Chapter 7 2022
Chapter 7 2022
Chapter 7 2022
Chapter 7
Fundamental Sampling
Distributions
Population
6
5
4
3
2
1
17 18 19 20 21 22 23 24 25
2 V(X) E(x 2
) [ E ( x)]2 446 212 5
2.236068 2.24
Dr. Yehya Mesalam 3
Distribution of Sample Means from Samples of Size n = 2
Sample # Scores Mean ( X )
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
Dr. Yehya Mesalam 4
Distribution of Sample Means from Samples of Size n = 2
μ x E( X) x.f(x) 21
x
μ x 21
2
x V( X) E( x ) [ E ( x )] 443.5 21 2.5
2 2 2
x 1.581139
2.236068
X 1.581139
n 2
Population
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9
2 V(X) E(x 2
) [ E ( x)]2 30 25 5
2.236068 2.24
Dr. Yehya Mesalam 8
Distribution of Sample Means from Samples of Size n = 2
Sample # Scores Mean ( X )
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
Dr. Yehya Mesalam 9
Distribution of Sample Means from Samples of Size n = 2
μ x E( X) x.f(x) 5
x
μx 5
2
x V( X) E( x ) [ E ( x )] 27.5 25 2.5
2 2
x 1.581139
2.236068
X 1.581139
n 2
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9
sample mean
6
= 5, = 2.24
5 Distribution of Sample Means
4
3
2
6 X = 5, X = 1.58
1
5
1 2 3 4 5 6 7 8 9
4
3
2
1
1 2 3 4 5 6 7 8 9
sample mean
6
= 5, = 2.24
5
4 Distribution of Sample Means
3
2
6 X = 5, X = 1.58
1
5
1 2 3 4 5 6 7 8 9 2.24
4
X 1.58
3 2
2
1
1 2 3 4 5 6 7 8 9
sample mean
24
22
X = 5
20 X = 1.29
18
16
14
12 2.24
X 1.29
10 3
8
6
4
2
1 2 3 4 5 6 7 8 9
sample mean
Dr. Yehya Mesalam 15
Distribution of Sample Means
6
5 Things to Notice
4
3 1. The sample means tend to pile
2 up around the population mean.
1
x μ
z 3. The distribution of sample means
σ has less variability than does the
n population distribution.
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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
n
3. will approach a normal distribution as n approaches
infinity
The mean of the sampling distribution
X
The standard deviation of sampling distribution
(“standard error of the mean”)
X
n
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Clarifying Formulas
Distribution of
Population Sample Sample Means
X X
X X
n
N
ss
s
ss X
N n 1 n
notice
2
2
X
n
1 .95
α α
.025 .025
2 2
We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms
E x μ
z α/2 . 2
n [ ]
E
Standard
Normal
(t with df = ∞)
t (df = 13)
t-distributions are bell-
shaped and symmetric, but
have ‘fatter’ tails than the t (df = 5)
normal
0 t
Dr. Yehya Mesalam 31
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 t α/2, t 24,.025 2.0639
The confidence interval is
S S
x t α/2, μ x t α/2,
n n
8 8
50 (2.0639) μ 50 (2.0639)
25 25
46.698 μ 53.302
Dr. Yehya Mesalam 34
Confidence Intervals for the Population
Variance
(n 1)s2 (n 1)s 2
σ 2
2
χ α/2,
2
χ1 - α/2,
probability probability
α/2 = .025 α/2 = .025
216
216 = 6.91 216 = 28.85
Dr. Yehya Mesalam 38
Solution
(n 1)s 2 (n 1)s 2
σ 2
χ α/2,
2
χ12- α/2,
2 2
16 * 0.78 16 * 0.78
σ
2
28.845 6.908
0.3374 σ 1.40912
σ12 σ 22 σ12 σ 22
(x1 x 2 ) z α/2 μ1 μ 2 (x1 x 2 ) z α/2
n1 n 2 n1 n 2
σ12 and σ22 Unknown and n1+n2 >30 use Z
The confidence interval for μ1 – μ2 is:
s12 s 22 s12 s 22
(x1 x 2 ) z α/2 μ1 μ 2 (x1 x 2 ) z α/2
n1 n 2 n1 n 2
Dr. Yehya Mesalam 45
Confidence Interval between (Two Means)
σ12 and σ22 Unknown and n1+n2 <=30 use t
(x1 x 2 ) (μ1 μ 2 )
t
1 1
Sp
n1 n 2
The confidence interval for μ1 – μ2 is:
1 1 1 1
(x1 x 2 ) t α/2, .s p μ1 μ 2 (x1 x 2 ) t α/2, .s p
n1 n 2 n1 n 2
Where
CPU1 CPU2
Number Tested 16 13
Sample mean 3004 2538
Sample std dev 74 56
Sp
n1 1S12 n 2 1S2 2
16 1742 13 1562 66.537
(n1 n 2 2) (16 13 2)
1 1 1 1
(x1 x 2 ) t α/2, .s p μ1 μ 2 (x1 x 2 ) t α/2, .s p
n1 n 2 n1 n 2
1 1 1 1
(3004 2538) (2.052) * 66.537 μ1 μ 2 (3004 2538) (2.052) * 66.537
16 13 16 13
416.69 μ1 μ 2 515.31
σ12 σ 22 σ12 σ 22
(x1 x 2 ) z α/2 μ1 μ 2 (x1 x 2 ) z α/2
n1 n 2 n1 n 2
- 7.85885 μ1 μ 2 1.858845
s 2p s 2p s 2p s 2p
(x1 x 2 ) t , α/2 μ1 μ1 (x1 x 2 ) t , α/2
n1 n 2 n1 n 2
sp 5.813
11 11 2
1 1 1 1
(72 75) 2.086 * 5.813 μ1 μ 2 (72 75) 2.086 * 5.813
11 11 11 11
- 8.1705 μ1 μ 2 2.1705