ST22: Statistics 2:
Advanced Statistical Methods
Sessions 1&2: Revision ST11: Statistics 1
Statistics and Probability
Revision ST11: Statistics 1: Statistics and Probability
Course content
Data & statistical thinking
Univariate analysis
Bivariate analysis
Basic principles in probability
Discrete random variables
Continuous random variables
Sampling distributions
Estimation (Point estimators and Confidence Intervals)
Hypothesis testing (one sample/two sample/chi-square)
Textbook
 Statistics for Business and Economics (New international edition 13/E), by McClave J., Benson G. & Sincich T Pearson.
Statistics for
Business and Economics
7th Edition
Chapter 5
Continuous Random Variables
and Probability Distributions
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-1
5.1
Continuous Probability Distributions
„
A continuous random variable is a variable that
can assume any value in an interval
„
„
„
„
„
thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value,
depending only on the ability to measure
accurately.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-5
Probability as an Area
Shaded area under the curve is the
probability that X is between a and b
f(x)
P (a ≤ x ≤ b)
= P (a < x < b)
(Note that the probability
of any individual value is
zero)
a
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
b
x
Ch. 5-9
The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Exponential
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-10
The Uniform Distribution
„
The uniform distribution is a probability
distribution that has equal probabilities for all
possible outcomes of the random variable
f(x)
Total area under the
uniform probability
density function is 1.0
xmin
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
xmax x
Ch. 5-11
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
if a ≤ x ≤ b
b−a
0
otherwise
where
f(x) = value of the density function at any x value
a = minimum value of x
b = maximum value of x
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-12
Properties of the
Uniform Distribution
„
The mean of a uniform distribution is
a+b
μ=
2
„
The variance is
2
(b
a)
2
σ =
12
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-13
Uniform Distribution Example
Example: Uniform probability distribution
over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = .25 for 2 ≤ x ≤ 6
f(x)
μ=
.25
a+b 2+6
=
=4
2
2
(b - a) 2 (6 - 2) 2
σ =
=
= 1.333
12
12
2
2
6
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x
Ch. 5-14
5.2
„
Expectations for Continuous
Random Variables
The mean of X, denoted μX , is defined as the
expected value of X
μX = E(X)
„
The variance of X, denoted σX2 , is defined as the
expectation of the squared deviation, (X - μX)2, of a
random variable from its mean
σ 2X = E[(X − μX )2 ]
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-15
Linear Functions of Variables
„
„
Let W = a + bX , where X has mean μX and
variance σX2 , and a and b are constants
Then the mean of W is
μW = E(a + bX) = a + bμX
„
the variance is
σ
„
2
W
= Var(a + bX) = b σ
2
2
X
the standard deviation of W is
σW = b σX
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-16
Linear Functions of Variables
(continued)
„
An important special case of the previous results is the
standardized random variable
X − μX
Z=
σX
„
which has a mean 0 and variance 1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-17
5.3
The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Exponential
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Ch. 5-18
The Normal Distribution
(continued)
‘Bell Shaped’
„ Symmetrical
f(x)
„ Mean, Median and Mode
are Equal
Location is determined by the
mean, μ
„
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+ ∞ to − ∞
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
σ
μ
x
Mean
= Median
= Mode
Ch. 5-19
The Normal Distribution
(continued)
„
„
„
„
The normal distribution closely approximates the
probability distributions of a wide range of random
variables
Distributions of sample means approach a normal
distribution given a “large” sample size
Computations of probabilities are direct and elegant
The normal probability distribution has led to good
business decisions for a number of applications
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-20
Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-21
The Normal Distribution Shape
f(x)
Changing μ shifts the
distribution left or right.
σ
Changing σ increases
or decreases the
spread.
μ
x
Given the mean μ and variance σ we define the normal
distribution using the notation
X ~ N(μ,σ 2 )
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-22
The Normal Probability
Density Function
„
The formula for the normal probability density
function is
1
−(x −μ)2 /2σ 2
f(x) =
e
2πσ
Where
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
x = any value of the continuous variable, −∞ < x < ∞
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-23
Cumulative Normal Distribution
„
For a normal random variable X with mean μ and
variance σ2 , i.e., X~N(μ, σ2), the cumulative
distribution function is
F(x 0 ) = P(X ≤ x 0 )
f(x)
P(X ≤ x 0 )
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x0
x
Ch. 5-24
Finding Normal Probabilities
The probability for a range of values is
measured by the area under the curve
P(a < X < b) = F(b) − F(a)
a
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
μ
b
x
Ch. 5-25
Finding Normal Probabilities
(continued)
F(b) = P(X < b)
a
μ
b
x
a
μ
b
x
a
μ
b
x
F(a) = P(X < a)
P(a < X < b) = F(b) − F(a)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-26
The Standardized Normal
„
Any normal distribution (with any mean and
variance combination) can be transformed into the
standardized normal distribution (Z), with mean 0
and variance 1
f(Z)
Z ~ N(0 ,1)
1
0
„
Z
Need to transform X units into Z units by subtracting the
mean of X and dividing by its standard deviation
X −μ
Z=
σ
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-27
Example
„
If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
X − μ 200 − 100
Z=
=
= 2.0
σ
50
„
This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-28
Comparing X and Z units
100
0
200
2.0
X
Z
(μ = 100, σ = 50)
( μ = 0 , σ = 1)
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (X) or in standardized units (Z)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-29
Finding Normal Probabilities
b −μ⎞
⎛ a −μ
<Z<
P(a < X < b) = P⎜
⎟
σ ⎠
⎝ σ
⎛ b −μ⎞ ⎛ a −μ⎞
= F⎜
⎟ − F⎜
⎟
⎝ σ ⎠ ⎝ σ ⎠
f(x)
a
a −μ
σ
µ
b
x
0
b −μ
σ
Z
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-30
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(X) P( −∞ < X < μ) = 0.5
0.5
P(μ < X < ∞ ) = 0.5
0.5
μ
X
P( −∞ < X < ∞) = 1.0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-31
Appendix Table 1
„
„
The Standardized Normal table in the textbook
(Appendix Table 1) shows values of the
cumulative normal distribution function
For a given Z-value a , the table shows F(a)
(the area under the curve from negative infinity to a )
F(a) = P(Z < a)
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
a
Z
Ch. 5-32
The Standardized Normal Table
ƒ Appendix Table 1 gives the probability F(a) for
any value a
.9772
Example:
P(Z < 2.00) = .9772
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
2.00
Z
Ch. 5-33
The Standardized Normal Table
(continued)
ƒ For negative Z-values, use the fact that the
distribution is symmetric to find the needed
probability:
.9772
.0228
Example:
P(Z < -2.00) = 1 – 0.9772
= 0.0228
0
2.00
Z
.9772
.0228
-2.00
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
Z
Ch. 5-34
General Procedure for Finding
Probabilities
To find P(a < X < b) when X is
distributed normally:
„
Draw the normal curve for the problem in
terms of X
„
Translate X-values to Z-values
„
Use the Cumulative Normal Table
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-35
Finding Normal Probabilities
„
„
Suppose X is normal with mean 8.0 and
standard deviation 5.0
Find P(X < 8.6)
X
8.0
8.6
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Ch. 5-36
Finding Normal Probabilities
(continued)
„
Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(X < 8.6)
Z=
X − μ 8.6 − 8.0
=
= 0.12
σ
5.0
μ=8
σ = 10
8 8.6
μ=0
σ=1
X
P(X < 8.6)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0 0.12
Z
P(Z < 0.12)
Ch. 5-37
Solution: Finding P(Z < 0.12)
Standardized Normal Probability
Table (Portion)
z
F(z)
.10
.5398
.11
.5438
.12
.5478
.13
.5517
P(X < 8.6)
= P(Z < 0.12)
F(0.12) = 0.5478
0.00
Z
0.12
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-38
Upper Tail Probabilities
„
„
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now Find P(X > 8.6)
X
8.0
8.6
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 5-39
Upper Tail Probabilities
(continued)
„
Now Find P(X > 8.6)…
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - 0.5478 = 0.4522
0.5478
1.000
Z
0
0.12
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
1.0 - 0.5478
= 0.4522
Z
0
0.12
Ch. 5-40
Statistics for
Business and Economics
7th Edition
Chapter 6
Sampling and
Sampling Distributions
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-1
6.1

