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Statistics
Hypothesis Tests
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Contents
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Developing Null and Alternative Hypotheses
Type I and Type II Errors
Population Mean: σ Known
Population Mean: σ Unknown
Population Proportion
Hypothesis Testing and Decision Making
Calculating the Probability of Type II Errors
Determining the Sample Size for
Hypothesis Tests About a Population Mean
Contents
Population Proportion
 Hypothesis Testing and Decision Making
 Calculating the Probability of Type II Errors
 Determining the Sample Size for
Hypothesis Tests About a Population Mean

STATISTICS in PRACTICE


John Morrell & Company is considered
the oldest continuously operating meat
manufacturer in the United States.
Market research at Morrell provides
management with up to- date information on the
company’s various products and how the
products compare with competing brands of
similar products.
STATISTICS in PRACTICE


One research question concerned whether
Morrell’s Convenient Cuisine Beef Pot Roast
was the preferred choice of more than 50% of
the consumer population. The hypothesis test
for the research question is Ho: p ≤ .50 vs.
Ha: p > .50
In this chapter we will discuss how to formulate
hypotheses and how to conduct tests like the
one used by Morrell.
Developing Null and Alternative
Hypotheses

Hypothesis testing can be used to determine
whether a statement about the value of a
population parameter should or should not be
rejected.

The null hypothesis, denoted by Ho , is a
tentative assumption about a population
parameter.
Developing Null and Alternative
Hypotheses

The alternative hypothesis, denoted by Ha,
is the opposite of what is stated in the null
hypothesis.

The alternative hypothesis is what the test is
attempting to establish.
Developing Null and Alternative
Hypotheses

Testing Research Hypotheses

Testing the Validity of a Claim

Testing in Decision-Making Situations
Developing Null and Alternative
Hypotheses

Testing Research Hypotheses

The research hypothesis should be expressed
as the alternative hypothesis.

The conclusion that the research hypothesis
is true comes from sample data that contradict
the null hypothesis.
Developing Null and Alternative
Hypotheses

Testing Research Hypotheses
Example:
Consider a particular automobile model that
currently attains an average fuel efficiency of
24 miles per gallon.
A product research group developed a new
fuel injection system specifically designed to
increase the miles-per-gallon rating.
Developing Null and Alternative
Hypotheses
The appropriate null and alternative hypotheses
for the study are:
Ho : μ ≤ 24
Ha : μ > 24
Developing Null and Alternative
Hypotheses

Testing Research Hypotheses
• In research studies such as these, the null and
alternative hypotheses should be formulated
so that the rejection of H0 supports the
research conclusion. The research hypothesis
therefore should be expressed as the
alternative hypothesis.
Developing Null and Alternative
Hypotheses

Testing the Validity of a Claim
• Manufacturers’ claims are usually given the
benefit of the doubt and stated as the null
hypothesis.
• The conclusion that the claim is false comes
from sample data that contradict the null
hypothesis.
Developing Null and Alternative
Hypotheses
Testing the Validity of a Claim
Example:
Consider the situation of a manufacturer of soft
drinks who states that it fills two-liter containers
of its products with an average of at least 67.6
fluid ounces.

Developing Null and Alternative
Hypotheses
Testing the Validity of a Claim
A sample of two-liter containers will be selected,
and the contents will be measured to test the
manufacturer’s claim.
The null and alternative hypotheses as follows.
Ho : μ ≥ 67. 6
Ha : μ < 67. 6

Developing Null and Alternative
Hypotheses
•
Testing the Validity of a Claim
In any situation that involves testing the validity
of a claim, the null hypothesis is generally based
on the assumption that the claim is true. The
alternative hypothesis is then formulated so that
rejection of H0 will provide statistical evidence
that the stated assumption is incorrect.
Action to correct the claim should be considered
whenever H0 is rejected.

Developing Null and Alternative
Hypotheses
Testing in Decision-Making Situations
• A decision maker might have to choose
between two courses of action, one
associated with the null hypothesis and
another associated with the alternative
hypothesis.

