Chapter 7 2022.pdf

Probability and Statistics
Chapter 7
Fundamental Sampling
Distributions
Dr. Yehya Mesalam 1

Population
17 18 19 20 21 22
1
2
3
4
5
6
23 24 25
Sampling Distribution of the Sample Mean
The sampling distribution of the sample mean is the probability
distribution of the population of the sample means obtainable
from all possible samples of size n from a population of size N
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DistributionofSampleMeansfromSamplesofSizen=2
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x f(x) x.f(x) x2.f(x)
18 0.25 4.5 81
20 0.25 5 100
22 0.25 5.5 121
24 0.25 6 144
sum 1 21 446
21
P(x)
x
E(X)
μ
x


 
(x
V(X) 2
2
5
21
446
)]
(
[
) 2
2





 x
E
E

2.24
2.236068



1 18, 18 18
2 18, 20 19
3 18, 22 20
4 18, 24 21
5 20, 18 19
6 20, 20 20
7 20, 22 21
8 20, 24 22
9 22, 18 20
10 22, 20 21
11 22, 22 22
12 22, 24 23
13 24, 18 21
14 24, 20 22
15 24, 22 23
16 24, 24 24
Sample # Scores Mean ( )
X
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𝒙 f f(x) 𝒙.f(x) 𝒙2.f(x)
18 1 0.0625 1.125 20.25
19 2 0.125 2.375 45.125
20 3 0.1875 3.75 75
21 4 0.25 5.25 110.25
22 3 0.1875 4.125 90.75
23 2 0.125 2.875 66.125
24 1 0.0625 1.5 36
sum 16 1 21 443.5

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21
f(x)
.
x
)
X
E(
μ
x
x 

 
x
(
)
X
V( 2
2
5
.
2
21
5
.
443
)]
(
[
) 2
2





 x
E
E
x

1.581139

x

1.581139
2
2.236068



n
X


21
μx 
 

Population
1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
Sampling Distribution of the Sample Mean
The sampling distribution of the sample mean is the probability
distribution of the population of the sample means obtainable
from all possible samples of size n from a population of size N
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x f(x) x.f(x) x2.f(x)
2 0.25 0.5 1
4 0.25 1 4
6 0.25 1.5 9
8 0.25 2 16
sum 1 5 30
5
P(x)
x
E(X)
μ
x


 
(x
V(X) 2
2
5
25
30
)]
(
[
) 2





 x
E
E

2.24
2.236068



1 2, 2 2
2 2,4 3
3 2,6 4
4 2,8 5
5 4,2 3
6 4,4 4
7 4,6 5
8 4,8 6
9 6,2 4
10 6,4 5
11 6,6 6
12 6,8 7
13 8,2 5
14 8,4 6
15 8.6 7
16 8.8 8
Sample # Scores Mean ( )
X
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𝒙 f f(x) 𝒙.f(x) 𝒙2.f(x)
2 1 0.0625 0.125 0.25
3 2 0.125 0.375 1.125
4 3 0.1875 0.75 3
5 4 0.25 1.25 6.25
6 3 0.1875 1.125 6.75
7 2 0.125 0.875 6.125
8 1 0.0625 0.5 4
sum 16 1 5 27.5

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5
f(x)
.
x
)
X
E(
μ
x
x 

 
x
(
)
X
V( 2
2
5
.
2
25
5
.
27
)]
(
[
) 2





 x
E
E
x

1.581139

x

1.581139
2
2.236068



n
X


5
μx 
 

1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
sample mean
We can use the distribution of sample means to
answer probability questions about sample
means
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1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
Distribution of Individuals in Population
Distribution of Sample Means
1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
sample mean
 = 5,  = 2.24
X = 5, X = 1.58
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1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
Distribution of Individuals in Population
1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
sample mean
 = 5,  = 2.24
X = 5, X = 1.58
58
.
1
2
24
.
2


X

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1 2 3 4 5 6
2
4
6
8
10
12
7 8 9
sample mean
14
16
18
20
22
24
Sampling Distribution (n = 3)
X = 5
X = 1.29
29
.
1
3
24
.
2