Tools of Business Statistics
Descriptive statistics


Collecting, presenting, and describing data
Inferential statistics

Drawing conclusions and/or making decisions
concerning a population based only on
sample data
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-2
Populations and Samples

A Population is the set of all items or individuals
of interest


Examples:
All likely voters in the next election
All parts produced today
All sales receipts for November
A Sample is a subset of the population

Examples:
1000 voters selected at random for interview
A few parts selected for destructive testing
Random receipts selected for audit
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-3
Population vs. Sample
Population
a b
Sample
cd
b
ef gh i jk l m n
o p q rs t u v w
x y
z
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c
gi
o
n
r
u
y
Ch. 6-4
Why Sample?

Less time consuming than a census

Less costly to administer than a census

It is possible to obtain statistical results of a
sufficiently high precision based on samples.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-5
Simple Random Samples




Every object in the population has an equal chance of
being selected
Objects are selected independently
Samples can be obtained from a table of random
numbers or computer random number generators
A simple random sample is the ideal against which
other sample methods are compared
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-6
Inferential Statistics

Making statements about a population by
examining sample results
Sample statistics
(known)
Population parameters
Inference
Sample
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
(unknown, but can
be estimated from
sample evidence)
Population
Ch. 6-7
Inferential Statistics
Drawing conclusions and/or making decisions
concerning a population based on sample results.

Estimation


e.g., Estimate the population mean
weight using the sample mean
weight
Hypothesis Testing

e.g., Use sample evidence to test
the claim that the population mean
weight is 120 pounds
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-8
6.2

Sampling Distributions
A sampling distribution is a distribution of
all of the possible values of a statistic for
a given size sample selected from a
population
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-9
Chapter Outline
Sampling
Distributions
Sampling
Distribution of
Sample
Mean
Sampling
Distribution of
Sample
Proportion
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Sampling
Distribution of
Sample
Variance
Ch. 6-10
Sampling Distributions of
Sample Means
Sampling
Distributions
Sampling
Distribution of
Sample
Mean
Sampling
Distribution of
Sample
Proportion
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Sampling
Distribution of
Sample
Variance
Ch. 6-11
Developing a
Sampling Distribution




Assume there is a population …
Population size N=4
A
B
C
D
Random variable, X,
is age of individuals
Values of X:
18, 20, 22, 24 (years)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-12
Developing a
Sampling Distribution
(continued)
Summary Measures for the Population Distribution:
X