Developing Null and Alternative
Hypotheses

Testing in Decision-Making Situations
Example:
Accepting a shipment of goods from a supplier
or returning the shipment of goods to the supplier.
Assume that specifications for a particular part
require a mean length of two inches per part.
The null and alternative hypotheses would be
formulated as follows.
H0 : μ = 2
Ha : μ ≠ 2
Developing Null and Alternative
Hypotheses

•
•
Testing in Decision-Making Situations
If the sample results indicate H0 cannot be
rejected and the shipment will be accepted.
If the sample results indicate H0 should be
rejected the parts do not meet specifications.
The quality control inspector will have
sufficient evidence to return the shipment to
the supplier.
Developing Null and Alternative
Hypotheses

Testing in Decision-Making Situations
• We see that for these types of situations,
action is taken both when H0 cannot be
rejected and when H0 can be rejected.
Summary of Forms for Null and
Alternative Hypotheses about a
Population Mean

The equality part of the hypotheses always
appears in the null hypothesis.(Follows Minitab)

In general, a hypothesis test about the value of
a population mean  must take one of the
following three forms (where 0 is the
hypothesized value of the population mean).
Summary of Forms for Null and
Alternative Hypotheses about a
Population Mean
H 0 :   0
H a :   0
H 0 :   0
H a :   0
H 0 :   0
H a :   0
One-tailed One-tailed Two-tailed
(lower-tail) (upper-tail)
Null and Alternative Hypotheses
Example: Metro EMS
A major west coast city provides
one of the most comprehensive
emergency medical services in
the world.
Operating in a multiple
hospital system with
approximately 20 mobile medical
units, the service goal is to respond to medical
emergencies with a mean time of 12 minutes or less.

Null and Alternative Hypotheses

Example: Metro EMS
The director of medical services
wants to formulate a hypothesis
test that could use a sample of
emergency response times to
determine whether or not the
service goal of 12 minutes or less
is being achieved.
Null and Alternative Hypotheses
H0:   The emergency service is meeting
the response goal; no follow-up
action is necessary.
Ha:  The emergency service is not
meeting the response goal;
appropriate follow-up action is
necessary.
where: μ = mean response time for the population
of medical emergency requests
Type I and Type II Errors

Because hypothesis tests are based on sample
data, we must allow for the possibility of errors.

A Type I error is rejecting H0 when it is true.
Applications of hypothesis testing that only
control the Type I error are often called
significance tests.

Type I and Type II Errors
Type I error
 The probability of making a Type I error
when the null hypothesis is true as an
equality is called the level of significance.
 The Greek symbol α(alpha) is used to
denote the level of significance, and
common choices for α are .05 and .01.
Type I and Type II Errors
Type I error
 If the cost of making a Type I error is high,
small values of α are preferred. If the cost of
making a Type I error is not too high, larger
values of α are typically used.
Type I and Type II Errors
Type II error
 A Type II error is accepting H0 when it is false.
 It is difficult to control for the probability of
making a Type II error. Most applications of
hypothesis testing control for the probability
of making a Type I error.
 Statisticians avoid the risk of making a Type II
error by using “do not reject H0” and not
“accept H0”.
Type I and Type II Errors
Population Condition
Conclusion
H0 True
( < 12)
H0 False
( > 12)
Not reject H0
Correct
(Conclude  < 12) Decision
Type II Error
Reject H0
Type I Error
(Conclude  > 12)
Correct
Decision
Steps of Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.
Step 2. Specify the level of significance α .
Step 3. Collect the sample data and compute
the test statistic.
p-Value Approach
Step 4. Use the value of the test statistic to compute
the p-value.
Step 5. Reject H0 if p-value < α.
Steps of Hypothesis Testing
Critical Value Approach
Step 4. Use the level of significance to
determine the critical value and the
rejection rule.
Step 5. Use the value of the test statistic and
the rejection rule to determine whether to
reject H0.
p-Value Approach to
One-Tailed Hypothesis Testing
The p-value is the probability, computed using
the test statistic, that measures the support
(or lack of support) provided by the sample
for the null hypothesis. The p-value is also called
the observed level of significance.
 If the p-value is less than or equal to the level
of significance α , the value of the test statistic
is in the rejection region.
 Reject H0 if the p-value < α .

Critical Value Approach to
One-Tailed Hypothesis Testing


The test statistic z has a standard normal
probability distribution.
We can use the standard normal probability
distribution table to find the z-value with an
area of α in the lower (or upper) tail of the
distribution.
Critical Value Approach to
One-Tailed Hypothesis Testing

The value of the test statistic that established
the boundary of the rejection region is called
the critical value for the test.

