© 1998 Prentice-Hall, Inc.
Statistics for Managers Using
Microsoft Excel
Fundamentals of Hypothesis Testing
Chapter 7

Learning Objectives




Describe the hypothesis testing process
Distinguish the types of hypotheses
Explain hypothesis testing errors
Solve hypothesis testing problems



7-1
One population mean
One population proportion
One & two-tailed tests
Statistics for Managers Using Microsoft Excel, 1/e
Statistical Methods
© 1998 Prentice-Hall, Inc.
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
7-2
Hypothesis
Testing
Statistics for Managers Using Microsoft Excel, 1/e
Hypothesis Testing
© 1998 Prentice-Hall, Inc.
Population


 


I believe the
population
mean age is 50
(hypothesis).
Reject
hypothesis!
Not close.

Random
sample
Mean 
X = 20
7-3
Statistics for Managers Using Microsoft Excel, 1/e
What’s a Hypothesis?
© 1998 Prentice-Hall, Inc.

A belief about a
population parameter

Parameter is
population mean,
proportion, variance

Must be stated
before analysis
I believe the mean GPA
of this class is 3.5!
© 1984-1994 T/Maker Co.
7-4
Statistics for Managers Using Microsoft Excel, 1/e
Null Hypothesis
© 1998 Prentice-Hall, Inc.

What is tested

Has serious outcome if incorrect
decision made

Always has equality sign: , or 

Designated H0


Example

7-5
Pronounced ‘H sub-zero’ or ‘H oh’
H0:   3
Statistics for Managers Using Microsoft Excel, 1/e
Alternative Hypothesis
© 1998 Prentice-Hall, Inc.

Opposite of null hypothesis

Always has inequality sign: , , or 

Designated H1

Example

7-6
H1:  < 3
Statistics for Managers Using Microsoft Excel, 1/e
Identifying Hypotheses in Problems
© 1998 Prentice-Hall, Inc.

Problem: Test that the population mean
is not 3

Steps


State the question statistically:   3
State the opposite statistically:  = 3


Select the null hypothesis:  = 3

7-7
Must be mutually exclusive & exhaustive
Has the =, , or sign
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Identifying Hypotheses Thinking
Challenge
What are the hypotheses?

7-8
Is the population average amount of TV
viewing 12 hours?
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Identifying Hypotheses Thinking
Challenge
What are the hypotheses?

Is the population average amount of TV
viewing 12 hours?

Is the population average amount of TV
viewing different from 12 hours?
7-9
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Identifying Hypotheses Thinking
Challenge
What are the hypotheses?

Is the population average amount of TV
viewing 12 hours?

Is the population average amount of TV
viewing different from 12 hours?

Is the average cost per hat less than or
equal to $20?
7 - 10
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Identifying Hypotheses Thinking
Challenge
What are the hypotheses?

Is the population average amount of TV
viewing 12 hours?

Is the population average amount of TV
viewing different from 12 hours?

Is the average cost per hat less than or
equal to $20?

Is the average amount spent in the
bookstore greater than $25?
7 - 11
Statistics for Managers Using Microsoft Excel, 1/e
Basic Idea
© 1998 Prentice-Hall, Inc.
Sampling Distribution
It is unlikely
that we would
get a sample
mean of this
value ...
... therefore,
we reject the
hypothesis
that  = 50.
... if in fact this were
the population mean
20
7 - 12
 = 50
Sample Mean
H0
Statistics for Managers Using Microsoft Excel, 1/e
Level of Significance
© 1998 Prentice-Hall, Inc.

