Transcript The Mean of the Sampling Distribution

Chapter Seven
Introduction to Sampling
Distributions
Section 2
The Central Limit Theorem
1
Key Points 7.2
• For a normal distribution, use mu and sigma to
construct the theoretical sampling distribution for
the statistic x – bar
• For large samples, use sample estimates to
construct a good approximate sampling
distribution for the statistic x-bar
• Learn the statement and underlying mean of the
central limit theorem well enough to explain it to a
friend who is intelligent, but (unfortunately) does
not know much about statistics
2
Let x be a random variable
with a normal distribution with
mean  and standard deviation
. Let x be the sample mean
corresponding to random
samples of size n taken from
the distribution.
3
Facts about sampling
distribution of the mean:
• The x distribution is a normal distribution.
• The mean of the x distribution is  (the
same mean as the original distribution).
• The standard deviation of the x
distribution is  n (the standard deviation
of the original distribution, divided by the
square root of the sample size).
4
We can use this theorem to
draw conclusions about means
of samples taken from normal
distributions.
If the original distribution is
normal, then the sampling
distribution will be normal.
5
The Mean of the Sampling
Distribution
x
6
The mean of the sampling
distribution is equal to the
mean of the original
distribution.
x  
7
The Standard Deviation of the
Sampling Distribution
x
8
The standard deviation of the
sampling distribution is equal to the
standard deviation of the original
distribution divided by the square
root of the sample size.

x 
n
9
The time it takes to drive between
cities A and B is normally distributed
with a mean of 14 minutes and a
standard deviation of 2.2 minutes.
• Find the probability that a trip between
the cities takes more than 15 minutes.
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
10
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that a trip between
the cities takes more than 15 minutes.
Find this
area
15  14
z
 0.45
14 15
2.2
P( z  0.45)  1.00  0.6736  0.3264
11
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
 x    14

2.2
x 

 0.73
n
9
12
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
Find this
area
15  14
z
 1.37
0.73
14 15
P( z  1.37 )  0.5  0.4147  0.0853
13
What if the Original
Distribution Is Not Normal?
Use the Central Limit Theorem!
14
MOVIE!
15
Central Limit Theorem
If x has any distribution with mean  and
standard deviation , then the sample
mean based on a random sample of size
n
will have a distribution that
approaches the normal distribution
(with mean  and standard deviation 
divided by the square root of n) as n
increases without bound.
16
How large should the sample
size be to permit the
application of the Central
Limit Theorem?
In most cases a sample size of
n = 30 or more assures that the
distribution will be approximately
normal and the theorem will apply.
17
Central Limit Theorem
• For most x distributions, if we use a
sample size of 30 or larger, the x
distribution will be approximately
normal.
18
Central Limit Theorem
• The mean of the sampling distribution is
the same as the mean of the original
distribution.
• The standard deviation of the sampling
distribution is equal to the standard
deviation of the original distribution
divided by the square root of the sample
size.
19
Central Limit Theorem
Formula
x  
20
Central Limit Theorem
Formula

x 
n
21
Central Limit Theorem
Formula
z 
x  

x
x
x  

 / n
22
Application of the Central
Limit Theorem
Records indicate that the packages shipped by a
certain trucking company have a mean weight of
510 pounds and a standard deviation of 90
pounds. One hundred packages are being shipped
today. What is the probability that their mean
weight will be:
a.
b.
c.
more than 530 pounds?
less than 500 pounds?
between 495 and 515 pounds?
23
Are we authorized to use the
Normal Distribution?
Yes, we are attempting to
draw conclusions about
means of large samples.
24
Applying the Central Limit
Theorem
What is the probability that their mean weight will
be more than 530 pounds?
Consider the distribution of sample means:
 x  510,  x  90 / 100  9
P( x > 530): z = 530 – 510 = 20 = 2.22
9
9
.0132
P(z > 2.22) = _______
25
Applying the Central Limit
Theorem
What is the probability that their mean weight
will be less than 500 pounds?
P( x < 500): z = 500 – 510 = –10 = – 1.11
9
9
.1335
P(z < – 1.11) = _______
26
Applying the Central Limit
Theorem
What is the probability that their mean weight
will be between 495 and 515 pounds?
P(495 < x < 515) :
for 495: z = 495 – 510 =  15 =  1.67
9
9
for 515: z = 515 – 510 = 5 = 0.56
9
9
.6648
P(  1.67 < z < 0.56) = _______
27
28
29
30
31
32
33
34
35