The rejection rule is:
• Lower tail: Reject H0 if z < - z α
• Upper tail: Reject H0 if z > z α
One-Tailed Hypothesis Testing


Population Mean: σ Known
One-tailed tests about a population mean
take one of the following two forms.
Lower Tail Test
Upper Tail Test
H 0 :   0
H a :   0
H 0 :   0
H a :   0
Lower-Tailed Test About a
Population Mean:s Known

p-Value < a ,
so reject H0.
p-Value Approach
a = .10
Sampling
distribution
x  0
of z 
/ n
p-value
7
z
z = -za =
-1.46 -1.28
0
Lower-Tailed Test About a
Population Mean : s Known

p-Value < a ,
so reject H0.
p-Value Approach
Sampling
distribution
x  0
of z 
/ n
a = .04
p-Value

z
0
za =
1.75
z=
2.29
Lower-Tailed Test About a
Population Mean : s Known

Critical Value Approach
Sampling
distribution
x  0
of z 
/ n
Reject H0
a 
Do Not Reject H0
z
za = 1.28
0
Upper-Tailed Test About a
Population Mean : s Known

Critical Value Approach
Sampling
distribution
x  0
of z 
/ n
Reject H0
Do Not Reject H0
a
z
0
za = 1.645
One-Tailed Hypothesis Testing
Example:
 The label on a large can of Hilltop Coffee
states that the can contains 3 pounds of
coffee. The Federal Trade Commission
(FTC) knows that Hilltop’s production
process cannot place exactly 3 pounds of
coffee in each can, even if the mean filling
weight for the population of all cans filled is
3 pounds per can.
One-Tailed Hypothesis Testing
Example:
 However, as long as the population mean
filling weight is at least 3 pounds per can,
the rights of consumers will be protected.
One-Tailed Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.
denoting the population mean filling weight and
the hypothesized value of the population mean is
μ0 = 3. The null and alternative hypotheses are
H0 : μ ≥ 3
Ha : μ < 3
Step 2. Specify the level of significance α.
we set the level of significance for the hypothesis
test at α= .01
One-Tailed Hypothesis Testing
Step 3. Collect the sample data and compute the
test statistic.
Test statistic:
z
x  o
/ n
Sample data: σ = .18 and sample size n=36,
x = 2.92 pounds.
One-Tailed Hypothesis Testing
Step 3. Collect the sample data and compute the
test statistic.
z
x  o
x
x 3

.03
We have z = -2.67
One-Tailed Hypothesis Testing
p-Value Approach
Step 4. Use the value of the test statistic to
compute the p-value.
Using the standard normal distribution table, the
area between the mean and z = - 2.67 is .4962.
The p-value is .5000 - .4962 = .0038,

Step 5. p-value = .0038 < α = .01
Reject H0.
One-Tailed Hypothesis Testing

p-value for The Hilltop Coffee Study when
sample mean x = 2.92 and z = -2 .67.
One-Tailed Hypothesis Testing
Critical Value Approach
Step 4. Use the level of significance to determine
the critical value and the rejection rule.
The critical value is z = -2.33.
One-Tailed Hypothesis Testing
Step 5. Use the value of the test statistic and
the rejection rule to determine whether to
reject H0.
We will reject H0 if z < - 2.33.
Because z = - 2.67 < 2.33, we can reject H0
and conclude that Hilltop Coffee is under
filling cans.
One-Tailed Tests About a
Population Mean : s Known
Example: Metro EMS
The response times for a random
sample of 40 medical emergencies
were tabulated. The sample mean
is 13.25 minutes. The population
standard deviation is believed to
be 3.2 minutes.

One-Tailed Tests About a
Population Mean : s Known
Example: Metro EMS
The EMS director wants to
perform a hypothesis test, with a
.05 level of significance, to determine
whether the service goal of 12 minutes or less
is being achieved.

One-Tailed Tests About a
Population Mean: s Known
 p -Value and Critical Value Approaches
1. Develop the hypotheses.
H0:  
Ha: 
2. Specify the level of significance. a = .05
3. Compute the value of the test statistic.
z
x
13.25  12

 2.47
 / n 3.2 / 40
One-Tailed Tests About a
Population Mean: s Known

p –Value Approach
4. Compute the p –value.
For z = 2.47, cumulative probability = .9932.
p–value = 1  .9932 = .0068
5. Determine whether to reject H0.
Because p–value = .0068 < α = .05, we reject H0.
We are at least 95% confident that Metro EMS
is not meeting the response goal of 12 minutes.
One-Tailed Tests About a
Population Mean: s Known

p –Value Approach
Sampling
distribution
x  0
of z 
/ n
a = .05
p-value

z
0
za =
1.645
z=
2.47
One-Tailed Tests About a
Population Mean: s Known

Critical Value Approach
4. Determine the critical value and rejection rule.
For α = .05, z.05 = 1.645
Reject H0 if z > 1.645
5. Determine whether to reject H0.
Because 2.47 > 1.645, we reject H0.
We are at least 95% confident that Metro EMS
is not meeting the response goal of 12 minutes.
Two-Tailed Hypothesis Testing

