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Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
1
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
Some parts of these slides were prepared based on
Managing Business Process Flow, Anupindi et al. 2012. Pearson.
Essentials of Modern Busines Statistics, Anderson et al. 2012. Cengage.
Before coming to class, please watch the following 3
repository lectures on youtube
Mean and Variance of Sample Mean.youtube.repository
https://www.youtube.com/watch?v=7mYDHbrLEQo
The Sampling Distribution of the Sample Mean.youtube.repository
https://www.youtube.com/watch?v=q50GpTdFYyI
Confidence Interval.youtube.repository
https://www.youtube.com/watch?v=lwpobQmUTd8
The link to the excel file
Sampling and Confidence Interval-exl
http://www.csun.edu/~aa2035/CourseBase/S-Sampling-CI/ArdiCh78.xlsx
X: , , y = 2x y = 2,
 y = 2
Past data on a specific stock shows that the return of this stock
has a mean of 0.05 and StdDev of 0.05. Therefore, if we invest $1,
our investment after one year will have an average of $0.05 and
standard deviation of $0.05.
Using simulation in excel show what is the mean and standard
deviation of 10,000 and 20,000 investments.
A random variable x with mean of  , and standard deviation of σ
is multiplied by 2 generates the random variable y=2x.
x: ( , σ)  y: (?,?)
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
4
$10,000 or $20,000 in One Stock
Probability is between 0 and 1.
rand() is between 0 and 1.
Therefore, it is valid if we assume that a rand() is a random
probability. Accoordingly
=NORM.S.INV(rand())
Will provide us with a random z  suppose it is z = 0.441475
x= µ + z = x = 5%+ 0.441475(5%)  x= 5%+2.21% = 7.21%
On the same line of reasoning,
=NORM.INV(probability, µ, )
=NORM.INV(probability, 5%, 5%)
Will directly generate a random x from N(µ, )
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
5
10,000 and 20,000 in One Stock
1
2
3
4
5
6
7
8
9
10
11
997
998
999
1000
1
0.029
0.032
0.054
0.030
0.070
0.116
0.064
0.056
0.068
0.042
0.112
0.081
-0.011
0.021
-0.023
10000
112
837
389
-545
411
946
824
1043
1133
122
-267
600
556
-114
364
20000
-683
841
850
1717
1067
986
1946
2384
1919
389
1762
684
2058
1361
1046
=10,000*NORM.INV(rand(), 5%, 5%)
=20,000*NORM.INV(rand(), 5%, 5%)
`
Min=
Max=
Mean=
Variance
StdDev=
CV
Mean/SrdDev
Sampling Distribution & Confidence Interval
10000
-1293
2708
490
250278
500
1.02
0.98
Ardavan Asef-Vaziri
20000
-2595
4259
964
1119770
1058
1.10
0.9110
Jan.-2016
1.97
4.47
2.12
6
X: , , y = 2x y = 2,
 y = 2
A random variable x (, σ).
A random variable y = 2x.
y = 2 
σy = 2σ
A random variable x (, σ).
A random variable y = nx.
y = n
σy = nσ
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
7
ROP; Variable R, Fixed L
Demand is fixed and is 50 units per day. From the time
that we order until the time we receive the order is
referred to as Lead Time. Suppose average lead time is
5 days and standard deviation of lead time is 1 day.
At what level of inventory should we place an order such
that the service level is 90% (Probability of demand
during the lead time exceeding inventory is 10%).
This point is known as Reorder point (ROP).
The difference between ROP and Average demand
during lead time is referred to as Safety Stock.
What is the average demand during the lead time?
What is standard deviation of demand during lead time?