Defines unlikely values of sample
statistic if null hypothesis is true


Designated (alpha)


Called rejection region of sampling
distribution
Typical values are .01, .05, .10
Selected by researcher at start
7 - 13
Statistics for Managers Using Microsoft Excel, 1/e
Rejection Region (One-Tail Test)
© 1998 Prentice-Hall, Inc.
Sampling Distribution
Level of Confidence
Rejection
Region

1-
Nonrejection
Region
Critical
Value
7 - 14
Ho
Value
Sample Statistic
Observed sample statistic
Statistics for Managers Using Microsoft Excel, 1/e
Rejection Region (One-Tail Test)
© 1998 Prentice-Hall, Inc.
Sampling Distribution
Level of Confidence
Rejection
Region

1-
Nonrejection
Region
Critical
Observed
sample statistic Value
7 - 15
Ho
Value
Sample Statistic
Statistics for Managers Using Microsoft Excel, 1/e
Rejection Regions (Two-Tailed Test)
© 1998 Prentice-Hall, Inc.
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1-
1/2 
1/2 
Nonrejection
Region
Critical
Value
7 - 16
Ho
Sample Statistic
Value Critical
Observed
Value
sample statistic
Statistics for Managers Using Microsoft Excel, 1/e
Rejection Regions (Two-Tailed Test)
© 1998 Prentice-Hall, Inc.
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1-
1/2 
1/2 
Nonrejection
Region
Critical
Value
7 - 17
Ho
Sample Statistic
Value Critical
Value
Observed
sample statistic
Statistics for Managers Using Microsoft Excel, 1/e
Rejection Regions (Two-Tailed Test)
© 1998 Prentice-Hall, Inc.
Sampling Distribution
Level of Confidence
Rejection
Region
Rejection
Region
1-
1/2 
1/2 
Nonrejection
Region
Observed
sample statistic
7 - 18
Critical
Value
Ho
Sample Statistic
Value Critical
Value
Statistics for Managers Using Microsoft Excel, 1/e
Risk of Errors in Making Decision
© 1998 Prentice-Hall, Inc.

Type I error




Reject true null hypothesis
Has serious consequences
Probability of Type I error is 
 Called level of significance
Type II error
7 - 19

Do not reject false null hypothesis

Probability of Type II error is  (Beta)
Statistics for Managers Using Microsoft Excel, 1/e
Decision Results
© 1998 Prentice-Hall, Inc.
H0: Innocent
Jury Trial
H0 Test
Actual Situation
Verdict
Innocent Guilty Decision H0 True
Innocent Correct
Guilty
7 - 20
Actual Situation
Error
Error
Correct
Do Not
Reject
H0
Reject
H0
1-
H0
False
Type II
Error
()
Type I
Power
Error () (1 - )
Statistics for Managers Using Microsoft Excel, 1/e
 &  Have an Inverse Relationship
© 1998 Prentice-Hall, Inc.
You can’t reduce both
errors simultaneously!


7 - 21
Statistics for Managers Using Microsoft Excel, 1/e
Factors Affecting 
© 1998 Prentice-Hall, Inc.

True value of population parameter
  increases when difference with
hypothesized parameter decreases

Significance level, 

Population standard deviation, 

Sample size, n
7 - 22
 Increases when decreases
 Increases when  increases
 Increases when n decreases
Statistics for Managers Using Microsoft Excel, 1/e
Hypothesis H0 Testing Steps
© 1998 Prentice-Hall, Inc.

State H0

State H1

Choose 

Choose n

Choose test
7 - 23
Statistics for Managers Using Microsoft Excel, 1/e
H0 Testing Steps
© 1998 Prentice-Hall, Inc.

State H0

Set up critical values

State H1

Collect data

Choose 

Compute test statistic

Choose n

Make statistical decision

Choose test

Express decision
7 - 24
Statistics for Managers Using Microsoft Excel, 1/e
One Population Tests
© 1998 Prentice-Hall, Inc.
One
Population

Unknown
Proportion
Z Test
t Test
Z Test
(1
(1 &
& 22
tail)
tail)
(1
(1 &
& 22
tail)
tail)
(1
(1 &
& 22
tail)
tail)

Known
7 - 25
Mean
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.