In hypothesis testing, the general form for a
two-tailed test about a population mean is as
follows:
H 0:    0
H a:    0
p-Value Approach to
Two-Tailed Hypothesis Testing
 Compute the p-value using the following three
steps:
1. Compute the value of the test statistic z.
2. If z is in the upper tail (z > 0), find the area
under the standard normal curve to the right
of z. If z is in the lower tail (z < 0), find the
area under the standard normal curve to the
left of z.
p-Value Approach to
Two-Tailed Hypothesis Testing
3. Double the tail area obtained in step 2 to obtain
the p –value.
 The rejection rule:
Reject H0 if the p-value < a
.
Critical Value Approach to
Two-Tailed Hypothesis Testing

The critical values will occur in both the lower
and upper tails of the standard normal curve.

Use the standard normal probability
distribution table to find za/2 (the z-value
with an area of a/2 in the upper tail of
the distribution).

The rejection rule is:
Reject H0 if z < -za/2 or z > za/2
Two-Tailed Hypothesis Testing


Example: The U.S. Golf Association
(USGA) establishes rules that
manufacturers of golf equipment must meet
if their products are to be acceptable for use
in USGA events.
MaxFlight produce golf balls with an
average distance of 295 yards.
Two-Tailed Hypothesis Testing

When the average distance falls below 295
yards, the company worries about losing
sales because the golf balls do not provide
as much distance as advertised.
Two-Tailed Hypothesis Testing


When the average distance passes 295 yards,
MaxFlight’s golf balls may be rejected by the
USGA for exceeding the overall distance
standard concerning carry and roll.
A hypothesized value of μ0= 295 and the
null and alternative hypotheses test are as
follows:
H0 : μ = 295 VS Ha : μ≠ 295
Two-Tailed Hypothesis Testing

If the sample mean x is significantly less
than
295 yards or significantly greater than 295
yards, we will reject H0.
Two-Tailed Hypothesis Testing
Step 1. Develop the null and alternative
hypotheses.
A hypothesized value of μ0 = 295 and
H0 : μ = 295
Ha : μ≠ 295
Step 2. Specify the level of significance α.
we set the level of significance for the
hypothesis test at α = .05
Two-Tailed Hypothesis Testing
Step 3. Collect the sample data and compute the
test statistic.
Test statistic:
x
z
o
/ n
Sample data: σ = 12 and sample size n=50,
x = 297.6 yards.
We have
x   o 297.6  295
z

 1.53
/ n
12 / 50
Two-Tailed Hypothesis Testing
p-Value Approach
Step 4. Use the value of the test statistic to
compute the p-value.
The two-tailed p-value in this case is given by
P(z < -1.53) + P(z > 1.53)= 2(.0630) = .1260.
Step 5. We do not reject H0 because the
p-value = .1260 > .05. No action will be taken
to adjust manufacturing process.

Two-Tailed Hypothesis Testing

p-Value for The Maxflight Hypothesis Test
Two-Tailed Hypothesis Testing
Critical Value Approach
Step 4. Use the level of significance to determine
the critical value and the rejection rule. The
critical values are - z.025 = -1.96 and z.025 = 1.96.
Two-Tailed Hypothesis Testing
Step 5. Use the value of the test statistic and
the rejection rule to determine whether to
reject H0.
We will reject H0 if z < - 1.96 or if z > 1.96
Because the value of the test statistic is
z=1.53, the statistical evidence will not
permit us to reject the null hypothesis
at the .05 level of significance.
Example: Glow Toothpaste
Two-Tailed Test About a Population Mean: s
Known
The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
will be selected in order to check the
filling process.

Example: Glow Toothpaste

Two-Tailed Test About a Population Mean: s
Known
Quality assurance procedures call for
the continuation of the filling process if the
sample results are consistent with the assumption
that the mean filling weight for the population of
toothpaste
tubes is 6 oz.; otherwise the process will be
adjusted.
Example: Glow Toothpaste
Two-Tailed Test About a Population Mean:
s Known
Assume that a sample of 30 toothpaste tubes
provides a sample mean of 6.1 oz. The population
standard deviation is believed to be 0.2 oz.

