Business research proposal
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100-9.pdf
9-1
Chapter 9
Introduction to Hypothesis Testing
9-2
Chapter Goals
After completing this chapter, you should be
able to:
Formulate null and alternative hypotheses involving a
single population mean or proportion
Know what Type I and Type II errors are
Formulate a decision rule for testing a hypothesis
Know how to use the test statistic, critical value, and
p-value approaches to test the null hypothesis
Compute the probability of a Type II error
9-3
What is Hypothesis Testing?
Statistical inference Estimating population parameters based on sample
statistics
An analytical method for making decisions
Through gathering statistical evidence a claim about a population can be accepted or rejected
Must have enough evidence to reject, otherwise you accept the claim
A procedure that incorporates sampling error We never actually 100% “prove” anything because of
sampling error
9-4
What is a Hypothesis?
A hypothesis is a claim (assumption) about a population parameter:
population mean
population proportion
Example: The mean monthly cell phone bill
of this city is µ = $42
Example: The proportion of adults in this
city with cell phones is π = 0.68
9-5
The Null Hypothesis, H0
States the assumption or default position
(numerical) to be tested
Example: The average number of TV sets in
U.S. Homes is at least three ( )
Is always about a population parameter,
not about a sample statistic
3μ:H 0
3μ:H 0
3x:H0
9-6
The Null Hypothesis, H0
Begin with the assumption that the null hypothesis is true
Similar to the notion of innocent until proven guilty
Refers to the status quo
Always contains “=” , “≤” or “” sign
May or may not be rejected
Based on the statistical evidence gathered
(continued)
9-7
The Alternative Hypothesis, HA
Is the opposite of the null hypothesis
e.g.: The average number of TV sets in U.S.
homes is less than 3 ( HA: µ < 3 )
Challenges the status quo
Never contains the “=” , “≤” or “” sign
May or may not be accepted
Is generally the hypothesis that is believed
(or needs to be supported) by the
researcher – a research hypothesis
9-8
Formulating Hypotheses
Example 1: Ford Motor Company has
worked to reduce road noise inside the cab
of the redesigned F150 pickup truck. It
would like to report in its advertising that
the truck is quieter. The average of the
prior design was 68 decibels at 60 mph.
What is the appropriate hypothesis test?
9-9
Formulating Hypotheses
Example 1: Ford Motor Company has worked to reduce road noise inside
the cab of the redesigned F150 pickup truck. It would like to report in its
advertising that the truck is quieter. The average of the prior design was
68 decibels at 60 mph.
What is the appropriate test?
H0: µ ≥ 68 (the truck is not quieter) status quo
HA: µ < 68 (the truck is quieter) wants to support
If the null hypothesis is rejected, Ford has sufficient evidence to support that the truck is now quieter.
9-10
Formulating Hypotheses
Example 2: The average annual income of
buyers of Ford F150 pickup trucks is
claimed to be $65,000 per year. An
industry analyst would like to test this
claim.
What is the appropriate hypothesis test?
9-11
Example 1: The average annual income of buyers of Ford F150 pickup trucks is claimed to be $65,000 per year. An industry analyst would like to test this claim.
What is the appropriate test?
H0: µ = 65,000 (income is as claimed) status quo HA: µ ≠ 65,000 (income is different than claimed)
The analyst will believe the claim unless sufficient evidence is found to discredit it.
Formulating Hypotheses
9-12
3 outcomes for a hypothesis test
1. No error
2. Type I error
3. Type II error
Errors in Making Decisions
9-13
Errors in Making Decisions
Type I Error
Rejecting a true null hypothesis
Considered a serious type of error
The probability of Type I Error is
Called level of significance of the test
Set by researcher in advance
(continued)
9-14
Errors in Making Decisions (continued)
Type II Error
Failing to reject (i.e., accept) a false null
hypothesis
The probability of Type II Error is β
β is a calculated value, the formula is
discussed later in the chapter
9-15
Population
Claim: the population mean age is 50.
Null Hypothesis:
REJECT
Suppose the sample
mean age is 20:
x = 20
Sample
Null Hypothesis
Is x = 20
likely if
µ = 50?
Hypothesis Testing Process
If not likely,
Now select a random sample:
H0: µ = 50
9-16
Sampling Distribution of x
μ = 50 If H0 is true
If it is unlikely that
we would get a
sample mean of
this value ...