μ
P(x)
i
N
18  20  22  24

 21
4
σ
2
(X
μ)

 i
N
.25
0
 2.236
18
20
22
24
A
B
C
D
x
Uniform Distribution
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-13
Developing a
Sampling Distribution
(continued)
Now consider all possible samples of size n = 2
1st
Obs
2nd Observation
18
20
22
24
18 18,18 18,20 18,22 18,24
16 Sample
Means
20 20,18 20,20 20,22 20,24
1st 2nd Observation
Obs 18 20 22 24
22 22,18 22,20 22,22 22,24
18 18 19 20 21
24 24,18 24,20 24,22 24,24
20 19 20 21 22
16 possible samples
(sampling with
replacement)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
22 20 21 22 23
24 21 22 23 24
Ch. 6-14
Developing a
Sampling Distribution
(continued)
Sampling Distribution of All Sample Means
Sample Means
Distribution
16 Sample Means
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
_
P(X)
.3
.2
.1
0
18 19
20 21 22 23
(no longer uniform)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
24
_
X
Ch. 6-15
Developing a
Sampling Distribution
(continued)
Summary Measures of this Sampling Distribution:
X

E(X) 
N
σX 

i
18  19  21   24

 21  μ
16
2
(
X

μ)
i

N
(18 - 21)2  (19 - 21)2    (24 - 21)2
 1.58
16
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-16
Comparing the Population with its
Sampling Distribution
Population
N=4
μ  21
σ  2.236
Sample Means Distribution
n=2
μX  21
σ X  1.58
_
P(X)
.3
P(X)
.3
.2
.2
.1
.1
0
0
18
20
22
24
A
B
C
D
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
X
18 19
20 21 22 23
24
_
X
Ch. 6-17
Expected Value of Sample Mean


Let X1, X2, . . . Xn represent a random sample from a
population
The sample mean value of these observations is
defined as
n
1
X   Xi
n i1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-18
Standard Error of the Mean


Different samples of the same size from the same
population will yield different sample means
A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
σ
σX 
n

Note that the standard error of the mean decreases as
the sample size increases
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-19
If sample values are
not independent
(continued)



If the sample size n is not a small fraction of the
population size N, then individual sample members
are not distributed independently of one another
Thus, observations are not selected independently
A correction is made to account for this:
σ2 N  n
Var( X ) 
n N 1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
or
σ
σX 
n
Nn
N 1
Ch. 6-20
If the Population is Normal

If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of X is also normally distributed with
μX  μ

and
σ
σX 
n
If the sample size n is not large relative to the population size N, then
μX  μ
and
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
σ Nn
σX 
n N 1
Ch. 6-21
Z-value for Sampling Distribution
of the Mean

Z-value for the sampling distribution of X :
( X  μ)
Z
σX
where:
X
μ
σx
= sample mean
= population mean
= standard error of the mean
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-22
Sampling Distribution Properties
Normal Population
Distribution
μx  μ
(i.e.
x is unbiased )
μ
x
μx
x
Normal Sampling
Distribution
(has the same mean)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-23
Sampling Distribution Properties
(continued)

For sampling with replacement:
As n increases,
Larger
sample size
σ x decreases
Smaller
sample size
μ
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x
Ch. 6-24
If the Population is not Normal

We can apply the Central Limit Theorem:


Even if the population is not normal,
…sample means from the population will be
approximately normal as long as the sample size is
large enough.
Properties of the sampling distribution:
μx  μ
and
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
σ
σx 
n
Ch. 6-25
Central Limit Theorem
As the
sample
size gets
large
enough…
n↑
the sampling
distribution
becomes
almost normal
regardless of
shape of
population
x
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-26
If the Population is not Normal
(continued)
Sampling distribution
properties:
Population Distribution
Central Tendency
μx  μ
Variation
σ
σx 
n
x
μ
Sampling Distribution
(becomes normal as n increases)
Larger
sample
size
Smaller
sample size
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
μx
x
Ch. 6-27
How Large is Large Enough?


For most distributions, n > 25 will give a
sampling distribution that is nearly normal
For normal population distributions, the
sampling distribution of the mean is always
normally distributed
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-28
Example


Suppose a large population has mean μ = 8
and standard deviation σ = 3. Suppose a
random sample of size n = 36 is selected.
What is the probability that the sample mean is
between 7.8 and 8.2?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-29
Example
(continued)
Solution:


Even if the population is not normally
distributed, the central limit theorem can be
used (n > 25)
… so the sampling distribution of
approximately normal
x
is

… with mean μx = 8

σ
3
…and standard deviation σ x  n  36  0.5
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-30
Example
(continued)
Solution (continued):


μX -μ
 7.8 - 8
8.2 - 8 
P(7.8  μ X  8.2)  P



3
σ
3


36
n
36


 P(-0.5  Z  0.5)  0.3830
Population
Distribution
???
??
?
?
?
?
?
?
μ8
Sampling
Distribution
Standard Normal
Distribution
Sample
?
X
.1915
+.1915
Standardize
7.8
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
μX  8
8.2
x
-0.5
μz  0
0.5
Z
Ch. 6-31
Acceptance Intervals

Goal: determine a range within which sample means are
likely to occur, given a population mean and variance