Two-Tailed Z Test for Mean
( Known)
Assumptions


Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)

Null hypothesis has = sign only

Z-test statistic
X  x
X 
Z 


x
n
7 - 26
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed Z Test Example
© 1998 Prentice-Hall, Inc.
Does an average box of
cereal contain 368 grams
of cereal? A random
sample of 25 boxes
showed
X = 372.5. The
company has specified 
to be 15 grams. Test at
the .05 level.
7 - 27
368 gm.
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed Z Test Solution
© 1998 Prentice-Hall, Inc.
H0:  = 368
Test Statistic:
H1:   368
X   372.5  368
Z

 1.50

15
n
25
  .05
n  25
Critical Value(s):
Reject H00
Reject H00
.025
.025
-1.96 0 1.96 Z
7 - 28
Decision:
Do not reject at  = .05
Conclusion:
No evidence
average is not 368
Statistics for Managers Using Microsoft Excel, 1/e
p-Value
© 1998 Prentice-Hall, Inc.

Probability of obtaining a test statistic
more extreme ( or than actual
sample value given H0 is true

Called observed level of significance


Used to make rejection decision


7 - 29
Smallest value of  H0 can be rejected
If p-value  , do not reject H0
If p-value < , reject H0
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed Z Test
p-Value Solution
© 1998 Prentice-Hall, Inc.
p-value is P(Z  -1.50 or Z  1.50) = .1336
1/2 p-Value
.0668
1/2 p-Value

.0668 .5000
- .4332
.0668
.4332
-1.50 0 1.50

7 - 30
From Z table:
lookup 1.50

Z
Z value of sample
statistic
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed Z Test
p-Value Solution
© 1998 Prentice-Hall, Inc.
(p-Value = .1336)  ( = .05).
Do not reject.
1/2 p-Value = .0668
1/2 p-Value = .0668
Reject
Reject
1/2  = .025
1/2  = .025
-1.50 0 1.50
Z
Test statistic is in ‘Do not reject’ region
7 - 31
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Two-Tailed Z Test Thinking
Challenge
You’re a Q/C inspector. You want to
find out if a new machine is making
electrical cords to customer
specification: average breaking
strength of 70 lb. with  = 3.5 lb.
You take a sample of 36 cords &
compute a sample mean of 69.7 lb.
At the .05 level, is there evidence
that the machine is not meeting the
average breaking strength?
7 - 32
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:  = 70
Test Statistic:
H1:   70
 = .05
n = 36
Critical Value(s):
Reject H0
Reject H0
.025
.025
-1.96 0 1.96 Z
7 - 33
X   69.7  70
Z

  .51

3.5
n
36
Decision:
Do not reject at  = .05
Conclusion:
No evidence
average is not 70
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.

Assumptions



One-Tailed Z Test for Mean
( Known)
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
Null hypothesis has  or  sign only
7 - 34
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.

One-Tailed Z Test for Mean
( Known)
Assumptions


Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)

Null hypothesis has  or  sign only

Z-test statistic
X  x
X  
Z 


x
n
7 - 35
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test for
Mean Hypotheses
© 1998 Prentice-Hall, Inc.
H0:0 H1: < 0
H0:0 H1: > 0
Reject H 0
Reject H 0


0
Must be significantly
below 
7 - 36
Z
0
Z
Small values satisfy
H0 . Don’t reject!
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test
Finding Critical Z
© 1998 Prentice-Hall, Inc.
What Is Z given  = .025?
.500
- .025
.475

=1
 = .025

0
7 - 37
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test
Finding Critical Z
© 1998 Prentice-Hall, Inc.
What Is Z given  = .025?
.500
- .025
.475

=1

Standardized Normal
Probability Table (Portion)
Z
.06
.07
1.6 .4505 .4515 .4525
 = .025

0 1.96 Z
1.7 .4599 .4608 .4616
1.8 .4678 .4686 .4693
 1.9
7 - 38
.05
.4744 .4750 .4756
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Tailed Z Test Example
Does an average box of
cereal contain more than
368 grams of cereal? A
random sample of 25
boxes showedX = 372.5.
The company has specified
 to be 15 grams. Test at
the .05 level.
368 gm.
7 - 39
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution
© 1998 Prentice-Hall, Inc.
H0:   368
Test Statistic:
H1:  > 368
X   372.5  368
Z