Perform a hypothesis test, at the .03
level of significance, to help determine
whether the filling process should continue
operating or be stopped and corrected.
Two-Tailed Tests About a
Population Mean:s Known
 p –Value and Critical Value Approaches
H0 :   6
1. Determine the hypotheses. H :   6
a
2. Specify the level of significance. α = .03
3. Compute the value of the test statistic.
x  0
6.1  6
z

 2.74
 / n .2 / 30
Two-Tailed Tests About a
Population Mean : s Known
p –Value Approach
4. Compute the p –value.

For z = 2.74, cumulative probability = .9969
p–value = 2(1 - .9969) = .0062
5. Determine whether to reject H0.
Because p–value = .0062 < α = .03, we reject H0.
We are at least 97% confident that the
mean filling weight of the toothpaste
tubes is not 6 oz.
Two-Tailed Tests About a
Population Mean : s Known

p-Value Approach
1/2
p -value
= .0031
1/2
p -value
= .0031
a/2 =
a/2 =
.015
.015
z
z = -2.74
-za/2 = -2.17
0
za/2 = 2.17
z = 2.74
Two-Tailed Tests About a
Population Mean : s Known

Critical Value Approach
4. Determine the critical value and rejection rule.
For a/2 = .03/2 = .015, z.015 = 2.17
Reject H0 if z < -2.17 or z > 2.17
5. Determine whether to reject H0.
Because 2.47 > 2.17, we reject H0.
We are at least 97% confident that the mean
filling weight of the toothpaste tubes is not 6 oz.
Two-Tailed Tests About a
Population Mean : σ Known
 Critical Value Approach
Sampling
distribution
x  0
of z 
/ n
Reject H0
Reject H0
Do Not Reject H0
a/2 = .015
-2.17
a/2 = .015
0
2.17
z
Summary of Hypothesis Tests about
a Population Mean: σ Known Case
Confidence Interval Approach to
Two-Tailed Tests About a
Population Meant
 Select a simple random sample from the
population and use the value of the sample
mean x to develop the confidence interval for
the population mean x .
(Confidence intervals are covered in Chapter 8.)
 If the confidence interval contains the
hypothesized value 0 do not reject H0.
Otherwise, reject H0.
Confidence Interval Approach to
Two-Tailed Tests About a
Population Mean
The 97% confidence interval for  is
x  za / 2

 6.1  2.17(.2 30)  6.1  .07924
n
or 6.02076 to 6.17924
Because the hypothesized value for the
population mean, 0 = 6, is not in this interval,
the hypothesis-testing conclusion is that the
null hypothesis, H0:  = 6, can be rejected.
Tests About a Population Mean:
 Unknown

Test Statistic
t
x  0
s/ n
This test statistic has a t distribution
with n - 1 degrees of freedom.
Tests About a Population Mean:
 Unknown

Rejection Rule: p -Value Approach
Reject H0 if p –value < a
 Rejection Rule: Critical Value Approach
H0:   Reject H0 if t  ta
H0:   
Reject H0 if t  ta
H0:  Reject H if t < - t or t > t
0
a
a
p -Values and the t Distribution



The format of the t distribution table provided in
most statistics textbooks does not have sufficient
detail to determine the exact p-value p-value for
a hypothesis test.
However, we can still use the t distribution table
to identify a range for the p-value.
An advantage of computer software packages is
that the computer output will provide the p-value
for the t distribution.
Tests About a Population Mean:
 Unknown


Example: A business travel magazine wants
to classify transatlantic gateway airports
according to the mean rating for the
population of business travelers.
A rating scale with a low score of 0 and a
high score of 10 will be used, and airports
with a population mean rating greater than 7
will be designated as superior service
airports.
Tests About a Population Mean:
 Unknown

The null and alternative hypotheses for this
upper tail test are as follows:
H0 : μ ≤ 7
Ha : μ > 7
Tests About a Population Mean:
 Unknown

Test Statistic
x  0
t
s/ n
The sampling distribution
of t has n – 1 df.
We have =7.25, s = 1.052, and n = 60, and
the value of the test statistic is

p-Value Approach
Using Table 2 in Appendix B, the t
distribution with 59 degrees of freedom.

We see that t 1.84 is between 1.671 and
2.001. Although the table does not provide
the exact p-value, the values in the “Area in
Upper Tail” row show that the p-value must
be less than .05 and greater than .025.

MINITAB OUTPUT

The test statistic t = 1.84, and the exact
p-value is .035 for the Heathrow rating
hypothesis test.
A p-value = .035 < .05 leads to the rejection
of the null hypothesis.


