Transcript PPT

The Nature of Hypothesis Testing
• Formal process for making an inference
• Consider many of the concepts of a hypothesis test
and look at several decision-making situations
• The entire process starts by identifying something
of concern and then formulating two hypotheses
about it
1
Hypothesis
Hypothesis: A statement that something is true
Statistical Hypothesis Test: A process by which a
decision is made between two opposing hypotheses. The
two opposing hypotheses are formulated so that each
hypothesis is the negation of the other. (That way one of
them is always true, and the other one is always false).
Then one hypothesis is tested in hopes that it can be shown
to be a very improbable occurrence thereby implying the
other hypothesis is the likely truth.
2
Null & Alternative Hypothesis
There are two hypotheses involved in making a decision:
Null Hypothesis, Ho: The hypothesis to be tested. Assumed
to be true. Usually a statement that a population parameter has
a specific value. The “starting point” for the investigation.
Alternative Hypothesis, Ha: A statement about the same
population parameter that is used in the null hypothesis.
Generally this is a statement that specifies the population
parameter has a value different, in some way, from the value
given in the null hypothesis. The rejection of the null
hypothesis will imply the likely truth of this alternative
hypothesis.
3
Notes
1. Basic idea: proof by contradiction
Assume the null hypothesis is true and look for evidence to
suggest that it is false
2. Null hypothesis: the status quo
A statement about a population parameter that is assumed to be
true
3. Alternative hypothesis: also called the research hypothesis
Generally, what you are trying to prove?
We hope experimental evidence will suggest the alternative
hypothesis is true by showing the unlikeliness of the truth of the
null hypothesis
4
Example
 Example: Suppose you are investigating the effects of a new pain
reliever. You hope the new drug relieves minor
muscle aches and pains longer than the leading pain
reliever. State the null and alternative hypotheses.
Solutions:
• Ho: The new pain reliever is no better than the leading pain
reliever
• Ha: The new pain reliever lasts longer than the leading pain
reliever
5
Example
 Example: You are investigating the presence of radon in homes
being built in a new development. If the mean level of
radon is greater than 4 then send a warning to all home
owners in the development. State the null and
alternative hypotheses.
Solutions:
• Ho: The mean level of radon for homes in the development is
4 (or less)
• Ha: The mean level of radon for homes in the development is
greater than 4
6
Hypothesis Test Outcomes
Decision
Fail to reject Ho
Reject Ho
Null Hypothesis
True
False
Type A correct decision
Type II error
Type I error
Type B correct decision
Type A correct decision: Null hypothesis true, decide in its favor
Type B correct decision: Null hypothesis false, decide in favor of
alternative hypothesis
Type I error: Null hypothesis true, decide in favor of alternative
hypothesis
Type II error: Null hypothesis false, decide in favor of null
hypothesis
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Example
 Example: A calculator company has just received a large
shipment of parts used to make the screens on graphing
calculators. They consider the shipment acceptable if
the proportion of defective parts is 0.01 (or less). If the
proportion of defective parts is greater than 0.01 the
shipment is unacceptable and returned to the
manufacturer. State the null and alternative
hypotheses, and describe the four possible outcomes
and the resulting actions that would occur for this test.
Solutions:
• Ho: The proportion of defective parts is 0.01 (or less)
• Ha: The proportion of defective parts is greater than 0.01
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Fail To Reject Ho
Null Hypothesis Is True:
Null Hypothesis Is False:
Type A correct decision
Type II error
Truth of situation: The
proportion of defective parts
is 0.01 (or less)
Truth of situation: The
proportion of defective parts
is greater than 0.01
Conclusion: It was
determined that the
proportion of defective parts
is 0.01 (or less)
Conclusion: It was
determined that the
proportion of defective parts
is 0.01 (or less)
Action: The calculator
company received parts with
an acceptable proportion of
defectives
Action: The calculator
company received parts with
an unacceptable proportion of
defectives
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Reject Ho
Null hypothesis is true:
Type I error
Truth of situation: The
proportion of defectives is
0.01 (or less)
Conclusion: It was
determined that the
proportion of defectives is
greater than 0.01
Action: Send the shipment
back to the manufacturer.
The proportion of defectives
is acceptable
Null hypothesis is false:
Type B correct decision
Truth of situation: The
proportion of defectives is
greater than 0.01
Conclusion: It was
determined that the
proportion of defectives is
greater than 0.01
Action: Send the shipment
back to the manufacturer.
The proportion of defectives
is unacceptable
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Errors
Notes:
1. The type II error sometimes results in what represents a lost
opportunity
2. Since we make a decision based on a sample, there is always the
chance of making an error
Probability of a type I error = a
Probability of a type II error = b
Error in Decision
Rejection of a true H o
Failure to reject a false H o
Type
I
II
Probability
a
b
Correct Decision
Failure to reject a true Ho
Rejection of a false Ho
Type
A
B
Probability
1-a
1-b
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Notes
1. Would like a and b to be as small as possible
2. a and b are inversely related
3. Usually set a (and don’t worry too much about b. Why?)
4. Most common values for a and b are 0.01 and 0.05
5. 1 - b : the power of the statistical test
A measure of the ability of a hypothesis test to reject a false
null hypothesis
6. Regardless of the outcome of a hypothesis test, we never
really know for sure if we have made the correct decision
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Interrelationship
Interrelationship between the probability of a type I error (a),
the probability of a type II error (b), and the sample size (n)
P (type I error)
a
P (type II error)
b
Sample
Size
n
13
Level of Significance & Test Statistic
Level of Significance, a : The probability of committing the
type I error
Test Statistic: A random variable whose value is calculated from
the sample data and is used in making the decision fail to reject Ho
or reject Ho
Notes:
 The value of the test statistic is used in conjunction with a
decision rule to determine fail to reject Ho or reject Ho

