Transcript Ch 8.1 Distribution of the Sample Mean Objective A : Shape, Center

Modular 13
Ch 8.1 to 8.2
Ch 8.1 Distribution of the Sample Mean
Objective A : Shape, Center, and Spread of the Distribution
of x
Objective B : Finding Probability of x that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective A : Shape, Center, and Spread of the Distribution
of p̂
Objective B : Finding Probability of p̂ that is Normally
Distributed
Ch 8.1 Distribution of the Sample Mean
Objective A : Shape, Center, and Spread of the Distribution
of x
A1. Sampling Distributions of Mean
Assume equal chances for each number to be selected.
S  {1, 2, 3}
1
1
1
P(1)  , P ( 2)  , P (3) 
3
3
3
P (x )
1
3
This population distribution of
x is normally distributed.
1
2
3
x
Sampling Probability distribution of mean x .
Let’s say we select two elements (n  2 ) from S  {1, 2, 3} with
replacement. (Independent case)
List out all possible combinations (sample space) and x for each
combination.
1
2
3
1
11
1 2
1 3
(
1
,
3
)
1
 1.5
x
2
(1, 2) x 
2 (1,1) x 
2
2
2
3
1
22
2 1
23
 2 (2, 3) x 
 1.5 (2, 2) x 
 2.5
2 ( 2,1) x 
2
2
2
3
1
2 (3,1) x  3  1  2 (3, 2) x  3  2  2.5 (3, 3) x  3  3  3
2
2
2
3
Probability distribution of x is summarized in the table shown below.
x
1
1.5
2
2.5
3
P (x )
1
9
2
9
3
9
2
9
1
9
Probability histogram for x .
P (x )
39
29
19
1
1.5
2 2.5
3
x
Let’s compare the distribution shape of
x and x .
P (x )
The population distribution of
uniformly distributed .
1
3
1
2
3
x is
x
P (x )
The sampling distribution of x is
normally distributed.
39
29
19
1 1.5 2 2.5 3
Is this by chance that
x
x is normally distributed?
A2. Central Limit Theorem
A. If the population distribution of x is normally distributed, the
sampling distribution of x is normally distributed regardless
of the sample size n .
If the population distribution x is not normally distributed, the
sampling distribution of x is guaranteed to be normally distributed
if n  30 .
B. Mean/standard deviation of a sampling distribution of x vs
mean/standard deviation of a population distribution of x .
The mean and standard deviation of population distribution
are  and  respectively.
The mean of the sampling distribution of x is  x .
x  
The standard deviation of the sampling distribution of x is  x .
x 

n
Example 1 : Determine  x and  x from the given parameters of
the population and the sample size.
  27,   6, n  15
 x    27
x 

6

 1.549
n
15
Example 2 : A simple random sample of is obtained from a
population with   64 and   18 .
(a) If the population distribution is skewed to the right,
what condition must be applied in order to guarantee
the sampling distribution of x is normally distributed?
Since the population distribution is not normally
distributed, the selected sample size must be greater
than or equal to 30 (i.e n  30 ).
(b) If the sample size is n  9, what must be true regarding
the distribution of the population in order to guarantee
the sampling distribution of x to be normally
distributed?
For small sample size, the population distribution must
be normally distributed in order to guarantee
the sampling distribution of x to be normally
distributed.
Ch 8.1 Distribution of the Sample Mean
Objective A : Shape, Center, and Spread of the Distribution
of x
Objective B : Finding Probability of x that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective A : Shape, Center, and Spread of the Distribution
of p̂
Objective B : Finding Probability of p̂ that is Normally
Distributed
Ch 8.1 Distribution of the Sample Mean
Objective B : Finding Probability of x that is Normally
Distributed
Standardize x to Z
Recall : Standardize x to Z : Z 
Now : Standardize x to Z :
Z
x

x  x
x
Example 1 : A simple random sample of size n  36 is obtained from
a population mean   64 and population standard
deviation   18 .
(a) Describe the sampling distribution x .
Since n  30 , x is normally distributed.
(b) What is P( x  62.6) ?
 x    64
Z
x  x
x
x 