 1.50

15
n
25
 = .05
n = 25
Critical Value(s):
Reject
.05
0 1.645 Z
7 - 40
Decision:
Do not reject at  = .05
Conclusion:
No evidence average
is more than 368
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Tailed Z Test
p-Value Solution
(p-Value = .0668)  ( = .05).
Do not reject.
p-Value = .0668
Reject
 = .05
0 1.50
Z
Test statistic is in ‘Do not reject’ region
7 - 41
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Tailed Z Test Thinking
Challenge
You’re an analyst for Ford. You
want to find out if the average
miles per gallon of Escorts is at
least 32 mpg. Similar models
have a standard deviation of
3.8 mpg. You take a sample of
60 Escorts & compute a sample
mean of 30.7 mpg. At the .01
level, is there evidence that the
miles per gallon is at least 32?
7 - 42
Statistics for Managers Using Microsoft Excel, 1/e
Solution Template
© 1998 Prentice-Hall, Inc.
H0:
Test Statistic:
H1:
=
n=
Critical Value(s):
Decision:
Conclusion:
0
7 - 43
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:
Test Statistic:
H1:
=
n=
Critical Value(s):
Decision:
Conclusion:
7 - 44
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   32
Test Statistic:
H1:  < 32
=
n=
Critical Value(s):
Decision:
Conclusion:
7 - 45
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   32
Test Statistic:
H1:  < 32
 = .01
n = 60
Critical Value(s):
Decision:
Conclusion:
7 - 46
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   32
Test Statistic:
H1:  < 32
 = .01
n = 60
Critical Value(s):
Decision:
Reject
.01
-2.33 0
7 - 47
Conclusion:
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:  32
Test Statistic:
H1:  < 32
 = .01
n = 60
Critical Value(s):
X   30.7  32
Z

 2.65

3.8
n
60
Decision:
Reject
.01
-2.33 0
7 - 48
Conclusion:
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   32
Test Statistic:
H1:  < 32
 = .01
n = 60
Critical Value(s):
X   30.7  32
Z

 2.65

3.8
n
60
Reject
Decision:
Reject at  = .01
.01
Conclusion:
-2.33 0
7 - 49
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   32
Test Statistic:
H1:  < 32
 = .01
n = 60
Critical Value(s):
Reject
.01
-2.33 0
7 - 50
Z
X   30.7  32
Z

 2.65

3.8
n
60
Decision:
Reject at  = .01
Conclusion:
There is evidence
average is less than 32
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
p-Value
Thinking Challenge
You’re an analyst for Ford. You
want to find out if the average
miles per gallon of Escorts is at
least 32 mpg. Similar models
have a standard deviation of 3.8
mpg. You take a sample of 60
Escorts & compute a sample
mean of 30.7 mpg. What is the
value of the observed level of
significance (p-Value)?
7 - 51
Statistics for Managers Using Microsoft Excel, 1/e
p-Value
Solution*
© 1998 Prentice-Hall, Inc.
p-Value is P(Z  -2.65) = .004.
p-Value < ( = .01). Reject H0.

Use
alternative
hypothesis
to find
direction

p-Value
.004
.5000
- .4960
.0040
.4960
-2.65 0

7 - 52
Z value of
sample statistic

Z
From Z table:
lookup 2.65
Statistics for Managers Using Microsoft Excel, 1/e
One Population Tests
© 1998 Prentice-Hall, Inc.
One
Population

Unknown
Proportion
Z Test
t Test
Z Test
(1 & 2
tail)
(1 & 2
tail)
(1 & 2
tail)

Known
7 - 53
Mean
Statistics for Managers Using Microsoft Excel, 1/e
t Test for Mean
© 1998 Prentice-Hall, Inc.