By the Central Limit Theorem, we know that the distribution of X
is approximately normal if n is large enough, with mean μ and
standard deviation σ X
Let zα/2 be the z-value that leaves area α/2 in the upper tail of the
normal distribution (i.e., the interval - zα/2 to zα/2 encloses
probability 1 – α)
Then
μ  z  /2 σ X
is the interval that includes X with probability 1 – α
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-32
6.4
Sampling Distributions of
Sample Variance
Sampling
Distributions
Sampling
Distribution of
Sample
Mean
Sampling
Distribution of
Sample
Proportion
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Sampling
Distribution of
Sample
Variance
Ch. 6-33
Sample Variance

Let x1, x2, . . . , xn be a random sample from a
population. The sample variance is
n
1
2
s2 
(x

x
)

i
n  1 i1


the square root of the sample variance is called
the sample standard deviation
the sample variance is different for different
random samples from the same population
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-34
Sampling Distribution of
Sample Variances

The sampling distribution of s2 has mean σ2
E(s2 )  σ 2

If the population distribution is normal, then
4
2σ
Var(s2 ) 
n 1

If the population distribution is normal then
(n - 1)s 2
σ2
has a 2 distribution with n – 1 degrees of freedom
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-35
The Chi-square Distribution


The chi-square distribution is a family of distributions,
depending on degrees of freedom:
d.f. = n – 1
0 4 8 12 16 20 24 28
d.f. = 1

2 0 4 8 12 16 20 24 28 2 0 4 8 12 16 20 24 28 2
d.f. = 5
d.f. = 15
Text Table 7 contains chi-square probabilities
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-36
Degrees of Freedom (df)
Idea: Number of observations that are free to vary
after sample mean has been calculated
Example: Suppose the mean of 3 numbers is 8.0
Let X1 = 7
Let X2 = 8
What is X3?
If the mean of these three
values is 8.0,
then X3 must be 9
(i.e., X3 is not free to vary)
Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2
(2 values can be any numbers, but the third is not free to vary
for a given mean)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-37
Chi-square Example

A commercial freezer must hold a selected
temperature with little variation. Specifications call
for a standard deviation of no more than 4 degrees
(a variance of 16 degrees2).
 A sample of 14 freezers is to be
tested
 What is the upper limit (K) for the
sample variance such that the
probability of exceeding this limit,
given that the population standard
deviation is 4, is less than 0.05?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-38
Finding the Chi-square Value
2
(n

1)s
χ2 
σ2

Is chi-square distributed with (n – 1) = 13
degrees of freedom
Use the the chi-square distribution with area 0.05
in the upper tail:
213 = 22.36 (α = .05 and 14 – 1 = 13 d.f.)
probability
α = .05
2
213 = 22.36
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-39
Chi-square Example
(continued)
213 = 22.36
So:
(α = .05 and 14 – 1 = 13 d.f.)
2


(n
1)s
2 
  0.05
 χ13
P(s2  K)  P
 16

(n  1)K
 22.36
16
or
so
K
(where n = 14)
(22.36)(16)
 27.52
(14  1)
If s2 from the sample of size n = 14 is greater than 27.52, there is
strong evidence to suggest the population variance exceeds 16.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 6-40
Statistics for
Business and Economics
7th Edition
Chapter 7
Estimation: Single Population
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-1
Confidence Intervals
Contents of this chapter:
 Confidence Intervals for the Population
Mean, μ




when Population Variance σ2 is Known
when Population Variance σ2 is Unknown
Confidence Intervals for the Population
Proportion, p̂ (large samples)
Confidence interval estimates for the
variance of a normal population
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-2
7.1

Definitions
An estimator of a population parameter is



a random variable that depends on sample
information . . .
whose value provides an approximation to this
unknown parameter
A specific value of that random variable is
called an estimate
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-3
Point and Interval Estimates


A point estimate is a single number,
a confidence interval provides additional
information about variability
Lower
Confidence
Limit
Point Estimate
Upper
Confidence
Limit
Width of
confidence interval
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-4
Point Estimates
We can estimate a
Population Parameter …
with a Sample
Statistic
(a Point Estimate)
Mean
μ
x
Proportion
P
p̂
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-5
Unbiasedness

A point estimator θ̂ is said to be an
unbiased estimator of the parameter  if the
expected value, or mean, of the sampling
distribution of θ̂ is ,
E(θˆ )  θ

Examples:
 The sample mean x is an unbiased estimator of μ
2 is an unbiased estimator of σ2
 The sample variance s
 The sample proportion p̂ is an unbiased estimator of P
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-6
Unbiasedness
(continued)

θ̂1 is an unbiased estimator, θ̂ 2 is biased:
θ̂1
θ̂ 2
θ
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
θ̂
Ch. 7-7
Bias


Let θ̂ be an estimator of 
The bias in θ̂ is defined as the difference
between its mean and 
Bias(θˆ )  E(θˆ )  θ

The bias of an unbiased estimator is 0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-8
Most Efficient Estimator



Suppose there are several unbiased estimators of 
The most efficient estimator or the minimum variance
unbiased estimator of  is the unbiased estimator with the
smallest variance
Let θ̂1 and θ̂ 2 be two unbiased estimators of , based on
the same number of sample observations. Then,


θ̂1 is said to be more efficient than θ̂ 2 if Var(θˆ 1 )  Var(θˆ 2 )
The relative efficiency of θ̂1 with respect to θ̂ 2 is the ratio
of their variances:
Var(θˆ 2 )
Relative Efficiency 
Var(θˆ )
1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-9
7.2
Confidence Intervals