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
8
μ and σ of the Lead Time and Fided demand per period
x: ( , σ)  y: (n , nσ)
1
2
Sampling Distribution & Confidence Interval
3
Ardavan Asef-Vaziri
4
Jan.-2016
5
9
μ and σ of L and Fixed R
If Lead time is variable and Demand is fixed
L: Lead Time
L: Average Lead Time
L: Standard deviation of Lead time
R: Demand per period
LTD: Average Demand during lead time
LTD = L × R
LTD: Standard deviation of demand during lead time
𝜎𝐿𝑇𝐷 =R𝜎𝐿
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
10
μ and σ of L and Fixed R
If Lead time is variable and Demand is fixed
L: Lead Time
L: Average Lead Time = 5 days
L: Standard deviation of Lead time = 1 day
R: Demand per period = 50 per day
LTD: Average Demand During Lead Time
LTD = 5 × 50 = 250
LTD: Standard deviation of demand during lead time
𝜎𝐿𝑇𝐷 =R𝜎𝐿
𝜎𝐿𝑇𝐷 =50(1) =50
=NORM.INV(0.9,250,50) =314.1
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
11
X: , , y = x1+X2, y = 2,
2 y= 22,  y= √2
A random variable x has mean of  , and standard deviation of σ.
A random variable y is equal to summation of 2 random variable
x.
y = x1+x2
x: ( , σ)  y: (?,?)
Mean(y) = y = Mean(x1)+ Mean(x1) =  +  = 2 
VAR(y) = σ y 2 = VAR(x1) + VAR(x1) = σ2 + σ2 = 2 σ2
StdDev(y) = σ y = 𝟐σ
Using simulation in excel show what is the mean and standard
deviation of inversing $20,000 in one stock or inversing two
$10,000 each in one stock.
Suppose past data shows that return of all these stocks are 0.05,
0.5.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
12
20,000 One Stock or 20,000 in Two Stocks
1
2
3
4
5
6
7
8
9
10
11
997
998
999
1000
20000
2786
2085
2718
43
372
221
1406
328
2003
1826
-156
377
1058
-189
961
10000
545
782
-178
626
1048
1162
424
874
1094
734
-109
747
1061
597
-48
10000
-153
-125
501
601
723
567
1411
1455
676
218
-315
730
531
972
390
2(10,000)
393
657
324
1227
1771
1729
1835
2328
1770
951
-424
1477
1592
1569
342
Sampling Distribution & Confidence Interval
`
Min=
Max=
Mean=
Variance
StdDev=
CV
Mean/SrdDev
Ardavan Asef-Vaziri
20000
-2136
3993
1001
983323
992
0.99
1.0091
Jan.-2016
2(10000)
-1384
3291
1021
515268
718
0.70
1.4220
1.02
0.52
0.72
13
20,000 One Stock or 20,000 in Two Stocks
4000
3000
2000
1000
0
-1000
-2000
-3000
-4000
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
14
X: , , y = x1+X2, y = 2,
2 y= 22,  y= √2
A random variable x has mean of  , and standard deviation of σ.
A random variable y is equal to summation of 2 random variable
x.
y = x1+x2
x: ( , σ)  y: (?,?)
y = 2
σy2 = 2σ2  σ y = 𝟐σ
y = x1+x2+ x3+ ……….+xn
x: ( , σ)  y = (?,?)
y = n
σy2 = n σ2  σy = 𝒏σ
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
15
100,000 in One Stock or 10,000 in Each of 10 stocks
Suppose there are 10 stocks and with high probability they all
have Normal pdf return with mean of 5% and standard
deviation of 5%. These stocks are your only options and no
more information is available. You have to invest $100,000. What
do you do?
100000
6268
-607
6237
18040
8487
7579
9032
839
2152
10000 10000
324
302
37
-418
243
798
-273
214
68
1166
470
-430
100
1431
-231
1699
1476
-276
10000 10000 10000 10000 10000 10000 10000 10000
138
-336
352
687
769
1246
935
922
404
243
210
-78
91
112
1010
478
872
394
181
1165
695
330
523
1657
-45
399
991
1435
661
630
238
191
-235
-100
-179
131
555
195
-138
-42
45
-400
504
661
453
607
717
31
519
1016
760
1031
776
1037
761
857
-172
-751
1496
602
-152
28
662
496
204
965
501
687
1143
251
-203
-290
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
100000
5340
2090
6859
4442
1421
2658
8290
3677
4459
16
100,000 vs. 10(10,000) investment in N(5%, 5%)
Investment=
Min=
Max=
Mean=
Variance
StdDev=
CV
Mean/SrdDev
1.00
-0.09
0.21
0.05
0.00
0.05
0.94
1.06
100000
-11950
25974
4746
26057541
5105
1.08
0.9298
Sampling Distribution & Confidence Interval
94721
11623364914
107812
Ardavan Asef-Vaziri
10(10000)