Critical Value Approach
The rejection rule is thus
Reject H0 if t > 1.671
With the test statistic t = 1.84 > 1.671, H0
is rejected.
Example: Highway Patrol
One-Tailed Test About a Population Mean: 
Unknown
A State Highway Patrol periodically samples
vehicle speeds at various locations
on a particular roadway.
The sample of vehicle speeds
is used H0: m < 65 to test the hypothesis
The locations where H0 is rejected are deemed
the best locations for radar traps.

Example: Highway Patrol

One-Tailed Test About a Population Mean: s
Unknown
At Location F, a sample of 64 vehicles shows a
mean speed of 66.2 mph with a
standard deviation of
4.2 mph. Use a = .05 to
test the hypothesis.
One-Tailed Test About a
Population Mean :  Unknown
 p –Value and Critical Value Approaches
1. Determine the hypotheses.
H0:  < 65
Ha:  > 65
2. Specify the level of significance. a = .05
3. Compute the value of the test statistic.
x  0 66.2  65
t

 2.286
s / n 4.2 / 64
One-Tailed Test About a
Population Mean :  Unknown
 p –Value Approach
4. Compute the p –value.
For t = 2.286, the p–value must be less
than .025 (for t = 1.998) and greater
than .01 (for t = 2.387).
.01 < p–value < .025
One-Tailed Test About a
Population Mean:  Unknown
 p –Value Approach
5. Determine whether to reject H0.
Because p–value < a = .05, we reject H0.
We are at least 95% confident that the
mean speed of vehicles at Location F is
greater than 65 mph.
One-Tailed Test About a
Population Mean:  Unknown
 Critical Value Approach
4. Determine the critical value and rejection rule.
For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669
Reject H0 if t > 1.669
One-Tailed Test About a
Population Mean:  Unknown
 Critical Value Approach
5. Determine whether to reject H0.
Because 2.286 > 1.669, we reject H0.
We are at least 95% confident that the
mean speed of vehicles at Location F is
greater than 65 mph. Location F is a good
candidate for a radar trap.
One-Tailed Test About a
Population Mean:  Unknown
Reject H0
Do Not Reject H0
0
a
ta =
1.669
t
Tests About a Population Mean:
 Unknown

Summary of Hypothesis Tests about a
Population Mean:  Unknown Case
A Summary of Forms for Null and
Alternative Hypotheses About a
Population Proportion

The equality part of the hypotheses always
appears in the null hypothesis.

In general, a hypothesis test about the value of
a population proportion p must take one of the
following three forms (where p0 is the
hypothesized value of the population
proportion).
A Summary of Forms for Null and
Alternative Hypotheses About a
Population Proportion
H 0 : p  p0
H a : p  p0
H 0 : p  p0
H a : p  p0
H 0 : p  p0
H a : p  p0
One-tailed One-tailed Two-tailed
(lower tail) (upper tail)
Tests About a Population
Proportion

Test Statistic
z
p  p0
p
where:
p 
p0 (1  p0 )
n
assuming np > 5 and n(1 – p) > 5
Tests About a Population Proportion

Rejection Rule: p –Value Approach
Reject H0 if p –value < α

Rejection Rule: Critical Value Approach
H:p  p
Reject H0 if z > z
H:p  p
Reject H0 if z < -z
H0:pp
Reject H0 if z  za orz  z
Upper Tail Test About a
Population Proportion


Example: Pine Creek golf course
Over the past year, 20% of the players at
Pine Creek were women.
A special promotion designed to attract
women golfers. One month after the
promotion was implemented.
The null and alternative hypotheses are as
follows:
H0: p ≤ .20
H : p > .20
Upper Tail Test About a
Population Proportion


level of significance of α = .05 be used.
Test statistics
Upper Tail Test About a
Population Proportion
level of significance of α = .05 be used.
 Suppose a random sample of 400 players was
selected, and that 100 of the players were
women.
The proportion of women golfers in the
sample is
and the value of the test statistic is

Upper Tail Test About a
Population Proportion
p-Value Approach
 the p-value is the probability that z is greater
than or equal to z = 2.50. Thus, the p-value for
the test is .5000 – .4938 = .0062.
Upper Tail Test About a
Population Proportion
A p-value = .0062 < .05 gives sufficient
statistical evidence to reject H0 at the .05
level of significance.
Critical Value Approach
 The critical value corresponding to an area
of .05 in the upper tail of a standard normal
distribution is z.05 = 1.645. To reject H0 if
z ≥ 1.645.
 Because z = 2.50 > 1.645, H0 is rejected.