How much uncertainty is associated with a
point estimate of a population parameter?
An interval estimate provides more
information about a population characteristic
than does a point estimate
Such interval estimates are called confidence
intervals
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-10
Confidence Interval Estimate

An interval gives a range of values:




Takes into consideration variation in sample
statistics from sample to sample
Based on observation from 1 sample
Gives information about closeness to
unknown population parameters
Stated in terms of level of confidence

Can never be 100% confident
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-11
Confidence Interval and
Confidence Level


If P(a <  < b) = 1 -  then the interval from a
to b is called a 100(1 - )% confidence
interval of .
The quantity (1 - ) is called the confidence
level of the interval ( between 0 and 1)


In repeated samples of the population, the true value
of the parameter  would be contained in 100(1 - )%
of intervals calculated this way.
The confidence interval calculated in this manner is
written as a <  < b with 100(1 - )% confidence
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-12
Estimation Process
Random Sample
Population
(mean, μ, is
unknown)
Mean
X = 50
I am 95%
confident that
μ is between
40 & 60.
Sample
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-13
Confidence Level, (1-)
(continued)



Suppose confidence level = 95%
Also written (1 - ) = 0.95
A relative frequency interpretation:


From repeated samples, 95% of all the
confidence intervals that can be constructed will
contain the unknown true parameter
A specific interval either will contain or will
not contain the true parameter

No probability involved in a specific interval
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-14
General Formula

The general formula for all confidence
intervals is:
Point Estimate  (Reliability Factor)(Standard Error)

The value of the reliability factor
depends on the desired level of
confidence
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-15
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2 Known
Population
Proportion
Population
Variance
σ2 Unknown
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-16
Confidence Interval for μ
(σ2 Known)
7.2

Assumptions




Population variance σ2 is known
Population is normally distributed
If population is not normal, use large sample
Confidence interval estimate:
x  z α/2
σ
σ
 μ  x  z α/2
n
n
(where z/2 is the normal distribution value for a probability of /2 in
each tail)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-17
Margin of Error

The confidence interval,
x  z α/2

σ
σ
 μ  x  z α/2
n
n
Can also be written as x  ME
where ME is called the margin of error
ME  z α/2

σ
n
The interval width, w, is equal to twice the margin of
error
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-18
Reducing the Margin of Error
ME  z α/2
σ
n
The margin of error can be reduced if

the population standard deviation can be reduced (σ↓)

The sample size is increased (n↑)

The confidence level is decreased, (1 – ) ↓
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-19
Finding the Reliability Factor, z/2

Consider a 95% confidence interval:
1    .95
α
 .025
2
Z units:
X units:
α
 .025
2
z = -1.96
Lower
Confidence
Limit
0
Point Estimate
z = 1.96
Upper
Confidence
Limit
 Find z.025 = 1.96 from the standard normal distribution table
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-20
Common Levels of Confidence

Commonly used confidence levels are 90%,
95%, and 99%
Confidence
Level
Confidence
Coefficient,
Z/2 value
.80
.90
.95
.98
.99
.998
.999
1.28
1.645
1.96
2.33
2.58
3.08
3.27
1 
80%
90%
95%
98%
99%
99.8%
99.9%
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-21
Intervals and Level of Confidence
Sampling Distribution of the Mean
1 
/2
Intervals
extend from
/2
x
μx  μ
x1
σ
LCL  x  z
n
x2
to
σ
UCL  x  z
n
Confidence Intervals
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
100(1-)%
of intervals
constructed
contain μ;
100()% do
not.
Ch. 7-22
Example


A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
Determine a 95% confidence interval for the
true mean resistance of the population.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-23
Example
(continued)


A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is .35 ohms.
Solution:
σ
x z
n
 2.20  1.96 (.35/ 11)
 2.20  .2068
1.9932  μ  2.4068
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-24
Interpretation


We are 95% confident that the true mean
resistance is between 1.9932 and 2.4068
ohms
Although the true mean may or may not be
in this interval, 95% of intervals formed in
this manner will contain the true mean
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-25
7.3
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2 Known
Population
Proportion
Population
Variance
σ2 Unknown
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-26
Student’s t Distribution


Consider a random sample of n observations
 with mean x and standard deviation s
 from a normally distributed population with mean μ
Then the variable
x μ
t
s/ n
follows the Student’s t distribution with (n - 1) degrees
of freedom
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-27
Confidence Interval for μ
(σ2 Unknown)



If the population standard deviation σ is
unknown, we can substitute the sample
standard deviation, s
This introduces extra uncertainty, since
s is variable from sample to sample
So we use the t distribution instead of
the normal distribution
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-28
Confidence Interval for μ
(σ Unknown)
(continued)

Assumptions





Population standard deviation is unknown
Population is normally distributed
If population is not normal, use large sample
Use Student’s t Distribution
Confidence Interval Estimate:
x  t n-1,α/2
S
S
 μ  x  t n-1,α/2
n
n
where tn-1,α/2 is the critical value of the t distribution with n-1 d.f.
and an area of α/2 in each tail:
P(t n1  t n1,α/2 )  α/2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-29
Margin of Error

The confidence interval,
x  t n-1,α/2

S
S
 μ  x  t n-1,α/2
n
n
Can also be written as
x  ME
where ME is called the margin of error:
ME  t n-1,α/2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
σ
n
Ch. 7-30
Student’s t Distribution


The t is a family of distributions
The t value depends on degrees of
freedom (d.f.)

