Transcript 9.3 Sample Means

Daniel S. Yates
The Practice of Statistics
Third Edition
Chapter 9:
9.3 Sample Means
Copyright © 2008 by W. H. Freeman & Company
Essential Questions
• Given the means and standard deviation of a
population, how do you calculate the means and
standard deviation for the sampling distribution
of a sample mean?
• What is the shape of the sampling distribution of
a sample mean drawn from a population that
has a Normal distribution?
• What is the central limit theorem?
• How do you use the central limit theorem to
solve probability problems for the sampling
distribution of a sample mean?
Sampling Distribution of
The sampling distribution of X is the distribution of the values of X in all
possible samples of the same size from the population.
Two key facts:
• Means of random samples are less variable than
individual observations
• Means of random samples are more Normal than
individual observations.
The behavior of X in repeated samples:
• The sample mean X is an unbiased estimator of population mean

• The values of X is less spread out for larger samples. The variability
decreases at the rate of n.



• The standard deviation of the sampling distribution of X , X
can only
n
be used when the population is at least 10 times as large as the sample.
• These facts about the means and standard deviation ofX
what the population distribution looks like.
are true regardless
The Shape of the Sampling
Distribution of X
Example – Young Women’s
Heights
• The height of young women varies
approximately according to the N(64.5, 2.5)
distribution.
• If we choose one young woman at random, the
heights we get in repeated choices follows this
distribution.
• This distribution of the population is also the
distribution of one observation chosen at
random.
• We think of the population distribution as a
distribution of probabilities.
Example – Young Women’s
Heights
• The height of young women varies
approximately according to the N(64.5, 2.5)
distribution.
• What is the probability that a randomly selected
young woman is taller than 66.5 inches?
Find P(X > 66.5)
  64.5
  2.5
First standardize the value of X.
X   66.5  64.5
Z

 0.80

2.5
Using Table A,
P(X > 66.5) = 1 – P(Z < 0.80) = 1 – 0.7661 = 0.2119
Using a calculator,
P(X > 66.5) = NORMCDF(0.80, 9999) = 0.2119
Example – Young Women’s
Heights
• The height of young women varies approximately according to the
N(64.5, 2.5) distribution.
• Now, let’s take a SRS of 10 young women from this population and
compute the means for this sample.
• Find the probability that the means of the sample will be greater than
66.5.
Since we are working with the means, x, we will use the
sampling distribution of x.
 x  64.5 and  x 
First standardize.
Z=

n

x  x
x
2.5
 0.79
10
S0, N(64.5, 0.79)
66.5  64.5

 2.53
0.79
Using Table A,
P(x > 66.5) = 1 - P(Z < 2.53) = 1 - 0.9943 = 0.0057
Fact: the average of several observations are less variable than
individual observation.
This fact leads to the central limit theorem.
The Central Limit Theorem and Nonnormal Populations
As sample size, n, increases the distribution of sample means becomes
more normal.
n 1
n2
n  10
n  25
More about the Central Limit
Theorem.
The Central Limit Theorem can safely
be applied when n exceeds 30.
If n is large or the population distribution
is normal, the standardized variable
x  X
x
z

X
 n
has (approximately) a standard normal
(z) distribution.
Example Servicing Air-conditioners
• The time that a technician requires to perform preventive
maintenance on a air-conditioner is governed by an
exponential density curve. The mean time is 1 hour and the
standard deviation is 1 hour. Your company has a contract
to maintain 70 units in an apartment building. You must
schedule technicians’ time for a visit to this building. Is it
safe to budget an average of 1.1 hours for each unit? Or
should you budget an average of 1.25 hours?
• To determine whether it is safe to budget 1.1 hours, on the
average, the probability we want is P( x  1.1)
x  1
Z
x  x
x
x 

n
1.1  1

 0.83
0.120

1
 0.120
70
N(1, 0.120)
P( x  1.1)  1  P(Z  0.83)  1  P(Z  0.83)
 1  0.7967  0.2033
Example Servicing Air-conditioners
• The time that a technician requires to perform preventive
maintenance on a air-conditioner is governed by an exponential
density curve. The mean time is 1 hour and the standard deviation
is 1 hour. Your company has a contract to maintain 70 units in an
apartment building. You must schedule technicians’ time for a visit
to this building. Is it safe to budget an average of 1.1 hours for each
unit? Or should you budget an average of 1.25 hours?
• To determine whether it is safe to budget 1.25 hours, on the
average, the probability we want is
Your turn – work this part of the problem.
Based on your analysis, what is your
decision?
Example
A food company sells “18 ounce” boxes
of cereal. Let x denote the actual amount
of cereal in a box of cereal. Suppose that
x is normally distributed with µ = 18.03
ounces and  = 0.05.
a) What is the probability that a box will
contain less than 18 ounces?
18  18.03 

P(x  18)  P  z 

0.05


 P(z  0.60)  0.2743
Example - continued
b) A case consists of 24 boxes of cereal.
What is the probability that the mean
amount of cereal (per box in a case)
is less than 18 ounces?
The central limit theorem states that the
distribution of x is normally distributed so

18  18.03 
P(x  18)  P  z 

0.05 24 

 P(z  2.94)  0.0016
The sampling distribution of a sample mean x has mean  and a
standard deviation of

n
.
If the population is Normal then the sampling distribution for x will
be Normal.
When the population is not Normal, then the sampling distribution for x will
be Normal only if the sample size is large.