The decision rule is established prior to collecting the data
and specifies how you will reach the decision
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The Conclusion
a. If the decision is reject Ho, then the conclusion should be worded
something like, “There is sufficient evidence at the a level of
significance to show that . . . (the meaning of the alternative
hypothesis)”
b. If the decision is fail to reject Ho, then the conclusion should be
worded something like, “There is not sufficient evidence at the a
level of significance to show that . . . (the meaning of the
alternative hypothesis)”
Notes:
 The decision is about Ho
 The conclusion is a statement about Ha
 There is always the chance of making an error
15
Hypothesis Test of Mean 
( known): A Probability-Value Approach
• The concepts and much of the reasoning behind
hypothesis tests are given in the previous sections
• Formalize the hypothesis test procedure as it
applies to statements concerning the mean  of a
population ( known): a probability-value
approach
16
The Assumption...
The assumption for hypothesis tests about a mean  using a
known  : The sampling distribution of x has a normal distribution
Recall:
1. The distribution of x has mean 
2. The distribution of x has standard deviation 
n
Hypothesis test:
1. A well-organized, step-by-step procedure used to make a decision
2. Probability-value approach (p-value approach): a procedure that
has gained popularity in recent years. Organized into five steps.
17
The Probability-Value Hypothesis Test
A Five-Step Procedure:
1. The Set-Up
a. Describe the population parameter of concern
b. State the null hypothesis (Ho) and the alternative hypothesis (Ha)
2. The Hypothesis Test Criteria
a. Check the assumptions
b. Identify the probability distribution and the test statistic formula to be used
c. Determine the level of significance, a
3. The Sample Evidence
a. Collect the sample information
b. Calculate the value of the test statistic
4. The Probability Distribution
a. Calculate the p-value for the test statistic
b. Determine whether or not the p-value is smaller than a
5. The Results
a. State the decision about Ho
b. State a conclusion about Ha
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Example
 Example: A company advertises the net weight of its cereal is 24
ounces. A consumer group suspects the boxes are
underfilled. They cannot check every box of cereal, so
a sample of cereal boxes will be examined. A decision
will be made about the true mean weight based on the
sample mean. State the consumer group’s null and
alternative hypotheses. Assume  = 0.2
Solution:
1. The Set-Up
a. Describe the population parameter of concern
The population parameter of interest is the mean , the
mean weight of the cereal boxes
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Solution Continued
b. State the null hypothesis (Ho) and the alternative
hypothesis (Ha)
Formulate two opposing statements concerning 
Ho:  = 24 (  ) (the mean is at least 24)
Ha:  < 24 (the mean is less than 24)
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Possible Statements of Null & Alternative Hypotheses
Null Hypothesis
1. greater than or equal to ()
2. less than or equal to ()
3. equal to ()
Alternative Hypothesis
less than ()
greater than ()
not equal to ()
Notes:

The null hypothesis will be written with just the equal sign (a value
is assigned)