X

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1 2 3 4 5 6
1
2
3
4
5
6
7 8 9
sample mean
Things to Notice
1. The sample means tend to pile
up around the population mean.
2. The distribution of sample means
is approximately normal in
shape, even though the
population distribution was not.
3. The distribution of sample means
has less variability than does the
population distribution.
16
n
σ
μ
x
z


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Central Limit Theorem
For any population with mean  and standard deviation ,
the distribution of sample means for sample size n …
1. will have a mean of 
2. will have a standard deviation of
3. will approach a normal distribution as n approaches
infinity

n
 The mean of the sampling distribution
 The standard deviation of sampling distribution
(“standard error of the mean”)

 
X
n
X

 
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Population Sample
Distribution of
Sample Means
Clarifying Formulas
N
X


 n
X
X

 
 
X
N
ss


1


n
ss
s n
X

 
n
X
2
2 
 
notice
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Confidence Level, (1-)
• Suppose confidence level = 95%
• Also written (1 - ) = 0.95
• A relative frequency interpretation:
– From repeated samples, 95% of all the
confidence intervals that can be constructed
will contain the unknown true parameter
• A specific interval either will contain or
will not contain the true parameter
– No probability involved in a specific interval
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Confidence Interval for μ
– Population variance σ2 is known use Z
• Confidence interval estimate:
(where z/2 is the normal distribution value for a probability of /2 in each
tail)
n
σ
z
x
μ
n
σ
z
x α/2
α/2 



n
σ
μ
x
z


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Finding the Reliability Factor, z/2
• Consider a 95% confidence interval:
z = -1.96 z = 1.96
.95
1 


.025
2
α
 .025
2
α

Point Estimate
Lower
Confidence
Limit
Upper
Confidence
Limit
Z units:
X units: Point Estimate
0
 Find z.025 = 1.96 from the standard normal distribution table
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Common Levels of Confidence
• Commonly used confidence levels are 90%,
95%, and 99%
Confidence
Level
Confidence
Coefficient, Z/2 value
1.28
1.645
1.96
2.33
2.58
3.08
3.27
.80
.90
.95
.98
.99
.998
.999
80%
90%
95%
98%
99%
99.8%
99.9%


1
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Example
• A sample of 11 circuits from a large normal population has a
mean resistance of 2.20 ohms. We know from past testing that
the population standard deviation is 0.35 ohms. Determine a
95% confidence interval for the true mean resistance of the
population.
• Solution:
2.4068
μ
1.9932
.2068
2.20
)
11
(.35/
1.96
2.20
n
σ
z
x







We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms
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Confidence Interval for μ
• If the population standard deviation σ is
unknown, and n>30 use Z
• N<=30 use t distribution
n
s
z
x
μ
n
s
z
x α/2
α/2 



n
S
t
x
μ
n
S
t
x α/2,
α/2, 
 



where tα/2,n-1 is the critical value of the t distribution with n-1 d.f. and an
area of α/2 in each tail:
n
s
μ
x
z


n
s
μ
x
t


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Choice of Sample Size
• To Calculate the sample size needed for (1-α )
is
• Where E the error
2
α/2
]
.
z
[
E
n


μ
x
E 

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Example
• Assuming the population standard deviation =
3, how large should a sample be to estimate the
population mean with a margin of error not
exceeding 0.5?
2
α/2
]
.
z
[
E
n


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Solution
• where = 0.05
• Then from table
• Error =E = 0.5
• Then
• n= [ 1.96*3 / 0.5]2 = 138.3
• we need a sample of size at least 139
2
α/2
]
.
z
[
E
n


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96
.
1
z 25
.
0
α/2 
 o
z

Student’s t Distribution
• Consider a random sample of n observations
– with mean x and standard deviation s
– from a normally distributed population with mean μ
• Then the variable
follows the Student’s t distribution with (n - 1) degrees of
freedom
n
s/
μ
x
t


d.f. = n - 1
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Student’s t Distribution
t
0
t (df = 5)
t (df = 13)
t-distributions are bell-
shaped and symmetric, but
have ‘fatter’ tails than the
normal
Standard
Normal
(t with df = ∞)
Note: t Z as n increases
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Example
A random sample of n = 25 has x = 50 and
s = 8. Form a 95% confidence interval for μ
• Solution
d.f. = n – 1 = 24, so
The confidence interval is
2.0639
t
t 24,.025
α/2, 