31
9388
4976
2409193
1552
0.31
3.2060
Jan.-2016
0.95
10.82
3.29
17
Risk Aversion Individual
20000
15000
10000
5000
0
-5000
-10000
-15000
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
18
Problem Game- The News Vendor Problem
Daily demand for your merchandise has mean of 20 and standard deviation of
5. Sales price is $100 per unit of product.
You have decided to close this business line in 60 days. Your supplier has also
decided to close this line immediately, but has agreed to provide your last
order at a cost of $60 per unit. Any unsold product will be disposed at cost of
$10 per unit. How many units do you order
LTD = R ×L =20 ×60 = 1200.
Should we order 1200 units or more or less?
It depends on our service level.
Underage cost = Cu = p – c = 100 – 60 = 40.
Overage cost = Co = 60-0+10 =70
SL = Cu/(Cu+Co) = 40/(40+70) = 0.3636.
Due to high overage cost, SL*< 50%.
Z(0.3636) = ？
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
19
μ and σ of demand per period and fixed L
x: ( , σ)  y: (n , 𝑛 σ)
1
2
Sampling Distribution & Confidence Interval
3
4
Ardavan Asef-Vaziri
5
Jan.-2016
20
μ and σ of demand per period and fixed L
If Demand is variable and Lead time is fixed
L: Lead Time
R: Demand per period (per day, week, month)
R: Average Demand per period (day, week, month)
R: Standard deviation of demand (per period)
LTD: Average Demand During Lead Time
LTD = L × R
LTD: Standard deviation of demand during lead time
𝜎𝐿𝑇𝐷 = 𝐿𝜎𝑅
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
21
μ and σ of demand per period and fixed L
If demand is variable and Lead time is fixed
L: Lead Time = 5 days
R: Demand per day
R: Average daily demand =50
R: Standard deviation of daily demand =10
LTD: Average Demand During Lead Time
LTD = L × R = 5 × 50 = 250
LTD: Standard deviation of demand during lead time
𝜎𝐿𝑇𝐷 = 𝐿𝜎𝑅
𝜎𝐿𝑇𝐷 = 5 10 = 22.4  25
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
22
Now It is Transformed
The Problem originally was: If average demand per day
is 50 units and standard deviation of demand is 10
per day, and lead time is 5 days. Compute ROP at
90% service level. Compute safety stock.
We transformed it to: The average demand during the
lead time is 250 and the standard deviation of
demand during the lead time is 22.4. Compute ROP
at 90% service level. Compute safety stock.
=NORM.INV(0.9,250,22.4)
=278.7  279
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
23
Comparing the two problems
1
2
3
4
Sampling Distribution & Confidence Interval
5
1
Ardavan Asef-Vaziri
2
3
Jan.-2016
4
5
24
The News Vendor Problem- Example
Daily demand for your merchandise has mean of 20 and standard
deviation of 5. Sales price is $100 per unit of product.
You have decided to close this business line in 64 days. Your
supplier has also decided to close this line immediately, but has
agreed to provide your last order at a cost of $60 per unit. Any
unsold product will be disposed at cost of $10 per unit. How
many units do you order
Underage cost = Cu = = 100 – 60 = 40.
Overage cost = Co = 60 +10 =70
SL = Cu/(Cu+Co) = 40/(40+70) = 0.3636.
Due to high overage cost, SL*< 50%.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
25
The News Vendor Problem- Extended
LTD = R ×L =20 ×64 = 1280.