When equal is paired with less than or greater than, the combined
symbol is written beside the null hypothesis as a reminder that all
three signs have been accounted for in these two opposing
statements.
21
Examples
 Example: An automobile manufacturer claims a new model gets
at least 27 miles per gallon. A consumer groups disputes this
claim and would like to show the mean miles per gallon is
lower. State the null and alternative hypotheses.
Solution: Ho:  = 27 ()
and Ha:  < 27
 Example: A freezer is set to cool food to 10 . If the temperature is
higher, the food could spoil, and if the temperature is lower,
the freezer is wasting energy. Random freezers are selected
and tested as they come off the assembly line. The assembly
line is stopped if there is any evidence to suggest improper
cooling. State the null and alternative hypotheses.
Solution: Ho:  = 10
and
Ha:   10
22
Common Phrases & Their Negations
H o : ()
at least
no less than
not less than
H a : ()
less than
less than
less than
H o : ()
at most
no more than
not greater than
more than
more than
greater than
H o : ( )
is
not different from
same as
H a : ()
is not
different from
not same as
H a : ( )
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Example Continued
 Example Continued: Weight of cereal boxes
Recall: Ho:  = 24 () (at least 24)
Ha:  < 24 (less than 24)
2. The Hypothesis Test Criteria
a. Check the assumptions
The weight of cereal boxes is probably mound shaped. A sample
size of 40 should be sufficient for the CLT to apply. The sampling
distribution of the sample mean can be expected to be normal.
b. Identify the probability distribution and the test statistic to be used
To test the null hypothesis, ask how many standard deviations
away from  is the sample mean
test statistic : z* 
x-
 n
24
Solution Continued
c. Determine the level of significance
Let a = 0.05
3. The Sample Evidence
a. Collect the sample information
A random sample of 40 cereal boxes is examined
x  23.95
and
n  40
b. Calculate the value of the test statistic ( = 0.2)
x -  23.95 - 24


 -1.5811
z*
 n 0.2 40
4. The Probability Distribution
a. Calculate the p-value for the test statistic
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Probability-Value or p-Value
Probability-Value, or p-Value: The probability that the test statistic
could be the value it is or a more extreme value (in the direction of
the alternative hypothesis) when the null hypothesis is true (Note:
the symbol P will be used to represent the p-value, especially in
algebraic situations)
P
- 1.58
0
z
P  P ( z  z*)  P ( z  -1.58)  P ( z  1.58)
 0.0571
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Solution Continued
b. Determine whether or not the p-value is smaller than a
The p-value (0.0571) is greater than a (0.05)
5. The Results
Decision Rule:
a. If the p-value is less than or equal to the level of significance a, then the
decision must be to reject Ho
b. If the p-value is greater than the level of significance a, then the decision
must be to fail to reject Ho
a. State the decision about Ho
Decision about Ho : Fail to reject Ho
b. Write a conclusion about Ha
There is not sufficient evidence at the 0.05 level of significance to
show that the mean weight of cereal boxes is less than 24 ounces
27
Notes

If we fail to reject Ho, there is no evidence to suggest the
null hypothesis is false. This does not mean Ho is true.

The p-value is the area, under the curve of the
probability distribution for the test statistic, that is more
extreme than the calculated value of the test statistic.

There are 3 separate cases for p-values. The direction
(or sign) of the alternative hypothesis is the key.
28
Finding p-Values
1. Ha contains > (Right tail)
p-value = P(z > z*)
0
z*
z
2. Ha contains < (Left tail)
p-value = P(z < z*)
z*
0
z
3. Ha contains  (Two-tailed)
p-value = P(z < -|z*|) + P(z > |z*|)
- | z* | 0
| z* |
z
29
Example
 Example: The mean age of all shoppers at a local jewelry store is 37
years (with a standard deviation of 7 years). In an attempt to
attract older adults with more disposable income, the store
launched a new advertising campaign. Following the
advertising, a random sample of 47 shoppers showed a mean
age of 39.3. Is there sufficient evidence to suggest the
advertising campaign has succeeded in attracting older
customers?
Solution:
1. The Set-Up
a. Parameter of concern: the mean age, , of all shoppers
b. The hypotheses:
Ho:  = 37 ()
Ha:  > 37
30
Solution Continued
2. The Hypothesis Test Criteria
a. The assumptions: The distribution of the age of shoppers
is unknown. However, the sample size is large enough
for the CLT to apply.
b. The test statistic: The test statistic will be z*
c. The level of significance: none given
We will find a p-value
3. The Sample Evidence
a. Sample information: n  47,
b. Calculated test statistic: z* 
x  39.3
x -  39.3 - 37

