n=
Sigma(Xbar)=
Upper Probability
0.95CI: Upper Z(0.975
0.95CI: Upper Z(0.025
+Side
-Side
UCL
LCL
NORM.S.INV
41100
4500
0.95
76
516.19
0.975
1.96
-1.96
1012
-1012
42112
40088
Using X
Xbar=
SigmaX=
CI=
n=
Sigma(Xbar)=
Upper Probability
Lower Probability
NORM.INV
41100
4500
0.95
40
711.5124735
0.975
0.025
UCL 42495
LCL 39705
18 3-Ways to Compute CI in Excel: σ is known
Using Conf CONFIDENCE.NORM
Xbar=
41100
SigmaX=
4500
CI=
0.95
n=
54
Mrgin
UCL
LCL
1200
42300
39900
19 Sampling Error
For the case of a sample mean estimating a population mean,
the sampling error is | x − µ|
There is .95 probability that
precision
x estimates µ with z.0975σ x
Sampling
distribution
of
x
.025
µ − z.0975σ x
0.95 of all
x values
µ
0.975
µ + z.0975σ x
20 Example: National Discount, Inc.
National Discount has 260 retail outlets throughout the United
States. National evaluates each potential location for a new
retail outlet in part on the mean annual income of the
individuals in the marketing area of the new location.
The purpose of this example is to show how sampling can be
used to develop an interval estimate of the mean annual income
for individuals in a potential marketing area for National
Discount.
Based on similar annual income surveys, the standard deviation
of annual incomes in the entire population is considered known
with σ = $5,000.
We will use a sample size of n = 64.
21 Example: National Discount, Inc.
Precision Statement
There is a 0.95 probability that the value of a sample mean for
National Discount will provide a sampling error of $????? or
less.
Z0.975 = 1.959964
StdDev = 5000
StdDevXbar = 5000/SQRT(64)
StdDevXbar = 5000/8 =626
(Z0.975)(StdDevXbar) = 1.96(625) = 1225
Precision Statement
There is a .95 probability that the value of a sample mean for
National Discount will provide a sampling error of $1,225 or
less.
22 Sample Size for an Interval Estimate of a Population Mean
Suppose that National’s management team wants an estimate of
the population mean such that there is a 0.95 probability that the
sampling error is $500 or less.
How large a sample size is needed to meet the required precision?
Recall that σ = 5,000
23 Interval Estimation of a Population Mean: σ is
unknown
Instead of population standard deviation σ, we have sample
standard deviation of s
Instead of normal distribution, we have t distribution
The t distribution is a family of similar probability
distributions.
A specific t distribution depends on a parameter known as the
degrees of freedom.
As the number of degrees of freedom increases, the difference
between the t distribution and the standard normal probability
distribution becomes smaller and smaller.
A t distribution with more degrees of freedom has less
dispersion.
The mean of the t distribution is zero
24 N:fx
-3.00
-2.75
-2.50
-2.25
-2.00
-1.75
-1.50
-1.25
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
-3.00
-2.75
-2.50
-2.25
-2.00
-1.75
-1.50
-1.25
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
z and t: n=2, 10, 20
T-fx
N:fx
T-fx
25 Interval Estimation of a Population Mean: σ
Unknown
The interval estimate is given by:
x ± tα /2
s
n
where 1 -α is the confidence coefficient
tα/2 is the t value providing an area of α/2 in the upper tail of a t
distribution with n - 1 degrees of freedom
s is the sample standard deviation
26 Example: Apartment Rents
A reporter for a student newspaper is writing an article on the cost of off-campus housing.
A sample of 64
one-bedroom units within a 5-mile of campus is given below
Provide a
95% confidence interval estimate of the mean rent per month for the
population of one-bedroom units within a half-mile of campus.
1150
630
950
1110
1130
860
1810
1060
Xbar
StdDev
StdDevXbar
t0.025
UCL
LCL
910
370
1220
790
1410
1230
1210
600
1006.563
241.36
30.17
2.00
1066.852
946.2726
770
1120
900
930
1040
990
910
1050
1100
800
1220
1070
860
1020
810
1200
1000
1130
900
1370
1110
990
1150
1220
1030
1120
520
990
760
900
1320
750
780
1010
790
670
1190
1420
930
1160
1430
1180
760
1120
1060
850
710
870
TRUE
27
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