Number of observations that are free to vary after
sample mean has been calculated
d.f. = n - 1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-31
Student’s t Distribution
Note: t
Z as n increases
Standard
Normal
(t with df = ∞)
t (df = 13)
t-distributions are bellshaped and symmetric, but
have ‘fatter’ tails than the
normal
t (df = 5)
0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
t
Ch. 7-32
Student’s t Table
Upper Tail Area
df
.10
.05
.025
1 3.078 6.314 12.706
Let: n = 3
df = n - 1 = 2
 = .10
/2 =.05
2 1.886 2.920 4.303
/2 = .05
3 1.638 2.353 3.182
The body of the table
contains t values, not
probabilities
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
2.920 t
Ch. 7-33
t distribution values
With comparison to the Z value
Confidence
t
Level
(10 d.f.)
t
(20 d.f.)
t
(30 d.f.)
Z
____
.80
1.372
1.325
1.310
1.282
.90
1.812
1.725
1.697
1.645
.95
2.228
2.086
2.042
1.960
.99
3.169
2.845
2.750
2.576
Note: t
Z as n increases
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-34
Example
A random sample of n = 25 has x = 50 and
s = 8. Form a 95% confidence interval for μ

d.f. = n – 1 = 24, so t n1,α/2  t 24,.025  2.0639
The confidence interval is
S
S
 μ  x  t n-1,α/2
x  t n-1,α/2
n
n
8
8
50  (2.0639)
 μ  50  (2.0639)
25
25
46.698  μ  53.302
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-35
7.5
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2 Known
Population
Proportion
Population
Variance
σ2 Unknown
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 7-36
Confidence Intervals for the
Population Variance
 Goal: Form a confidence interval for the
population variance, σ2


The confidence interval is based on the
sample variance, s2
Assumed: the population is normally
distributed
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 8-37
Confidence Intervals for the
Population Variance
(continued)
The random variable

2
n 1
(n  1)s

σ2
2
follows a chi-square distribution with (n – 1)
degrees of freedom
Where the chi-square value
 n21,  denotes the number for which
P( χ n21  χn21, α )  α
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 8-38
Confidence Intervals for the
Population Variance
(continued)
The (1 - )% confidence interval for the
population variance is
2
(n  1)s2
(n

1)s
2

σ
 2
2
χ n1, α/2
χ n1, 1 - α/2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 8-39
Example
You are testing the speed of a batch of computer
processors. You collect the following data (in Mhz):
Sample size
Sample mean
Sample std dev
17
3004
74
Assume the population is normal.
Determine the 95% confidence interval for σx2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 8-40
Finding the Chi-square Values


n = 17 so the chi-square distribution has (n – 1) = 16
degrees of freedom
 = 0.05, so use the the chi-square values with area
0.025 in each tail:
2
χ n21, α/2  χ16
, 0.025  28.85
2
χ n21, 1 - α/2  χ16
, 0.975  6.91
probability
α/2 = .025
probability
α/2 = .025
216 = 6.91
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
216 = 28.85
216
Ch. 8-41
Calculating the Confidence Limits

The 95% confidence interval is
2
(n  1)s 2
(n

1)s
2

σ
 2
2
χ n1, α/2
χ n1, 1 - α/2
2
(17  1)(74) 2
(17
1)(74)

 σ2 
28.85
6.91
3037  σ 2  12683
Converting to standard deviation, we are 95%
confident that the population standard deviation of
CPU speed is between 55.1 and 112.6 Mhz
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 8-42
Statistics for
Business and Economics
7th Edition
Chapter 9
Hypothesis Testing:
Single Population
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-1
9.1

What is a Hypothesis?
A hypothesis is a claim
(assumption) about a
population parameter:

population mean
Example: The mean monthly cell phone bill
of this city is μ = $42

population proportion
Example: The proportion of adults in this
city with cell phones is p = .68
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-2
The Null Hypothesis, H0

States the assumption (numerical) to be
tested
Example: The average number of TV sets in
U.S. Homes is equal to three ( H0 : μ  3 )

Is always about a population parameter,
not about a sample statistic
H0 : μ  3
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
H0 : X  3
Ch. 9-3
The Null Hypothesis, H0
(continued)




Begin with the assumption that the null
hypothesis is true
 Similar to the notion of innocent until
proven guilty
Refers to the status quo
Always contains “=” , “≤” or “” sign
May or may not be rejected
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-4
The Alternative Hypothesis, H1

Is the opposite of the null hypothesis





e.g., The average number of TV sets in U.S.
homes is not equal to 3 ( H1: μ ≠ 3 )
Challenges the status quo
Never contains the “=” , “≤” or “” sign
May or may not be supported
Is generally the hypothesis that the
researcher is trying to support
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-5
Hypothesis Testing Process
Claim: the
population
mean age is 50.
(Null Hypothesis:
H0: μ = 50 )
Population
Is X 20 likely if μ = 50?
If not likely,
REJECT
Null Hypothesis
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Suppose
the sample
mean age
is 20: X = 20
Now select a
random sample
Sample
Ch. 9-6
Reason for Rejecting H0
Sampling Distribution of X
20
If it is unlikely that
we would get a
sample mean of
this value ...
μ = 50
If H0 is true
... if in fact this were
the population mean…
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
X
... then we
reject the null
hypothesis that
μ = 50.
Ch. 9-7
Level of Significance, 