LTD = ( 𝐿)*(R)
LTD = ( 64)* (5) = 40
SL* = 0.3636
The optimal Q = LTD + z σLTD
=NORM.INV(probability, mean, standard_dev)
=NORM.INV(0.3636,1280,40)
=1266.0459
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
26
Sampling
 Element. Any unit of data defined for processing is a data




element; for example, ACCOUNT NUMBER, NAME,
ADDRESS and CITY. 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. A sampling frame is the source material or device from
which a sample is drawn. It is a list of all those within a
population who can be sampled, and may include individuals,
households or institutions 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.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
27
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.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
28
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.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
29
X: , , y = (x1+X2)/2, y = ,
2 y= 2/2,  y=  / 2
A random variable x has mean of  , and standard
deviation of σ. A random variable y is equal to the
average of 2 random variables x.
y = (x1+x2)/2
x: ( , σ)  y: (?,?)
If it was x1+x2 then x1+x2: (2 , 𝟐σ)
Since it is (x1+x2)/2 or 1/2(x1+x2): 1/2(2 , 𝟐σ)
That is:  , ( 𝟐/2)σ
That is:  , 𝟐/( 𝟐 𝟐)σ
y = (x1+x2)/2:  , σ/ 𝟐
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
30
20 Samples of size 25
100.0 Puplulation Mean
35.0 Population StdDev
Samples 1
1
80
2
148
3
91
4
112
21
145
22
75
23
94
2
40
55
96
79
42
101
195
3
66
118
106
91
74
108
111
4
43
146
87
94
76
50
139
5
81
112
110
51
67
136
69
6
160
65
136
87
86
114
113
7
60
46
114
121
121
182
53
8
99
83
51
119
119
58
101
9
121
103
104
149
143
106
145
10
131
88
140
68
15
99
118
11
40
108
87
28
128
98
147
12
99
85
58
92
53
123
117
13
68
124
66
78
92
118
56
14
111
80
107
63
106
111
101
15
85
116
117
141
113
86
81
16
94
97
71
111
36
96
86
17
108
69
102
76
172
62
119
18
110
63
111
102
128
101
72
19
93
94
133
71
167
164
184
20
154
89
144
80
67
115
118
24
63
81
154
126
53
58
104
129
52
135
110
68
130
88
121
90
59
75
97
97
25
188
52
83
129
141
89
102
113
179
48
126
53
124
117
112
81
143
106
100
123
96.5
33.3
96.6
33.9
96.0
33.8
97.8
31.4
106.9
40.9
88.2
35.1
106.8
36.8
96.2
36.1
104.3
37.8
84.4
20.9
97.4
43.7
97.5
28.8
104.7
38.2
95.0
28.2
Mean
StdDev
102.3 102.5
32.8 41.6
99.6 Mean (n=25)
6.1 StdDev( n=25)
106.0 105.1
33.6 26.6
Xbar(25)
10
16
22
28
34
40
46
52
58
64
70
76
82
88
94
100
106
112
118
124
130
136
142
148
154
160
166
172
178
184
x
Sampling Distribution & Confidence Interval
102.8 104.8
43.5 30.1
78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120
Ardavan Asef-Vaziri
Jan.-2016
31
Sampling from an Infinite Population
A random variable x1: ( and )
A random variable x2: ( and )
……………………………………..
A random variable xn: ( and )
A random variable y = x1+x2+…..+xn: (?,?)
Mean(y) = Mean(x1)+ Mean(x2)+ ……. + Mean(xn)
(y) = (x1)+ (x2)+ ……. + (xn)
 (y) = n (x)
Var(y) = Var(x1)+ Var(x2)+ ……. Var(xn)
2(y) = 2 + 2 + ……. 2
2(y) = n 2 (x)
(y) =
𝑛 (x)
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
32
Sampling from an Infinite Population
A random Variable y= SUMx: (n , 𝑛 )
A random Variable SUMx/n = 𝑋: (?, ?)