Two-Tailed Test About a
Population Proportion

Example: National Safety Council
For a Christmas and New Year’s week, the
National Safety Council estimated that
500 people would be killed and 25,000
injured on the nation’s roads. The
NSC claimed that 50% of the
accidents would be caused by
drunk driving.
Two-Tailed Test About a
Population Proportion

Example: National Safety Council
A sample of 120 accidents showed that
67 were caused by drunk driving. Use
these data to test the NSC’s claim with
α = .05.
Two-Tailed Test About a
Population Proportion
 p –Value and Critical Value Approaches
1. Determine the hypotheses.
2. Specify the level of significance.
a = .05
Two-Tailed Test About a
Population Proportion
 p –Value and Critical Value Approaches
3. Compute the value of the test statistic.
p0 (1  p0 )
.5(1  .5)
p 

 .045644
n
120
z
a common
error is p
using in
this formula
p  p0
p

(67 /120)  .5
 1.28
.045644
Two-Tailed Test About a
Population Proportion
 p -Value Approach
4. Compute the p -value.
For z = 1.28, cumulative probability = .8997
p–value = 2(1 - .8997) = .2006
5. Determine whether to reject H0.
Because p–value = .2006 > a = .05, we
cannot reject H0.
Two-Tailed Test About a
Population Proportion
 Critical Value Approach
4. Determine the critical value and rejection
rule. For a/2 = .05/2 = .025, z = 1.96
.025
Reject H0 if z < -1.96 or z > 1.96
5. Determine whether to reject H0.
Because 1.278 > -1.96 and < 1.96, we cannot
reject H0.
Tests About a Population
Proportion

Summary of Hypothesis Tests about a Population
Proportion
Hypothesis Testing and Decision
Making
In many decision-making situations the
decision maker may want, and in some cases
may be forced, to take action with both the
conclusion do not reject H0 and the conclusion
reject H0.
 In such situations, it is recommended that
the hypothesis-testing procedure be extended
to include consideration of making a Type II
error.

Hypothesis Testing and Decision Making

Example:
A quality control manager must decide to
accept a shipment of batteries from a
supplier or to return the shipment because
of poor quality.
Suppose the null and alternative hypotheses
about the population mean follow.
H0 : μ ≥ 120
Ha : μ < 120
Hypothesis Testing and Decision
Making



If H0 is rejected, the appropriate action is to
return the shipment to the supplier.
If H0 is not rejected, the decision maker must
still determine what action should be taken.
In such decision-making situations, it is
recommended to control the probability of
making a Type II error. Because knowledge
of the probability of making a Type II error
will be helpful.
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean
1. Formulate the null and alternative hypotheses.
2. Using the critical value approach, use the level
of significance to determine the critical value
and the rejection rule for the test.
3. Using the rejection rule, solve for the value of
the sample mean corresponding to the critical
value of the test statistic.
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean
4. Use the results from step 3 to state the values
of the sample mean that lead to the acceptance
of H0; this defines the acceptance region.
5. Using the sampling distribution of x for a value
of μ satisfying the alternative hypothesis, and
the acceptance region from step 4, compute the
probability that the sample mean will be in the
acceptance region. (This is the probability of
making a Type II error at the chosen
level of μ.)
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean




Example: Batteries’ Quality
Suppose the null and alternative hypotheses
are
H0 : μ ≥ 120
Ha : μ < 120
level of significance of α = .05.
Sample size n=36 and σ = 12 hours.
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean

The critical value approach and z.05 = 1.645,
the rejection rule is
Reject H0 if z ≤ -1.645
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean

The rejection rule indicates that we will reject H0
x  120
if
z
 1.645
12 / 36
That indicates that we will reject H0 if
 12 
x  120  1.645
  116.71
 36 
and accept the shipment whenever x > 116.71.
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean

Compute probabilities associated with
making a Type II error (whenever the true
mean is less than 120 hours and we make
the decision to accept H0: μ ≥ 120.)
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean


If μ =112 is true, the probability of making a
Type II error is the probability that the sample
mean x is greater than 116.71 when μ = 112,
that is, P( x ≥ 116.71 | μ = 112) =?
The probability of making a Type II error (β )
when μ = 112 is .0091 .
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean

We can repeat these calculations for other
values of μ less than 120.
Calculating the Probability of a Type
II Error in Hypothesis Tests About a
Population Mean


The power of the test = the probability of
correctly rejecting H0 when it is false. For
any particular value of μ, the power is 1-β.
Power Curve for The Lot-Acceptance
Hypothesis Test
Calculating the Probability
of a Type II Error

Example: Metro EMS (revisited)
Recall that the response times for
a random sample of 40 medical
emergencies were tabulated. The
sample mean is 13.25 minutes.
The population standard deviation
is believed to be 3.2 minutes.
Calculating the Probability
of a Type II Error
Example: Metro EMS (revisited)
The EMS director wants to
perform a hypothesis test, with a
.05 level of significance, to determine
whether or not the service goal of 12
minutes or less is being achieved.