Defines the unlikely values of the sample
statistic if the null hypothesis is true


Defines rejection region of the sampling
distribution
Is designated by  , (level of significance)

Typical values are .01, .05, or .10

Is selected by the researcher at the beginning

Provides the critical value(s) of the test
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-8
Level of Significance
and the Rejection Region
Level of significance =
H0: μ = 3
H1: μ ≠ 3
/2
Two-tail test
/2
Represents
critical value
Rejection
region is
shaded
0
H0: μ ≤ 3
H1: μ > 3

Upper-tail test
H0: μ ≥ 3
H1: μ < 3

0

Lower-tail test
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
Ch. 9-9
Errors in Making Decisions

Type I Error
 Reject a true null hypothesis
 Considered a serious type of error
The probability of Type I Error is 

Called level of significance of the test

Set by researcher in advance
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-10
Errors in Making Decisions
(continued)

Type II Error
 Fail to reject a false null hypothesis
The probability of Type II Error is β
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-11
Outcomes and Probabilities
Possible Hypothesis Test Outcomes
Actual Situation
Key:
Outcome
(Probability)
Decision
H0 True
Do Not
Reject
H0
No Error
(1 -  )
Type II Error
(β)
Reject
H0
Type I Error
()
No Error
(1-β)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
H0 False
Ch. 9-12
Type I & II Error Relationship
 Type I and Type II errors can not happen at
the same time

Type I error can only occur if H0 is true

Type II error can only occur if H0 is false
If Type I error probability (  )
, then
Type II error probability ( β )
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-13
Hypothesis Tests for the Mean
Hypothesis
Tests for 
 Known
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
 Unknown
Ch. 9-14
Test of Hypothesis
for the Mean (σ Known)
9.2

Convert sample result ( x ) to a z value
Hypothesis
Tests for 
σ Known
σ Unknown
Consider the test
H0 : μ  μ0
H1 : μ  μ0
(Assume the population is normal)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
The decision rule is:
Reject H0 if z 
x  μ0
 zα
σ
n
Ch. 9-15
Decision Rule
x  μ0
Reject H0 if z 
 zα
σ
n
H0: μ = μ0
H1: μ > μ0
Alternate rule:

Reject H0 if x  μ0  Z ασ/ n
Z
x
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Do not reject H0
0
μ0
zα
μ0  z α
Reject H0
σ
n
Critical value x c
Ch. 9-16
p-Value Approach to Testing

p-value: Probability of obtaining a test
statistic more extreme ( ≤ or  ) than the
observed sample value given H0 is true


Also called observed level of significance
Smallest value of  for which H0 can be
rejected
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-17
p-Value Approach to Testing
(continued)


Convert sample result (e.g., x ) to test statistic (e.g., z
statistic )
Obtain the p-value
x - μ0
 For an upper
p - value  P(z 
, given that H0 is true)
σ/ n
tail test:
x - μ0
 P(z 
| μ  μ0 )
σ/ n

Decision rule: compare the p-value to 

If p-value <  , reject H0

If p-value   , do not reject H0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-18
Example: Upper-Tail Z Test
for Mean ( Known)
A phone industry manager thinks that
customer monthly cell phone bill have
increased, and now average over $52 per
month. The company wishes to test this
claim. (Assume  = 10 is known)
Form hypothesis test:
H0: μ ≤ 52 the average is not over $52 per month
H1: μ > 52
the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-19
Example: Find Rejection Region
(continued)

Suppose that  = .10 is chosen for this test
Find the rejection region:
Reject H0
 = .10
Do not reject H0
0
1.28
Reject H0
x  μ0
Reject H0 if z 
 1.28
σ/ n
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-20
Example: Sample Results
(continued)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following
results: n = 64, x = 53.1 ( = 10 was assumed known)

Using the sample results,
x  μ0
53.1  52

 0.88
z
σ
10
n
64
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-21
Example: Decision
(continued)
Reach a decision and interpret the result:
Reject H0
 = .10
Do not reject H0
0
1.28
Reject H0
z = 0.88
Do not reject H0 since z = 0.88 < 1.28
i.e.: there is not sufficient evidence that the
mean bill is over $52
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-22
Example: p-Value Solution
Calculate the p-value and compare to 
(continued)
(assuming that μ = 52.0)
p-value = .1894
Reject H0
 = .10
0
Do not reject H0
1.28
Z = .88
Reject H0
P(x  53.1| μ  52.0)
53.1 52.0 

 P z 

10/ 64 

 P(z  0.88)  1 .8106
 .1894
Do not reject H0 since p-value = .1894 >  = .10
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-23
One-Tail Tests

In many cases, the alternative hypothesis
focuses on one particular direction
H0: μ ≤ 3
H1: μ > 3
H0: μ ≥ 3
H1: μ < 3
This is an upper-tail test since the
alternative hypothesis is focused on
the upper tail above the mean of 3
This is a lower-tail test since the
alternative hypothesis is focused on
the lower tail below the mean of 3
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-24
Upper-Tail Tests

There is only one
critical value, since
the rejection area is
in only one tail
H0: μ ≤ 3
H1: μ > 3