A random Variable SUMx: (n, 𝑛)
A random Variable 𝑋 = SUMx/n = (1/n) SUMx
Mean (𝑋) = n/n = 
StdDev (𝑋) = (1/n)* 𝑛  = / 𝑛
x: ( and )
𝑿: (, / 𝒏)
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
33
Standard Deviation of 𝑋 for N=25 and N=50
Samples 1
1
80
2
148
3
91
4
112
21
145
22
75
23
94
2
40
55
96
79
42
101
195
3
66
118
106
91
74
108
111
4
43
146
87
94
76
50
139
5
81
112
110
51
67
136
69
6
160
65
136
87
86
114
113
7
60
46
114
121
121
182
53
8
99
83
51
119
119
58
101
9
121
103
104
149
143
106
145
10
131
88
140
68
15
99
118
11
40
108
87
28
128
98
147
12
99
85
58
92
53
123
117
13
68
124
66
78
92
118
56
14
111
80
107
63
106
111
101
15
85
116
117
141
113
86
81
16
94
97
71
111
36
96
86
17
108
69
102
76
172
62
119
18
110
63
111
102
128
101
72
19
93
94
133
71
167
164
184
20
154
89
144
80
67
115
118
24
63
81
154
126
53
58
104
129
52
135
110
68
130
88
121
90
59
75
97
97
25
188
52
83
129
141
89
102
113
179
48
126
53
124
117
112
81
143
106
100
123
96.5
33.3
96.6
33.9
96.0
33.8
97.8
31.4
106.9
40.9
88.2
35.1
106.8
36.8
96.2
36.1
104.3
37.8
84.4
20.9
97.4
43.7
106.0 105.1
33.6 26.6
97.5
28.8
104.7
38.2
95.0
28.2
102.8 104.8
43.5 30.1
Mean
StdDev
102.3 102.5
32.8 41.6
Mean
102.4
96.6
96.9
97.5
101.5
94.4
101.7
101.3
99.8
103.8
StdDev
37.1
33.3
32.3
38.9
36.5
31.8
38.8
27.7
33.6
37.0
100.0 Puplulation Mean
35.0 Population StdDev
99.6 Mean (n=25)
6.1 StdDev( n=25)
2.9
StdDev (n»50)
Sampling Distribution & Confidence Interval
1
4
7
10
13
16
19
22
25
28
31
34
37
40
43
46
49
52
55
58
61
64
67
70
73
76
79
82
85
88
91
94
97
100
103
106
109
112
115
118
121
124
127
130
133
136
139
142
145
148
151
154
157
160
163
166
169
172
175
178
99.6 Mean (n50 )
x
Ardavan Asef-Vaziri
Xbar(25)
Jan.-2016
Xbar(50)
34
Sampling from an Infinite Population
CENTRAL LIMIT THEOREM: In selecting random samples of
size n from a population, the sampling distribution of the
sample mean 𝑋 can be approximated by a normal distribution
as the sample size becomes large.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
35
Interval Estimation vs Point Estimation
𝑋 provides a point estimate for 
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 𝑋 , where none of the
𝑋𝑠 is equal to . But they are all unbiased estimates of .
Unbiased means their expected value is equal to .
x
 Point estimate;  = x
 Interval Estimate ;
with a certain
confidence
(probability)
x a μ  x a
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
36
𝑋 Distribution: (1-) Probability / 𝒏
35/sqrt(25)=7
x
78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
37
Relationship between μ and x
1-α
0.95
X    z(10.05 / 2 ) X
X    z(1 / 2 ) X
X    z(10.05 / 2 ) X
X    z(1 / 2 ) X
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
38
Relationship between μ and x
X    z(1 / 2 ) X
X    z(1 / 2 ) X
X  z(1 / 2 ) X   X  z(1 / 2 ) X  
X  z(1 / 2 ) X    X  z(1 / 2 ) X
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
39
Relationship between μ and x
X  z(1 / 2 ) X    X  z(1 / 2 ) X
X  z(10.05 / 2 ) X    X  z(10.05 / 2 ) X
X  z0.975 X    X  z0.975 X
X  1.96 X    X  1.96 X
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
40
𝑋 Distribution
35/sqrt(25)=7
x
78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
41
Interval Estimation of a Population Mean:  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  = 25
We want to find an interval estimate for 
This month interview scors are given below
Find the interval estimate for  with 95% confidence interval.