Calculating the Probability
of a Type II Error
1. Hypotheses are: H0: μ and Ha:μ 
2. Rejection rule is: Reject H0 if z > 1.645
3. Value of the sample mean that identifies
the rejection region:
4. We will accept H0 when x < 12.8323
Calculating the Probability
of a Type II Error
5. Probabilities that the sample mean will be
in the acceptance region:
Values of b1-b
14.0
13.6
13.2
12.8323
12.8
12.4
12.0001
-2.31
-1.52
-0.73
0.00
0.06
0.85
1.645
.0104
.0643
.2327
.5000
.5239
.8023
.9500
.9896
.9357
.7673
.5000
.4761
.1977
.0500
Calculating the Probability
of a Type II Error

Calculating the Probability of a Type II Error
Observations about the preceding table:

When the true population mean m is close to
the null hypothesis value of 12, there is a high
probability that we will make a Type II error.
Example:  = 12.0001,b = .9500
Calculating the Probability
of a Type II Error
Calculating the Probability of a Type II
Error
 When the true population mean m is far
above the null hypothesis value of 12,
there is a low probability that we will
make a Type II error.

Example: = 14.0, b = .0104
Power of the Test

The probability of correctly rejecting H0
when it is false is called the power of the test.

For any particular value of μ, the power is
1 – β.

We can show graphically the power
associated with each value of μ; such a graph
is called a power curve. (See next slide.)
Power Curve
Probability of Correctly
Rejecting Null Hypothesis
1.00
0.90
0.80
H0 False
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
11.5

12.0
12.5
13.0
13.5
14.0
14.5
Determining the Sample Size for a
Hypothesis Test About a Population
Mean


The specified level of significance
determines the probability of making a Type
I error.
By controlling the sample size, the probability
of making a Type II error is controlled.
Determining the Sample Size for a
Hypothesis Test About a Population
Mean


Example: how a sample size can be
determined for the lower tail test about a
population mean.
H0 : μ ≥ μ 0
Ha : μ < μ0
Let α be the probability of a Type I error
and zα and zβ are the z value corresponding
to an area of α and β, respectively in the
upper tail of the standard normal
distribution.
Determining the Sample Size for a
Hypothesis Test About a Population
Mean
Determining the Sample Size for a
Hypothesis Test About a Population
Mean

we compute c using the following formulas
Determining the Sample Size for a
Hypothesis Test About a Population
Mean

To determine the required sample size, we
solve for the n as follows.
Determining the Sample Size for a
Hypothesis Test About a Population
Mean

 
x
n
Reject
H0
c
H0:   
Ha: 
a
x
Sampling
Distribution
of x when
H0 is true
and m = m0

Sampling
distribution
of x when
H0 is false
and a > 0

b
c
a
x
Determining the Sample Size for a
Hypothesis Test About a Population
Mean
n
( za  zb ) 2  2
( 0   a )2
where
za = z value providing an area of a in the tail
zb = z value providing an area of b in the tail
 = population standard deviation
0 = value of the population mean in H0
a = value of the population mean used for the
Type II error
Note: In a two-tailed hypothesis test, use za /2 not za
Relationship Among a,b, and n
Once two of the three values are known, the
other can be computed.
 For a given level of significance a,
increasing the sample size n will reduce b.
 For a given sample size n, decreasing a will
increase b, whereas increasing a will
decrease b

Determining the Sample Size for a
Hypothesis Test About a Population
Mean
Let’s assume that the director of medical
services makes the following statements
about the allowable probabilities for the
Type I and Type II errors:
• If the mean response time is  = 12 minutes, I
am willing to risk an a = .05 probability of
rejecting H0.
• If the mean response time is 0.75 minutes over
the specification ( = 12.75), I am willing to risk
ab = .10 probability of not rejecting H0.

Determining the Sample Size for a
Hypothesis Test About a Population
Mean
Given
a = .05, b = .10
za = 1.645, zb = 1.28
0 = 12, a = 12.75
 = 3.2
( za  zb )2 2 (1.645  1.28)2 (3.2)2
n

 155.75  156
2
2
( 0  a )
(12  12.75)