53.302
μ
46.698
25
8
(2.0639)
50
μ
25
8
(2.0639)
50
n
S
t
x
μ
n
S
t
x α/2,
α/2,









 

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Confidence Intervals for the Population
Variance
The random variable
2
2
2
1
n
σ
1)s
(n



follows a chi-square distribution with (n – 1) degrees of
freedom
Where the chi-square value denotes the number
for which
2
,
1
n 
 
α
χ
χ 
 
 )
P( 2
α
,
1
n
2
1
n
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Confidence Intervals for the Population
Variance
The (1 - )% confidence interval for the population
variance is
2
/2,
-
1
2
2
2
/2,
2
1)s
(n
σ
1)s
(n

 α
α χ
χ




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Example
You are testing the speed of a batch of computer
processors. You collect the following data (in Mhz):
Sample size 17
Sample mean 3004
Sample std dev 74
Assume the population is normal.
Determine the 95% confidence interval for σ2
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Solution
• n = 17 so the chi-square distribution has (n – 1) =
16 degrees of freedom
•  = 0.05, so use the chi-square values with area
0.025 in each tail:
probability
α/2 = .025
2
16
2
16
= 28.85
6.91
28.85
2
0.975,16
2
/2,
-
1
2
0.025,16
2
/2,




χ
χ
χ
χ
α
α


2
16 = 6.91
probability
α/2 = .025
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Solution
• The 95% confidence interval is
Converting to standard deviation, we are 95% confident
that the population standard deviation of CPU speed is
between 55.1 and 112.6 Mhz
2
/2,
-
1
2
2
2
/2,
2
1)s
(n
σ
1)s
(n

 α
α χ
χ




6.91
1)(74)
(17
σ
28.85
1)(74)
(17 2
2
2




12683
σ
3037 2


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Example
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.
The lapping process which is used to grind certain
silicon wafers to the proper thickness is acceptable only
if the population standard deviation of the thickness of
dice cut from the wafers is at most 0.50 mil. If the
thicknesses of 17 dice cut from such wafers have a
standard deviation of 0.78 mil. Find 95% confidence
limits on .

Solution
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.
2
/2,
-
1
2
2
2
/2,
2
1)s
(n
σ
1)s
(n

 α
α χ
χ




6.908
0.78
*
6
1
σ
845
.
28
0.78
*
6
1 2
2
2


4091
.
1
σ
3374
.
0 2



ConfidenceIntervalbetween(TwoMeans)
σ1
2 and σ2
2 known use Z
2
2
2
1
2
1
α/2
2
1
2
1
2
2
2
1
2
1
α/2
2
1
n
σ
n
σ
z
)
x
x
(
μ
μ
n
σ
n
σ
z
)
x
x
( 








σ1
2 and σ2
2 Unknown and n1+n2 >30 use Z
The confidence interval for μ1 – μ2 is:
2
2
2
1
2
1
α/2
2
1
2
1
2
2
2
1
2
1
α/2
2
1
n
s
n
s
z
)
x
x
(
μ
μ
n
s
n
s
z
)
x
x
( 








2
2
2
1
2
1
2
1
2
1
n
σ
n
σ
)
μ
(μ
)
x
x
(
Z





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Where
2
n
n
1)s
(n
1)s
(n
s
2
1
2
2
2
2
1
1
p






Is the pooled variance
σ1
2 and σ2
2 Unknown and n1+n2 <=30 use t
2
1
2
1
2
1
n
1
n
1
)
μ
(μ
)
x
x
(
t





p
S
46
ConfidenceIntervalbetween(TwoMeans)
Dr. Yehya Mesalam
2
1
p
α/2,
2
1
2
1
2
1
p
α/2,
2
1
n
1
n
1
s
.
t
)
x
x
(
μ
μ
n
1
n
1
.s
t
)
x
x
( 







 