 2.25
 n
7 47
31
Solution Continued
4. The Probability Distribution
a. The p-value:
p - value  P( z  z*)
 P( z  2.25)
 1.0000 - 0.9878
 0.0122
0
2.25
z
b. Determine whether or not the p-value is smaller than a
A comparison is not possible, no a given
5. The Results
Because the p-value is so small (P < 0.05), there is evidence to
suggest the mean age of shoppers at the jewelry store is greater
than 37
32
p-Value
The idea of the p-value is to express the degree of belief in the null
hypothesis:
1. When the p-value is minuscule (like 0.0001), the null hypothesis would
be rejected by everyone because the sample results are very unlikely for
a true Ho
2. When the p-value is fairly small (like 0.01), the evidence against Ho is
quite strong and Ho will be rejected by many
3. When the p-value begins to get larger (say, 0.02 to 0.08), there is too
much probability that data like the sample involved could have occurred
even if Ho were true, and the rejection of Ho is not an easy decision
4. When the p-value gets large (like 0.15 or more), the data is not at all
unlikely if the Ho is true, and no one will reject Ho
33
p-Value Advantages & Disadvantage
Advantages of p-value approach:
1. The results of the test procedure are expressed in terms of a
continuous probability scale from 0.0 to 1.0, rather than simply on
a reject or fail to reject basis
2. A p-value can be reported and the user of the information can
decide on the strength of the evidence as it applies to his/her own
situation
3. Computers can do all the calculations and report the p-value, thus
eliminating the need for tables
Disadvantage:
1. Tendency for people to put off determining the level of
significance
34
Example
 Example: The active ingredient for a drug is manufactured using
fermentation. The standard process yields a mean of 26.5
grams (assume  = 3.2). A new mixing technique during
fermentation is implemented. A random sample of 32
batches showed a sample mean 27.1. Is there any evidence
to suggest the new mixing technique has changed the yield?
Solution:
1. The Set-Up
a. The parameter of interest is the mean yield of active
ingredient, 
b. The null and alternative hypotheses:
H0:  = 26.5
Ha:   26.5
35
Solution Continued
2 The Hypothesis Test Criteria
. a. Assumptions: A sample of size 32 is large enough to
satisfy the CLT
b. The test statistic: z
c. The level of significance: find a p-value
3. The Sample Evidence
a. From the sample: n  32,
x  27.1
b. The calculated test statistic: z 
x -  27.1 - 26.5

 1.06
 n
3.2 32
36
Solution Continued
4. The Probability Distribution
a. The p-value:
p - value  2  P( z | z* |)
 2  P( z  1.06)
 2  (1.0000 - 0.8554)
 2  0.1446  0.2892
0 1.06
z
b. The p-value is large
There is no a given in the statement of the problem
5. The Results
Because the p-value is large (P = 0.2892), there is no evidence to
suggest the new mixing technique has changed the mean yield
37
Hypothesis Test of mean  ( known):
A Classical Approach
• Concepts and reasoning behind hypothesis testing
• Formalize the hypothesis test procedure as it applies
to statements concerning  of a population with
known  : a classical approach
38
The Assumption...
The assumption for hypothesis tests about mean  using a
known  : The sampling distribution of x has a normal distribution
Recall:
1. The distribution of x has mean 
2. The distribution of x has standard deviation 
n
Hypothesis Test: A well-organized, step-by-step procedure used to
make a decision. The classical approach is the hypothesis test
process that has enjoyed popularity for many years.
39
The Classical Hypothesis Test
A Five-Step Procedure:
1. The Set-Up
a. Describe the population parameter of concern
b. State the null hypothesis (Ho) and the alternative hypothesis (Ha)
2. The Hypothesis Test Criteria
a. Check the assumptions
b. Identify the probability distribution and the test statistic to be used
c. Determine the level of significance, a
3. The Sample Evidence
a. Collect the sample information
b. Calculate the value of the test statistic
4. The Probability Distribution
a. Determine the critical region(s) and critical value(s)
b. Determine whether or not the calculated test statistic is in the critical region
5. The Results
a. State the decision about Ho
b. State the conclusion about Ha
40
Example
 Example: A company advertises the net weight of its cereal is 24
ounces. A consumer group suspects the boxes are
underfilled. They cannot check every box of cereal, so
a sample of cereal boxes will be examined. A decision
will be made about the true mean weight based on the
sample mean. State the consumer group’s null and
alternative hypotheses. Assume  = 0.2
Solution:
1. The Set-Up
a. Describe the population parameter of concern
The population parameter of interest is the mean, , the
mean weight of the cereal boxes
41
Solution Continued
b. State the null hypothesis (Ho) and the alternative
hypothesis (Ha)
Formulate two opposing statements concerning the :
Ho:  = 24 (  ) (the mean is at least 24)
Ha:  < 24 (the mean is less than 24)
42
Possible Statements of Null & Alternative Hypotheses
Null Hypothesis
1. greater than or equal to ()
2. less than or equal to ()
3. equal to ()
Alternative Hypothesis
less than ()
greater than ()
not equal to ()
Notes:

The null hypothesis will be written with just the equal sign
(a value is assigned)

When equal is paired with less than or greater than, the combined
symbol is written beside the null hypothesis as a reminder that all
three signs have been accounted for in these two opposing
statements
43
Examples
 Example: An automobile manufacturer claims a new model gets at
least 27 miles per gallon. A consumer groups disputes this
claim and would like to show the mean miles per gallon is
lower. State the null and alternative hypotheses.
Solution: Ho:  = 27 ()
and
Ha:  < 27
 Example: A freezer is set to cool food to 10 . If the temperature is
higher, the food could spoil, and if the temperature is
lower, the freezer is wasting energy. Random freezers
are selected and tested as they come off the assembly line.
The assembly line is stopped if there is any evidence to
suggest improper cooling. State the null and alternative
hypotheses.
Solution: Ho:  = 10
and
Ha:   10
44
Common Phrases & Their Negations
H o : ()
H a : ()
at least
no less than
not less than
less than
less than
less than
H o : ()
H a : ( )
at most
no more than
not greater than
more than
more than
greater than
H o : ( )
H a : ()
is
not different from
same as
is not
different from
not same as
45
Example Continued
 Example (continued): Weight of cereal boxes
Recall: Ho:  = 24 (>) (at least 24) Ha:  < 24 (less than 24)
Solution Continued:
2. The Hypothesis Test Criteria
a. Check the assumptions
The weight of cereal boxes is probably mound shaped. A sample size of 40
should be sufficient for the CLT to apply. The sampling distribution of the
sample mean can be expected to be normal.
b. Identify the probability distribution and the test statistic to be used
To test the null hypothesis, ask how many standard deviations away from 
is the sample mean
x-
test statistic : z* 
 n
46
Solution Continued
c. Determine the level of significance
Consider the four possible outcomes and their consequences
Let a = 0.05
3. The Sample Evidence
a. Collect the sample information
A random sample of 40 cereal boxes is examined
x  23 .95 and n  40
b. Calculate the value of the test statistic ( = 0.2)
x -  23 .95 - 24


 - 1 .5811
z*
 n
0.2 40
4. The Probability Distribution
a. Determine the critical region(s) and critical value(s)
47
Critical Region & Critical Value(s)
Critical Region: The set of values for the test statistic that
will cause us to reject the null hypothesis. The set of values
that are not in the critical region is called the noncritical region
(sometimes called the acceptance region).
Critical Value(s): The first or boundary value(s) of the critical
region(s)
48
Critical Region & Critical Value(s)
Illustration:
0.05
- 1.65
Critical
Region
0
z
Critical Value
49
Solution Continued
4. The Probability Distribution (Continued)
b. Determine whether or not the calculated test statistic is in the critical
region
*
- 1.65
0
z
The calculated value of z, z* = -1.58, is in the noncritical region
5. The Results
We need a decision rule
50
Decision Rule
Decision Rule:
a. If the test statistic falls within the critical region, we will reject Ho
(the critical value is part of the critical region)
b. If the test statistic is in the noncritical region, we will fail
to reject Ho
a. State the decision about Ho
Decision: Fail to reject Ho
b. State the conclusion about Ha
Conclusion: There is not enough evidence at the 0.05 level of
significance to show that the mean weight of cereal
boxes is less than 24
51
Notes
1. The null hypothesis specifies a particular value of a population
parameter
2. The alternative hypothesis can take three forms. Each form dictates a
specific location of the critical region(s)
3. For many hypothesis tests, the sign in the alternative hypothesis points
in the direction in which the critical region is located
Sign in the
Alternative