18

3
n
36
62.6  64  1.4


 0.47
3
3
P( x  62.6)  P( Z  0.47)
P(Z  0.47)  0.3192
From Table V
0.07
0.3192
 0.4
 0.47
0
Z
0.3192
Example 2: The upper leg of 20 to 29 year old males is normally
distributed with a mean length of 43.7cm and a standard
deviation of 4.2cm.
(a) What is the probability that a random sample of 12 males who are
20 to 29 years old results in a mean upper leg length that is
between 42cm and 48cm?
Population is normally distributed.
Since the population distribution is normally distributed, x is
normally distributed for any sample size.
P(42  x  48)   43.7   4.2
 x    43.7
x 

n  12
4.2

 1.2124
n
12
x  42
Z
x  48
x  x
Z
x
42  43.7
 1.7


 1.40
1.2124
1.2124
x  x
48  43.7
4.3


 3.55
1.2124
1.2124
P(1.40  Z  3.55)
From Table V
1.40  0.0808
3.55  0.9998
 0.9998  0.0808
 0.919 ( the whole green area )
x

0.9998
0.0808
 1.40
0
3.55 Z
(b) A random sample of 15 males who are 20 to 29 years old results in
a mean upper leg length greater than 46 cm. Do you find the
result unusual? Why?
In order to know if it is unusual or not, we need to find the
probability.
P( x  46) n  15   43.7   4.2
 x    43.7
Z
x  x
x
4.2

 1.0844
n
15
46  43.7

 2.12
1.0844
P( Z  2.12)
From Table V
2.12  0.9830
 1 0.9830
 0.017
x 


0.017
0.9830
0
2.12
Since 0.017 is smaller than 0.05, this result is unusual.
Z
Ch 8.1 Distribution of the Sample Mean
Objective A : Shape, Center, and Spread of the Distribution
of x
Objective B : Finding Probability of x that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective A : Shape, Center, and Spread of the Distribution
of p̂
Objective B : Finding Probability of p̂ that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective A : Shape, Center, and Spread of the Distribution
of p̂
Distribution of the Sample Proportions
x
A. Sampling distribution of sample proportion p̂ , where pˆ  .
n
The shape of the sampling distribution of p̂ is approximately
normally provided by,
np (1  p )  10
or npq  10,
where q  1  p
B. Finding the mean and standard deviation of p̂
 pˆ  p
 pˆ 
p(1  p)
n
Ch 8.1 Distribution of the Sample Mean
Objective A : Shape, Center, and Spread of the Distribution
of x
Objective B : Finding Probability of x that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective A : Shape, Center, and Spread of the Distribution
of p̂
Objective B : Finding Probability of p̂ that is Normally
Distributed
Ch 8.2 Distribution of the Sample Proportion
Objective B : Finding Probability of p̂ that is Normally
Distributed
Standardize p̂ to Z
Z
pˆ   pˆ
 pˆ
where  pˆ  p and  pˆ 
p(1  p)
n
provided p̂ is approximately normally distributed.
Example 1: A nationwide study indicated that 80% of college students
who use a cell phone, send and receive text messages on
their phone. A simple random sample of n  200 college
students using a cell phone is obtained.
(a) Describe sampling distribution of p̂ .
p  0.8, n  200
Check to see if np (1  p )  10
(200)(0.8)(1  0.8)  10
(200)(0.8)(0.2)  10
32  10
Since npq  10 , p̂ is normally distributed.
(b) What is the probability that 154 or fewer college students in the
sample send and receive text messages on the cell phone? Is this
unusual?
x 154
 0.77
Sample proportion : pˆ  
Looking for probability :
n 200
P( pˆ  0.77)
Standardize p̂ to Z
p(1  p)
(0.8)(1  0.8)

 0.028284
 pˆ  p  0.8  pˆ 
n
200
pˆ   pˆ 0.77  0.8
 0.03
Z


 1.06
 pˆ
0.028284 0.028284
P( Z  1.06)
From Table V
1.06  0.1446
 0.1446
0.1446
 1.06
0
Since 0.1446 is bigger than 0.05, this result is usual.
Z