Do not reject H0
Z
0
x
μ
zα
Reject H0
Critical value x c
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-25
Lower-Tail Tests

There is only one
critical value, since
the rejection area is
in only one tail
H0: μ ≥ 3
H1: μ < 3

Reject H0
-z
Do not reject H0
0
Z
μ
x
Critical value x c
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-26
Two-Tail Tests


In some settings, the
alternative hypothesis does
not specify a unique direction
There are two
critical values,
defining the two
regions of
rejection
H0: μ = 3
H1: μ  3
/2
/2
x
3
Reject H0
Do not reject H0
-z/2
Lower
critical value
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
Reject H0
+z/2
z
Upper
critical value
Ch. 9-27
Hypothesis Testing Example
Test the claim that the true mean # of TV
sets in US homes is equal to 3.
(Assume σ = 0.8)



State the appropriate null and alternative
hypotheses
 H0: μ = 3 , H1: μ ≠ 3
(This is a two tailed test)
Specify the desired level of significance
 Suppose that  = .05 is chosen for this test
Choose a sample size
 Suppose a sample of size n = 100 is selected
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-28
Hypothesis Testing Example
(continued)



Determine the appropriate technique
 σ is known so this is a z test
Set up the critical values
 For  = .05 the critical z values are ±1.96
Collect the data and compute the test statistic
 Suppose the sample results are
n = 100, x = 2.84 (σ = 0.8 is assumed known)
So the test statistic is:
z 
X  μ0
2.84  3
 .16


 2.0
σ
0.8
.08
n
100
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-29
Hypothesis Testing Example
(continued)

Is the test statistic in the rejection region?
Reject H0 if
z < -1.96 or
z > 1.96;
otherwise
do not
reject H0
 = .05/2
Reject H0
-z = -1.96
 = .05/2
Do not reject H0
0
Reject H0
+z = +1.96
Here, z = -2.0 < -1.96, so the
test statistic is in the rejection
region
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-30
Hypothesis Testing Example
(continued)

Reach a decision and interpret the result
 = .05/2
Reject H0
-z = -1.96
 = .05/2
Do not reject H0
0
Reject H0
+z = +1.96
-2.0
Since z = -2.0 < -1.96, we reject the null hypothesis
and conclude that there is sufficient evidence that the
mean number of TVs in US homes is not equal to 3
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-31
Example: p-Value

Compare the p-value with 

If p-value <  , reject H0

If p-value   , do not reject H0
Here: p-value = .0456
 = .05
Since .0456 < .05, we
reject the null
hypothesis
/2 = .025
(continued)
/2 = .025
.0228
.0228
-1.96
-2.0
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
0
1.96
Z
2.0
Ch. 9-32
9.3

t Test of Hypothesis for the Mean
(σ Unknown)
Convert sample result ( x ) to a t test statistic
Hypothesis
Tests for 
σ Known
σ Unknown
Consider the test
H0 : μ  μ0
The decision rule is:
H1 : μ  μ0
Reject H0 if t 
(Assume the population is normal)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
x  μ0
 t n-1, α
s
n
Ch. 9-33
t Test of Hypothesis for the Mean
(σ Unknown)
(continued)

For a two-tailed test:
Consider the test
H0 : μ  μ0
H1 : μ  μ0
(Assume the population is normal,
and the population variance is
unknown)
The decision rule is:
x  μ0
x  μ0
Reject H0 if t 
 t n-1, α/2 or if t 
 t n-1, α/2
s
s
n
n
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-34
Example: Two-Tail Test
( Unknown)
The average cost of a
hotel room in Chicago is
said to be $168 per
night. A random sample
of 25 hotels resulted in
x = $172.50 and
s = $15.40. Test at the
 = 0.05 level.
H0: μ = 168
H1: μ  168
(Assume the population distribution is normal)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-35
Example Solution:
Two-Tail Test
H0: μ = 168
H1: μ  168




= 0.05
/2=.025
Reject H0
-t n-1,α/2
-2.0639
n = 25
 is unknown, so
use a t statistic
t n1 
Critical Value:
t24 , .025 = ± 2.0639
/2=.025
Do not reject H0
Reject H0
0
1.46
t n-1,α/2
2.0639
x μ
172.50  168

 1.46
s
15.40
n
25
Do not reject H0: not sufficient evidence that
true mean cost is different than $168
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Ch. 9-36
9.6
Hypothesis Tests of
one Population Variance
 Goal: Test hypotheses about the
population variance, σ2
 If the population is normally distributed,

2
n 1
(n  1)s

σ2
2
has a chi-square distribution with (n – 1) degrees
of freedom
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 11-37
Hypothesis Tests of
one Population Variance
(continued)
The test statistic for hypothesis tests
about one population variance is
χ n21
(n  1)s 2

2
σ0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 11-38
Decision Rules: Variance
Population variance
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: σ2  σ02
H1: σ2 < σ02
H0: σ2 ≤ σ02
H1: σ2 > σ02
H0: σ2 = σ02
H1: σ2 ≠ σ02


χ n21,1
χ n21,
Reject H0 if
χ
2
n 1
χ
2
n 1,1
Reject H0 if
χ n21  χ n21,
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
/2
/2
χ n21,1 / 2
χ n21, / 2
Reject H0 if
or
χ n21  χ n21, / 2
χ n21  χ n21,1 / 2
Chap 11-39