Hypothesis
Critical Region One region
Left side
One-tailed test


Two regions
One region
Half on each side Right side
Two-tailed test One-tailed test
4. Significance level: a
52
Example
 Example: The mean water pressure in the main water pipe from a town
well should be kept at 56 psi. Anything less and several
homes will have an insufficient supply, and anything greater
could burst the pipe. Suppose the water pressure is checked
at 47 random times. The sample mean is 57.1. (Assume
 = 7). Is there any evidence to suggest the mean water
pressure is different from 56? Use a = 0.01
Solution:
1. The Set-Up
a. Describe the parameter of concern:
The mean water pressure in the main pipe
b. State the null and alternative hypotheses
Ho:  = 56
Ha:   56
53
Solution Continued
2. The Hypothesis Test Criteria
a. Check the assumptions:
A sample of n = 47 is large enough for the CLT to apply
b. Identify the test statistic
The test statistic is z
c. Determine the level of significance: a = 0.01 (given)
3. The Sample Evidence
a. The sample information: x  57.1, n  47
b. Calculate the value of the test statistic:
z 
x -  57.1 - 56

 1.077
 n
7 47
54
Solution Continued
4. The Probability Distribution
a. Determine the critical regions and the critical values
0.005
0.005
*
- 2.58
0
2.58
z
b. Determine whether or not the calculated test statistic is in the
critical region
The calculated value of z = 1.077, is in the noncritical region
55
Solution Continued
5. The Results
a. State the decision about Ho:
Fail to reject Ho
b. State the conclusion about Ha:
There is no evidence to suggest the water pressure is different
from 56 at the 0.01 level of significance
56
Example
 Example: An elementary school principal claims students receive
no more than 30 minutes of homework each night. A
random sample of 36 students showed a sample mean of
36.8 minutes spent doing homework (assume  = 7.5).
Is there any evidence to suggest the mean time spent on
homework is greater than 30 minutes? Use a = 0.01
Solution:
1. The Set-Up
The parameter of concern: , the mean time spent doing
homework each night
Ho:  = 30 ()
Ha:  > 30
57
Solution Continued
2. The Hypothesis Test Criteria
a. The sample size is n = 36, the CLT applies
b. The test statistic is z
c. The level of significance is given: a = 0.01
3. The Sample Evidence
x  36.8,
n  36
x -  36.8 - 30

 5.44
z 
 n 7.5 36
58
Solution Continued
4. The Probability Distribution
0.01
*
0
2.33
z
The calculated value of z = 5.44, is in the critical region
59
Solution Continued
5. The Results
Decision: Reject Ho
Conclusion: There is sufficient evidence at the 0.01 level of
significance to conclude the mean time spent on
homework by the elementary students is more than
30 minutes
Note: Suppose we took repeated sample of size 36. What would you
expect to happen?
60
Hypothesis-Testing Procedure
1. The t-statistic is used to complete a hypothesis test about a
population mean 
2. The test statistic:
t
x-
s n
with df  n - 1
3. The calculated t is the number of estimated standard errors
x is from the hypothesized mean 
4. Probability-Value or Classical Approach
61
Example
 Example: A random sample of 25 students registering for classes
showed the mean waiting time in the registration line was
22.6 minutes and the standard deviation was 8.0 minutes. Is
there any evidence to support the student newspaper’s claim
that registration time takes longer than 20 minutes? Use
a = 0.05 and assume waiting time is approximately normal.
Solution:
1. The Set-up
a. Population parameter of concern: the mean waiting time
spent in the registration line
b. State the null and alternative hypotheses:
Ho:  = 20 () (no longer than)
Ha:  > 20 (longer than)
62
Solution Continued
2. The Hypothesis Test Criteria
a. Check the assumptions:
The sampled population is approximately normal
b. Test statistic: t with df = n - 1 = 24
c. Level of significance: a = 0.05
3. The Sample Evidence
a. Sample information: n  25,
x  22.6, and
s 8
b. Calculate the value of the test statistic:
x -  22.6 - 20 2.6
t
s
n

8
25

1.6
 1.625
63
Solution Continued
Using the p-Value Procedure:
4. The Probability Distribution
a. The p-value: P  P (t  1.625, with df  24 )
Notes:

If this hypothesis test is done with the aid of a computer, most likely the
computer will compute the p-value for you