( Unknown)
Assumptions


Population is normally distributed
If not normal, only slightly skewed &
large sample (n  30) taken

Parametric test procedure

t test statistic
7 - 54
X  
t 
S
n
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2 

 /2 = .05
-2.920 0 2.920 t
 /2 = .05
7 - 55
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Two-Tailed t Test
Finding Critical t Values
Given: n = 3;  = .10
df = n - 1 = 2 

Upper Tail Area

 /2 = .05
7 - 56
df
.25
.10
.05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
-2.920 0 2.920 t
 /2 = .05
Critical Values of t Table
(Portion)
3 0.765 1.638 2.353

Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Two-Tailed t Test Example
Does an average box of
cereal contain 368
grams of cereal? A
random sample of 36
boxes had a mean of
372.5 & a standard
deviation of 12 grams.
Test at the .05 level.
368 gm.
7 - 57
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed t Test Solution
© 1998 Prentice-Hall, Inc.
H0:  = 368
Test Statistic:
H1:   368
 = .05
df = 36 - 1 = 35
Critical Value(s):
Decision:
Conclusion:
7 - 58
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed t Test Solution
© 1998 Prentice-Hall, Inc.
H0:  = 368
Test Statistic:
H1:   368
X   372.5  368
t

 2.25
S
12
n
36
 = .05
df = 36 - 1 = 35
Critical Value(s):
Reject H0
Reject H0
.025
.025
-2.0301 0 2.0301
7 - 59
t
Decision:
Reject at  = .05
Conclusion:
There is evidence pop.
average is not 368
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
Two-Tailed t Test
Thinking Challenge
You work for the FTC. A
manufacturer of detergent
claims that the mean weight
of detergent is 3.25 lb. You
take a random sample of 64
containers. You calculate the
sample average to be 3.238 lb.
with a standard deviation of
.117 lb. At the .01 level, is the
manufacturer correct?
7 - 60
3.25 lb.
Statistics for Managers Using Microsoft Excel, 1/e
Two-Tailed t Test Solution*
© 1998 Prentice-Hall, Inc.
H0:  = 3.25
Test Statistic:
H1:   3.25
X   3.238  3.25
t

  .82
S
.117
n
64
  .01
df  64 - 1 = 63
Critical Value(s):
Reject H0
Reject H0
.005
.005
-2.6561 0 2.6561
7 - 61
t
Decision:
Do not reject at  = .01
Conclusion:
There is no evidence
average is not 3.25
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed t Test Example
© 1998 Prentice-Hall, Inc.
Is the average capacity of
batteries at least 140
ampere-hours? A random
sample of 20 batteries had
a mean of 138.47 & a
standard deviation of 2.66.
Assume a normal
distribution. Test at the .05
level.
7 - 62
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed t Test Solution
© 1998 Prentice-Hall, Inc.
H0:   140
Test Statistic:
H1:  < 140
X   138.47  140
t

 2.57
S
2.66
 = .05
n
20
df = 20 - 1 = 19
Critical Value(s):
Decision:
Reject at  = .05
Reject
.05
-1.7291 0
7 - 63
t
Conclusion:
There is evidence pop.
average is less than 140
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Tailed t Test
Thinking Challenge
You’re a marketing analyst for
Wal-Mart. Wal-Mart had teddy
bears on sale last week. The
weekly sales ($ 00) of bears
sold in 10 stores was: 8 11 0
4 7 8 10 5 8 3.
At the .05 level, is there
evidence that the average
bear sales per store is more
than 5 ($ 00)?
7 - 64
Statistics for Managers Using Microsoft Excel, 1/e
One-Tailed t Test Solution*
© 1998 Prentice-Hall, Inc.
H0:   5
Test Statistic:
H1:  > 5
X   6.4  5
t

 1.31
S
3.373
n
10
 = .05
df = 10 - 1 = 9
Critical Value(s):
Reject
.05
0 1.8331
7 - 65
t
Decision:
Do not reject at  = .05
Conclusion:
There is no evidence
average is more than 5
Statistics for Managers Using Microsoft Excel, 1/e
Data Types
© 1998 Prentice-Hall, Inc.
Data
Numerical
(Quantitative)
Discrete
7 - 66
Categorical
(Qualitative)
Continuous
Statistics for Managers Using Microsoft Excel, 1/e
Categorical Data
© 1998 Prentice-Hall, Inc.

Categorical random variables yield
responses that classify

e.g., gender (male, female)

Measurement reflects # in category

Nominal or ordinal scale

Examples


7 - 67
Do you own savings bonds?
Do you live on-campus or off-campus?
Statistics for Managers Using Microsoft Excel, 1/e
Proportions
© 1998 Prentice-Hall, Inc.

