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BA 1605
Chapter 9 Sampling Distributions of Statistics/Estimators
 Because population parameters are rarely known, we rely on estimators to give us
approximations of their values. For example sample means, medians and modes
estimate  and sample standard deviations estimate . An estimator then is a
statistic, computed from sample values.
 An estimator is a random variable since its value depends on the sample drawn.
Because of this, an estimator has a probability distribution known as a sampling
distribution with a mean and a standard deviation (also known as the standard
error) of its own.
 A sampling distribution is a probability distribution of a statistic.
Section 9.1 Sampling Distribution of the Mean
 
 Mean of X is  X  E X   X  E  X 
 Standard error (deviation) of X is  X 
X
n
Example
Consider a population that consists of :2, 6, 4, 8
a) Find  and 
5, 2.236
b) List all possible samples of size 2 taken with replacement and find the mean of
each sample.
c) Construct a probability distribution of the sample means.
d) Find the mean and standard deviation of X
5, 1.581
Note:
 Random samples are drawn from populations, some normally distributed, and
some non normal. In this section we look examine the distribution of the sample
mean, X . When a random sample is drawn from a population with mean  and
standard deviation , then the mean of X is  and the standard deviation of X is

, whether the population is normal or not. Note that the standard deviation of
n
X as well as other estimators is often called the standard error of X .
 If the population is finite (ie. If n > 0.05N) then the standard error of the mean

N n

becomes
, by multiplying by what is known as the finite population
N 1
n
2




correction factor. Note that when population size N is much larger than sample
size n, then the finite population correction factor has a value close to 1.
When the original population is normally distributed, X has a normal distribution
When the original population does not have a normal distribution and n  30 ,
then we do not know the distribution of X .
When the original population does not have a normal distribution and n > 30 then
X has an approximately normal distribution. This is what is known as the
Central Limit Theorem and is one of the most powerful statistical theorems. It
enables us to use the sample mean in robust statistical inferential procedures on
the population mean .
In cases where X is normal, we can find probabilities associated with X using
value  mean
the Z transform. Recall that Z 
. Applying this to X we will use
std .deviation
X 
.
Z

n
Example:
A.C. Neilsen reported that children between the ages of 2 and 5 watch an average of 25
hours of television per week. Assume the variable is normally distributed and the
standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly
selected, find the probability that the mean number of hours they watch television will be
greater than 26.3 hours.
0.0262
Example:
The average age of a vehicle registered in Canada is 8 years or 96 months. Assume the
standard deviation is 16 months. If a random sample of 36 vehicles is selected, find the
probability that the mean of their ages is between 90 and 100 months.
0.921
Example:
The average number of pounds of meat a person consumes a year is 218.4 pounds.
Assume that the standard deviation is 25 pounds and the distribution is approximately
normal.
a)
Find the probability that a person selected at random consumes less than 224
pounds per year.
0.5871
b)
If a sample of 40 individuals is selected, find the probability that the mean of
the sample will be no less than 224 pounds per year.
0.9222
Exercises pages 289 – 290
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Section 9.2 Sampling Distribution of a Proportion
X
, where X is the number of successes
n
in a sample of size n. This is actually a binomial experiment and X is a binomially
distributed random variable. For example, when n=12 , P  pˆ  0.5  P  X  6
The estimate of the probability of success is pˆ 
 Exact binomial probabilities can be found by using a probability table or the
binomial probability distribution function (formula). But each method has its
limitation. The table for example only includes certain values of n or p. The
formula can take too long when calculating certain probabilities. For example, if
X has a binomial pd with n = 200 and p=.34, think about how tedious it would be
to calculate P(X > 100)
 In these cases, we use the normal approximation to the binomial.
value  mean
 Recall that Z 
. When using the normal distribution to
std .deviation
X  0.5  np
approximate the binomial, we would use Z 
. We can apply when
npq
np  5, nq  5 .
  0.5 is a correction for continuity that is employed when a continuous
distribution is used to approximate a discrete distribution.
Summary of the Normal Approximation to the Binomial Distribution
Binomial
Normal
When finding
Use
1. P(X=a)
Pa  0.5  X  a  0.5
2. P X  a 
P X  a  0.5
3. P (X > a)
P( X > a +0.5)
P( X < a + 0.5)
4. P X  a 
5. P( X < a)
P( X < a – 0.5)
Example:
A magazine reported that 6 % of North American drivers used the cell phone while
driving. If 300 drivers are selected at random, find the probability that exactly 25 say they
use the cell phone while driving.
0.0227
Example:
Of the members of a bowling league, 10 % are widowed. If 200 bowling league members
are selected at random, find the probability that at least 10 are widowed.
0.9934
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Example:
If a baseball player’s batting average is 0.320 (32 %), find the probability that the player
will get at most 26 hits in 100 times at bat.
0.1190
Exercises, pages 295 – 296
Section 9.3 Sampling Distribution of the Difference between Two Means
Given two populations that are normally distributed with means 1 and  2 and standard
deviations  1 and  2 , then:
 The difference between 2 sample means X 1  X 2 is normally distributed with an
expected value or mean of 1  2 and a standard deviation (or standard error) of
 12
n

 22
n2
 We use the following Z transform to find probabilities: Z 
X
1
 X 2    1  2 
 12
n1
Exercises pages 298 - 299

 22
n2