Using Table C: read the p-value directly from the table for many
situations:
Using Table C: 0.05 < P < 0.10
b. The p-value is not smaller than the level of significance, a
64
Solution Continued
Using the Classical Procedure:
4. The Probability Distribution
a. The critical value: t(24, 0.05) = 1.71
b. t is not in the critical region
5. The Results
a. Decision: Fail to reject Ho
b. Conclusion: There is insufficient evidence to show the mean
waiting time is greater than 20 minutes at the 0.05 level
of significance
65
Example
 Example: A new study indicates that higher than normal (220)
cholesterol levels are a good indicator of possible heart
attacks. A random sample of 27 heart attack victims showed
a mean cholesterol level of 231 and a standard deviation of
20. Is there any evidence to suggest the mean cholesterol
level is higher than normal for heart attack victims?
Use a = 0.01
Solution:
1. The Set-up
a. Population parameter of concern: The mean cholesterol
level of heart attack victims
b. State the the null and alternative hypothesis:
Ho:  = 220 () (mean is not greater than 220)
Ha:  > 220 (mean is greater than 220)
66
Solution Continued
2. The Hypothesis Test Criteria
a. Assumptions: We will assume cholesterol level is at
least approximately normal.
b. Test statistic: t ( unknown), df = n - 1 = 26
c. Level of significance: a = 0.01 (given)
3. The Sample Evidence
a. Sample information: n  27,
x  231,
and
s  20
b. Calculate the value of the test statistic:
t 
x-
231 - 220
11


 2.858
3.849
s
n
20
27
67
Solution Continued
4. The Probability Distribution
a. The critical value: t(26, 0.01) = 2.48
0.01
*
0
t(26, 0.01)
t
2.48
b. t* falls in the critical region
5. The Results
a. Decision: Reject H0
b. Conclusion: At the 0.01 level of significance, there is
sufficient evidence to suggest the mean cholesterol
level in heart attack victims is higher than normal
68
Inferences About the
Probability of Success & Proportion
• Possibly the most common inference of all
• Many examples of situations in which we are
concerned about something either happening or not
happening
• Two possible outcomes, and multiple independent
trials
69
Background
1. p: the binomial parameter, the probability of success on a
single trial
2. p ': the observed or sample binomial probability
x x represents the number of successes that
p' 
n occur in a sample consisting of n trials
3. For the binomial random variable x:
  np,
  npq ,
where
q  1- p
4. The distribution of x is approximately normal if n is larger than 20
and if np and nq are both larger than 5
70
Sampling Distribution of p'
Sampling Distribution of p': If a sample of size n is randomly
selected from a large population with p = P(success), then the
sampling distribution of p' has:
1. a mean  p ' equal to p,
2. a standard error  p ' equal to ( pq) / n , and
3. an approximately normal distribution if n is sufficiently large
In practice, use of the following guidelines will ensure normality:
1. The sample size is greater than 20
2. The sample consists of less than 10% of the population
3. The products np and nq are both larger than 5
71
The Assumptions...
The assumptions for inferences about the binomial parameter p:
The n random observations forming the sample are selected
independently from a population that is not changing during the
sampling
Confidence Interval Procedure:
The unbiased sample statistic p' is used to estimate the population
proportion p
The formula for the 1 - a confidence interval for p is:
p ' - z(a/2) 
p' q'
n
to
p ' + z(a/2) 
p' q'
n
where p'  x / n and q'  1 - p'
72
Example
 Example: A recent survey of 300 randomly selected fourth graders
showed 210 participate in at least one organized sport during
one calendar year. Find a 95% confidence interval for the
proportion of fourth graders who participate in an organized
sport during the year.
Solution:
1. Describe the population parameter of concern
The parameter of interest is the proportion of fourth graders who
participate in an organized sport during the year
2. Specify the confidence interval criteria
a. Check the assumptions
The sample was randomly selected
Each subject’s response was independent
73
Solution Continued
b. Identify the probability distribution
z is the test statistic
p' is approximately normal
n  300  20
np'  300(210 / 300)  210  5
nq'  300(90 / 300)  90  5
c. Determine the level of confidence: 1 - a  0.95
3. Collect and present sample evidence
Sample information: n = 300, and x = 210
The point estimate: p'  x / n  210 / 300  0.70
74
Solution Continued
4. Determine the confidence interval
a. Determine the confidence coefficients:
Using Table 4, Appendix B: z(a /2) = z(0.025) = 1.96
b. The maximum error of estimate:
p' q' 
(0.70)(0.30)
E  z(a /2) 
1.96
n
300
 (1.96) 0.0007  (1.96)(0.0265)  0.0519
c. Find the lower and upper confidence limits:
p '- E
0.70 - 0.0519
0.6481
to
to
to
p '+ E
0.70 + 0.0519
0.7519
75
Solution Continued
d. The Results
0.6481 to 0.7519 is a 95% confidence interval for the
true proportion of fourth graders who participate in an
organized sport during the year
z(a /2) ] 2  p * q *
[
Sample Size Determination: n 
E2
E: maximum error of estimate
1 - a: confidence level
p*: provisional value of p (q* = 1 - p*)
If no provisional values for p and q are given use p* = q* = 0.5
(Always round up)
76
Hypothesis-Testing Procedure
Hypothesis-Testing Procedure: For hypothesis tests concerning the
binomial parameter p, use the test statistic z*:
p'- p
x
z
where p' 
n
pq
n
 Example: (Probability-Value Approach) A hospital
administrator believes that at least 75% of all adults have
a routine physical once every two years. A random
sample of 250 adults showed 172 had physicals within
the last two years. Is there any evidence to refute the
administrator's claim? Use a = 0.05
77
Solution
1. The Set-up
a. Population parameter of concern: the proportion of
adults who have a physical every two years
b. State the null and alternative hypotheses:
Ho: p = 0.75 (>)
Ha: p < 0.75
2. The Hypothesis Test Criteria
a. Assumptions: 250 adults independently surveyed
b. Test statistic: z
n = 250
np = (250)(0.75) = 187.5 > 5
nq = (250)(0.25) = 62.5 > 5
c. Level of significance: a = 0.05
78
Solution Continued
3. The Sample Evidence
a. Sample information:
n  250,
x  172,
and
p'  172 / 250  0.688
p '- p
0.688 - 0.75