Involve categorical variables

Fraction or % of population in a category

If two categorical outcomes, binomial
distribution

7 - 68
Possess or don’t possess characteristic
Statistics for Managers Using Microsoft Excel, 1/e
Proportions
© 1998 Prentice-Hall, Inc.

Involve categorical variables

Fraction or % of population in a category

If two categorical outcomes, binomial
distribution


7 - 69
Possess or don’t possess characteristic
Sample proportion (ps)
X
num ber of successes
ps 

n
sam ple size
Statistics for Managers Using Microsoft Excel, 1/e
One Population Tests
© 1998 Prentice-Hall, Inc.
One
Population

Unknown
Proportion
Z Test
t Test
Z Test
(1
(1 &
& 22
tail)
tail)
(1
(1 &
& 22
tail)
tail)
(1
(1 &
& 22
tail)
tail)

Known
7 - 70
Mean
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.

Assumptions



7 - 71
One-Sample Z Test
for Proportion
Two categorical outcomes
Population follows binomial distribution
Normal approximation can be used
 n·p  5 & n·(1 - p)  5
Statistics for Managers Using Microsoft Excel, 1/e
One-Sample Z Test
for Proportion
© 1998 Prentice-Hall, Inc.

Assumptions




Two categorical outcomes
Population follows binomial distribution
Normal approximation can be used
 n·p  5 & n·(1 - p)  5
Z-test statistic for proportion
Z 
7 - 72
ps  P
P  (1  P )
n
Hypothesized
population proportion
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Proportion Z Test Example
The present packaging
system produces 10%
defective cereal boxes.
Using a new system, a
random sample of 200
boxes had 11 defects.
Does the new system
produce fewer defects?
Test at the .05 level.
7 - 73

n·p  5
n·(1 - p)  5
Statistics for Managers Using Microsoft Excel, 1/e
One-Proportion Z Test Solution
© 1998 Prentice-Hall, Inc.
H0: p  .10
Test Statistic:
H1: p < .10
 = .05
n = 200
Critical Value(s):
Decision:
Reject
.05
-1.645 0
7 - 74
Conclusion:
Z
Statistics for Managers Using Microsoft Excel, 1/e
One-Proportion Z Test Solution
© 1998 Prentice-Hall, Inc.
H0: p  .10
H1: p < .10
 = .05
Z
n = 200
Critical Value(s):
Reject
.05
-1.645 0
7 - 75
Z
Test Statistic:
11
 .10
ps  p
 200
 2.12
p  (1  p )
.10  (1 .10)
n
200
Decision:
Reject at  = .05
Conclusion:
There is evidence new
system < 10% defective
Statistics for Managers Using Microsoft Excel, 1/e
© 1998 Prentice-Hall, Inc.
One-Proportion Z Test Thinking
Challenge
You’re an accounting
manager. A year-end audit
showed 4% of transactions
had errors. You implement
new procedures. A random
sample of 500 transactions
had 25 errors. Has the
proportion of incorrect
transactions changed at the
.05 level?
7 - 76

n·p  5
n·(1 - p)  5
Statistics for Managers Using Microsoft Excel, 1/e
One-Proportion Z Test Solution*
© 1998 Prentice-Hall, Inc.
H0: p = .04
H1: p  .04
 = .05
Z
n = 500
Critical Value(s):
Reject H0
Reject H0
.025
.025
-1.96 0 1.96 Z
7 - 77
Test Statistic:
25

.
04
ps  p
 500
 114
.
p  (1  p )
.04  (1 .04)
n
500
Decision:
Do not reject at  = .05
Conclusion:
There is evidence
proportion is still 4%
Statistics for Managers Using Microsoft Excel, 1/e
Conclusion
© 1998 Prentice-Hall, Inc.

Described the hypothesis testing
process

Distinguished the types of hypotheses

Explained hypothesis testing errors

Solved hypothesis testing problems



7 - 78
One population mean
One population proportion
One & two-tailed tests
Statistics for Managers Using Microsoft Excel, 1/e