Example
You are testing two computer processors for speed.
Form a confidence interval for the difference in CPU
speed. You collect the following speed data (in Mhz):
CPU1 CPU2
Number Tested 16 13
Sample mean 3004 2538
Sample std dev 74 56
Assume both populations are normal with
equal variances, and use 95% confidence
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Solution
        4427.03
2)
13
(16
56
1
13
74
1
16
2)
n
(n
S
1
n
S
1
n
S
2
2
2
1
2
2
2
2
1
1
2
p 












The pooled variance is:
The t value for a 95% confidence interval is:
2.052
t
t 0.025,27
α/2, 


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        537
.
6
6
2)
13
(16
56
1
13
74
1
16
2)
n
(n
S
1
n
S
1
n
S
2
2
2
1
2
2
2
2
1
1
p 













Solution
• The 95% confidence interval is
13
1
16
1
66.537
*
(2.052)
2538)
(3004
μ
μ
13
1
16
1
66.537
*
(2.052)
2538)
(3004 2
1 








515.31
μ
μ
416.69 2
1 


We are 95% confident that the mean difference in CPU
speed is between 416.69 and 515.31 Mhz.
2
1
p
α/2,
2
1
2
1
2
1
p
α/2,
2
1
n
1
n
1
s
.
t
)
x
x
(
μ
μ
n
1
n
1
.s
t
)
x
x
( 







 

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Example
50
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Method 1 71 75 65 69 73 66 69 75 74 87 68
Method 2 72 77 84 78 69 70 77 81 65 77 75
As part of an industrial training program, some trainees are
instructed by Method 1, which is straight teaching-machine
instruction, and some are instructed by Method 2, which
also involves the personal attention of an instructor. If
random samples are taken from large groups of trainees
instructed by each of these two methods and the scores
with standard deviation are 6.06, and 5.58 respectively;
The score obtained in an appropriate achievement test
are
  %
100
1 

2
1 
 
•Use the 0.05 level of significance to find
confidence limits on
.

Solution
51
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.
SX SX2 Mean Variance S.D
A 792 52351 72 36.8 6.0663
B 825 62183 75 30.8 5.549775
2
2
2
1
2
1
α/2
2
1
2
1
2
2
2
1
2
1
α/2
2
1
n
σ
n
σ
z
)
x
x
(
μ
μ
n
σ
n
σ
z
)
x
x
( 








11
58
.
5
1
1
6.06
1.96
75)
(72
μ
μ
11
58
.
5
1
1
6.06
1.96
75)
(72
2
2
2
1
2
2









1.858845
μ
μ
7.85885
- 2
1 



Example
52
Dr. Yehya Mesalam
Method 1 71 75 65 69 73 66 69 75 74 87 68
Method 2 72 77 84 78 69 70 77 81 65 77 75
As part of an industrial training program, some trainees
are instructed by Method 1, which is straight teaching-
machine instruction, and some are instructed by Method
2, which also involves the personal attention of an
instructor. If random samples are taken from large
groups of trainees instructed by each of these two
methods and the scores which they obtained in an
appropriate achievement test are
  %
100
1 

2
1 
 
•Use the 0.05 level of significance to find
confidence limits on
.

Solution
53
Dr. Yehya Mesalam
.
SX SX2 Mean Variance S.D
A 792 52351 72 36.8 6.0663
B 825 62183 75 30.8 5.549775
2
2
p
1
2
p
α/2
,
2
1
1
1
2
2
p
1
2
p
α/2
,
2
1
n
s
n
s
t
)
x
x
(
μ
μ
n
s
n
s
t
)
x
x
( 







 

2
n
n
1)s
(n
1)s
(n
s
2
1
2
2
2
2
1
1
p






813
.
5
2
1
1
1
1
5.5497
*
0
1
6.0663
*
10
s
2
2
p 





Solution
54
Dr. Yehya Mesalam
.
11
1
1
1
1
813
.
5
*
086
.
2
75)
(72
μ
μ
11
1
1
1
1
813
.
5
*
086
.
2
75)
(72 2
1 








1705
.
2
μ
μ
8.1705
- 2
1 



Chapter 7 2022.pdf

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Chapter 7 2022.pdf