pq
(0.75)(0.25)
n
250
- 0.062
- 0.062


 -2.26
0.00075 0.02738
b. The test statistic: z* 
4. The Probability Distribution
a. The p-value:
Use Table A or use a computer
79
Solution Continued
p-value
- 2.26
P  P ( z  -2.26) 
0
z
 0.0119
b. The p-value is smaller than the level of significance, a
5. The Results
a. Decision: Reject Ho
b. Conclusion: There is evidence to suggest the proportion of
adults who have a routine physical exam every
two years is less than 0.75 at the 0.05 level of
significance
80
Example
 Example: (Classical Procedure) A university bookstore employee in
charge of ordering texts believes 65% of all students sell
their statistics books back to the bookstore at the end of the
class. To test this claim, 200 statistics students are selected
at random and 141 plan to sell their texts back to the
bookstore. Is there any evidence to suggest the proportion is
different from 0.65? Use a = 0.01
Solution:
1. The Set-up
a. Population parameter of concern: p = the proportion of
students who sell their statistics books back to the bookstore
b. The null and alternative hypotheses:
Ho: p = 0.65
Ha: p  0.65
81
Solution Continued
2. The Hypothesis Test Criteria
a. Assumptions: Sample randomly selected. Each subject’s response
was independent of other responses.
b. Test statistic: z
n = 200
np = (200)(0.65) = 130 > 5 ;
nq = (200)(0.35) = 70 > 5
c. Level of significance: a = 0.01
3. The Sample Evidence
a. n  200, x  141,
p'  141 / 200  0.705
b. Calculate the value of the test statistic:
p '- p
0.705 - 0.65

pq
(0.65)(0.35)
n
200
0.055
0.055


 1.63
0.0011375 0.03373
z
82
Solution Continued
4. The Probability Distribution
a. The critical value: z(0.005) = 2.58
0.005
0.005
*
- 2.58
0
2.58
z
b. z is not in the critical region
5. The Results
a. Decision: Do not reject Ho
b. Conclusion: There is no evidence to suggest the true proportion of
students who sell their statistics texts back to the
bookstore is different from 0.65 at the 0.05 level of
significance
83
Notes
1. There is a relationship between confidence intervals and twotailed hypothesis tests when the level of confidence and the level
of significance add up to 1
2. The confidence interval and the width of the noncritical region are
the same
3. The point estimate is the center of the confidence interval, and the
hypothesized mean is the center of the noncritical region
4. If the hypothesized value of p is contained in the confidence
interval, then the test statistic will be in the noncritical region
5. If the hypothesized value of p does not fall within the confidence
interval, then the test statistic will be in the critical region
84