Download Sampling Distribution of a Sample Mean

Section 5.1 Review
The Binomial Setting
1) There are a fixed number n of observations
2) The n observations are independent
3) Each observation falls into just one of two categories,
which, for convenience, we call “success” and “failure”.
4) The probability of a success, call it p, is the same for
each observation
Binomial Distributions
The distribution of the count X of successes in the binomial
setting is called the binomial distribution with parameters
n and p.
The parameter n is the number of observations
The parameter p is is the probability of a success on any
one observation.
The possible values of X are the whole numbers from 0
to n.
We say that X is B(n,p)
Binomial Probability Tables
(Table C, page T-7 to T-10)
Consider the previous transistor problem using the B(10,.1)
distribution.
Q: What is the probability that we
p
will pull an SRS from this distribution
n K
0.10
with exactly one bad transistor?
10
0
1
2
3
4
5
6
7
8
9
10
.3487
.3874
.1937
.0574
.0112
.0015
.0001
.0000
.0000
.0000
.0000
A: P(X = 1) = 0.3874
Binomial Mean and Standard Deviation
Q: If a count X is B(n,p), what are the mean X and the
standard deviation X ?
A: If a count X has the B(n,p) distribution, then :
X = np
X =
np (1-p)
Sample Proportions
In statistical sampling, we often want to estimate the
proportion p of “successes” in a population.
Our estimator is the sample proportion of successes :
p =
=
Count of successes in sample
Size of sample
X
n
Mean and Standard Deviation of a Sample Proportion
Let p be the sample proportion of successes in an SRS
of size n drawn from a large population having population
proportion p of successes.
The mean and standard deviation of p are the following :
p = p
p =
p(1-p)
n
Normal Approximations for Counts and Proportions
• Draw an SRS of size n from a large population having
population proportion p of successes.
• Let X be the count of successes in the sample, and
p = X / n be the sample proportion of successes.
• When n is large, the sampling distributions of these
statistics are approximately normal.
• X is approximately N( np,
• p is approximately N
(
np(1-p) )
p,
p(1-p)
n
)
Note: Use this when np  10 and n(1-p)  10
End O’ Review
The Sampling Distribution of a Sample Mean
Recall :
1) Counts and proportions are discrete random variables
that describe categorical data.
2) Measured data is usually described with continuous
random variables.
The Sampling Distribution of a Sample Mean
Recall :
3) Averages are less variable than individual observations
The Sampling Distribution of a Sample Mean
Recall :
4) Averages are more normal than individual observations
1
obs.
10
obs.
2
obs.
25
obs.
Sample Mean
Q: What is a sample mean?
A: This is an estimate of the mean of the underlying population
Q: How do we find the sample mean?
A: 1) Select an SRS of size n from a population
2) Measure a variable X on each individual in the sample
3) Let each observation be labeled X1, X2, … , Xn
4) The sample mean is the average of the Xi’s
Note: If the sample is large, each Xi can be thought of as an
independent random variable
Mean and Standard Deviation
of a Sample Mean
Let x be the mean of an SRS of size n from a population
having mean  and standard deviation . The mean and
standard deviation of x are:
x = 

x =
n
Mean and Standard Deviation
of a Sample Mean
Example: The height of a randomly chosen Jedi varies
according to the N(73, 2.8) distribution. If Yoda asked
the height of an SRS of 100 Jedi, what is the mean and
height of this sampling distribution?
 x =  = 73 inches

2.8
x =
=
n
100
=
2.8
10
=
0.28
The Sampling Distribution of x
Q: What is the shape of the distribution?
A: It depends on the population. So, if the population
distribution is normal, then so is the distribution of the
sample mean.
This leads to the following:
If a population has the N(, ) distribution, then the
sample mean has the N( , / n ) distribution.
In the previous example, the sample mean of the heights
of 100 Jedi has the N(73, 0.28) distribution.
The Sampling Distribution of x
• So, the sample mean from a normal distribution is normal.
• We can extend these ideas to the following :
Any linear combination of independent normal
random variables is also normally distributed.
• In other words, if X and Y are independent normal random
variables, then so is aX + bY.
• This means the following are normal:
2X + 7Y, 3X - 78Y, OX + 2Y,
X - Y, etc.
Example: Ben and Anakin are playing in the local Jedi golf
tournament. Ben’s golf game X has the N(60, 4) distribution.
Anakin’s golf game Y has the N(76, 12) distribution.
Q: What is the probability that Anakin will score lower than
Ben in the tourney? In other words, what is P(Y < X) ?
A: Notice that Y - X is a linear combination of two
independent random normal variables, so Y - X is normal.
What is the mean of the variable Y - X ?
Y-X = Y - X
= 76 - 60 = 16
Example: Ben and Anakin are playing in the local Jedi golf
tournament. Ben’s golf game X has the N(60, 4) distribution.
Anakin’s golf game Y has the N(76, 12) distribution.
Q: What is the probability that Anakin will score lower than
Ben in the tourney? In other words, what is P(Y < X) ?
A: Notice that Y - X is a linear combination of two
independent random normal variables, so Y - X is normal.
What is the standard deviation of the variable Y - X ?
2Y-X = 2Y + 2X = 122 + 42 =
Y-X = 160 = 12.65
160
So, Y - X has the N(16, 12.65) distribution.
Example: Ben and Anakin are playing in the local Jedi golf
tournament. Ben’s golf game X has the N(60, 4) distribution.
Anakin’s golf game Y has the N(76, 12) distribution.
Q: What is the probability that Anakin will score lower than
Ben in the tourney? In other words, what is P( Y < X) ?
So, Y - X has the N(16, 12.65) distribution.
P(Y < X) = P( Y - X < 0)
=P
(
(Y - X) - 16
12.65
<
0 - 16
12.65
)
= P ( Z < -1.26 ) = .1038
So, Anakin will score lower about 10% of the time.
Central Limit Theorem
• Draw an SRS of size n from any population with mean
and finite standard deviation .When n is large, the
sampling distribution of the sample mean is :

approximately N (, / n )
Notes:
• The distribution of a sum or average of many small random
quantities is close to normal.
• This is true if the quantities are not independent or even
if they have different distributions.
• How large does n have to be depends on the distribution.
The Sampling Distribution of x
(Revisited)
How do we find the sampling distribution ?
1) Take repeated random samples of size n from a
population with mean

2) Find the sample mean for each sample
3) Collect all the sample means and display their distribution.
The Sampling Distribution of x
(Revisited)
Homework
28, 31, 32, 37, 45