102
86
97
56
140
106
96
74
132
105
116
92
81
76
75
106
115
73
77
122
79
111
92
89
73
77
102
91
68
102
70
112
80
108
87
47
89
67
111
121
94
115
89
89
125
84
83
88
79
73
80
111
86
82
37
63
109
102
50
77
83
54
139
32
116
70
70
37
62
101
68
125
73
123
37
62
49
131
50
80
Sampling Distribution & Confidence Interval
87
84
92
29
91
122
126
86
64
25
75
109
78
53
98
56
160
57
68
55
Xbar
StdDevXbar
Z0.975
UCL
LCL
Ardavan Asef-Vaziri
Jan.-2016
78.69
2.5
1.96
83.58991
73.79009
TRUE
42
Example: National Discount, Inc.
National Discount has 260 retail outlets throughout the U.S.
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  = $5,000. We will use a sample size of n = 64.
Question. There is a 0.95 probability that the value of a sample
mean for National Discount will provide a sampling error of
$????? or less.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
43
Example: National Discount, Inc.
𝜎𝑋 =
𝜎
𝑛
=
5000
64
= 626
Z0.975 = 1.959964 
(Z0.975)(𝜎𝑋 ) = 1.96(625) = 1225
Answer. There is a 0.95 probability that the value of a sample
mean for National Discount will provide a sampling error of
$1,225 or less.
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
44
Example: National Discount, Inc.
Question. 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 (n) is
needed to meet the required precision? Recall that  = 5,000.
(Z0.975)(𝜎𝑋 ) = 1.96(𝜎𝑋 ) ≤ 500
𝜎𝑋 ≤ 500/1.96
𝜎𝑋 ≤ 500/1.96 = 255.2
𝜎𝑋 =
5000
𝑛
𝜎
𝑛
≤ 255.2
≤ 255.2
5000
≤ 𝑛
255.2
n ≥ 385
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
45
2-Ways to Compute CI in Excel:  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
22
959.40
0.975
1.96
-1.96
1880
-1880
42980
39220
Sampling Distribution & Confidence Interval
Using X
Xbar=
SigmaX=
CI=
n=
Sigma(Xbar)=
Upper Probability
Lower Probability
Ardavan Asef-Vaziri
NORM.INV
41100
4500
0.95
40
711.5124735
0.975
0.025
UCL 42495
LCL 39705
Jan.-2016
46
Third Way to Compute CI in Excel:  is known
Using Conf CONFIDENCE.NORM
Xbar=
41100
SigmaX=
4500
CI=
0.95
n=
34
Mrgin
UCL
LCL
Sampling Distribution & Confidence Interval
1513
42613
39587
Ardavan Asef-Vaziri
Jan.-2016
47
Interval Estimation of a Population Mean:  is
unknown
 Instead of population standard deviation , 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
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
48
-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
N:fx
T-fx
N:fx
Sampling Distribution & Confidence Interval
T-fx
Ardavan Asef-Vaziri
Jan.-2016
49
Interval Estimation of a Population Mean: 
Unknown
 The interval estimate is given by:
x  t / 2
s
n
The confidence level is 1 -
t/2 is the t value providing an area of /2 in the upper tail of a t
distribution with n - 1 degrees of freedom
s is the sample standard deviation
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
50
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
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
51
Sampling from an Infinite Population
A population has a mean of 300 and standard deviation of 60.
Given a random sample size of 100, what this the probability that
the sample mean will be within +9 of population mean.
-9 ≤ 𝑋-µ ≤ 9
|𝑋-µ| ≤ 9
|𝑋-µ|/𝑋 = 9/𝑋
𝑋 = σ/ 100  𝑋 = 60/10 = 6.
z = 9/6 = 1.5
=NORM.S.DIST(1.5,1)- =NORM.S.DIST(-1.5,1) =
=0.933193 -0.066807 = 86.6%
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
52
Sampling from an Infinite Population
What sample size do we need to have such that with 90%
confidence the margin of error is within +9 from mean.
Margin of Error = z(CL)*σ/ 𝑛
z(90%) =NORM.S.INV(0.95) = 1.644854
Margin of Error = 1.644854 *60/ 𝑛 = 98.7/ 𝑛
98.7/ 𝑛 ≤ 9
98.7/9 ≤ 𝑛
10.966 ≤ 𝑛
121 ≤ n
Sampling Distribution & Confidence Interval
Ardavan Asef-Vaziri
Jan.-2016
53