... then we
reject the null
hypothesis that
μ = 50
Reason for Rejecting H0
20
... if in fact this were
the population mean…
x
9-17
Outcomes and Probabilities
State of Nature
Decision
Do Not
Reject
H 0
No error
(1 - )
Type II Error
( β )
Reject
H 0
Type I Error
( )
Possible Hypothesis Test Outcomes
H0 False H0 True
Key:
Outcome
(Probability) No Error
( 1 - β )
9-18
Type I & II Error Relationship
Type I and Type II errors cannot happen at the same time
Type I error can only occur if H0 is true
Type II error can only occur if H0 is false
If Type I error probability ( ) , then
Type II error probability ( β )
9-19
Factors Affecting Type II Error
All else equal,
β when the difference between
hypothesized parameter and its true value
β when
β when σ
β when n
The formula used to
compute the value of β
is discussed later in the
chapter
9-20
Level of Significance,
Defines unlikely values of sample statistic if
null hypothesis is true
Defines rejection region of the sampling
distribution
Is designated by , (level of significance)
Typical values are 0.01, 0.05, or 0.10
Is selected by the researcher at the beginning
Provides the critical value(s) of the test
9-21
Hypothesis Tests for the Mean
σ Known σ Unknown
Hypothesis
Tests for
Assume first that the population
standard deviation σ is known
9-22
Level of Significance and the Rejection Region
H0: μ ≥ 3
HA: μ < 3
0
H0: μ ≤ 3
HA: μ > 3
H0: μ = 3
HA: μ ≠ 3
/2
Lower tail test
Level of significance =
0
/2
Upper tail test Two tailed test
0
-zα zα -zα/2 zα/2
Reject H0 Reject H0 Reject H0 Reject H0 Do not
reject H0
Do not
reject H0
Do not
reject H0
Example: Example: Example:
9-23
Reject H0 Do not reject H0
The cutoff value, or ,
is called a critical value
-zα
xα
-zα xα
0
µ=3
H0: μ ≥ 3
HA: μ < 3
n
σ zμx
Critical Value for Lower Tail Test
based on
9-24
Reject H0 Do not reject H0
Critical Value for Upper Tail Test
zα
xα
0
H0: μ ≤ 3
HA: μ > 3
n
σ zμx
µ=3
The cutoff value, or ,
is called a critical value
zα xα
9-25
Do not reject H0 Reject H0 Reject H0
There are two cutoff
values (critical values):
or
Critical Values for Two Tailed Tests
/2
-zα/2
xα/2
± zα/2
xα/2
0
H0: μ = 3
HA: μ 3
zα/2
xα/2
n
σ zμx
/2/2
Lower
Upper xα/2
Lower Upper
/2
µ=3
9-26
z-units: For given , find the critical z value(s):
-zα , zα ,or ±zα/2
Convert the sample mean x to a z test statistic:
Reject H0 if z is in the rejection region,
otherwise do not reject H0
x units: Given , calculate the critical value(s)
xα , or xα/2(L) and xα/2(U)
The sample mean is the test statistic. Reject H0 if x is in the
rejection region, otherwise do not reject H0
Two Equivalent Approaches to Hypothesis Testing
n
σ
μx z
9-27
1. Specify population parameter of interest
2. Formulate the null and alternative hypotheses
3. Specify the desired significance level, α
4. Define the rejection region
5. Take a random sample and determine whether
or not the sample result is in the rejection
region
6. Reach a decision and draw a conclusion
Process of Hypothesis Testing
9-28
Hypothesis Testing Example
Test the claim that the true mean # of TV
sets in US homes is at least 3. (Assume σ = 0.8)
1. Specify the population value of interest
The mean number of TVs in US homes
2. Formulate the appropriate null and alternative
hypotheses
H0: μ 3 HA: μ < 3 (This is a lower tail test)
3. Specify the desired level of significance
Suppose that = 0.05 is chosen for this test
9-29
Reject H0 Do not reject H0
4. Determine the rejection region
= .05
-zα= -1.645 0
This is a one-tailed test with = 0.05.
Since σ is known, the cutoff value is a z value:
Reject H0 if z < z = -1.645 ; otherwise do not reject H0
Hypothesis Testing Example (continued)
9-30
5. Obtain sample evidence and compute the
test statistic
Suppose a sample is taken with the following
results: n = 100, x = 2.84 ( = 0.8 is assumed known)
Then the test statistic is:
2.0 0.08
.16
100
0.8
32.84
n
σ
μx z
Hypothesis Testing Example
9-31
Reject H0 Do not reject H0
= .05
-1.645 0
6. Reach a decision and interpret the result
-2.0
Since z = -2.0 < -1.645, we reject the null
hypothesis that the mean number of TVs in US
homes is at least 3. There is sufficient evidence
that the mean is less than 3.
Hypothesis Testing Example (continued)
z
9-32
Reject H0
= .05
2.8684
Do not reject H0
3
An alternate way of constructing rejection region:
2.84
Since x = 2.84 < 2.8684,
we reject the null
hypothesis
Hypothesis Testing Example (continued)
x
Now
expressed
in x, not z
units
2.8684 100
0.8 1.6453
n
σ zμx αα
Not enough statistical evidence to conclude
that the number of TVs is at least 3
9-33
p-Value Approach to Testing
Convert Sample Statistic ( ) to Test Statistic
(a z value, if σ is known)
Determine the p-value from a table or
computer
Compare the p-value with
If p-value < , reject H0
If p-value , do not reject H0
x
9-34
p-Value Approach to Testing
p-value: Probability of obtaining a test
statistic more extreme ( ≤ or ) than the
observed sample value given H0 is true
Also called observed level of significance
Smallest value of for which H0 can be
rejected
(continued)
9-35
Adds a degree of significance to the result
of the hypothesis test
More than just a simple “reject”
Can now determine how strongly you
“reject” or “accept”
The further the p-value is from , the
stronger the decision
p-Value Approach to Testing (continued)
9-36
Example: How likely is it to see a sample mean
of 2.84 (or something further below the mean) if
the true mean is = 3.0?
p-value =0.0228
= 0.05
p-value Example
2.8684 3
2.84
x .022802.0)P(z
100 0.8
3.02.84 zP
3.0)μ|2.84xP(
0 -1.645 -2.0
z
9-37
Compare the p-value with
If p-value < , reject H0
If p-value , do not reject H0
Here: p-value = 0.0228 = 0.05
Since 0.0228 < 0.05, we
reject the null hypothesis
(continued)
p-value Example
p-value =0.0228
= 0.05
2.8684 3
2.84
9-38
Example: Upper Tail z Test for Mean ( Known)
A phone industry manager thinks that
customer monthly cell phone bill have
increased, and now average over $52 per
month. The company wishes to test this
claim. (Assume = 10 is known)
H0: μ ≤ 52 the average is not over $52 per month
HA: μ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim)
Form hypothesis test:
9-39
Reject H0 Do not reject H0
Suppose that = 0.10 is chosen for this test
Find the rejection region:
= 0.10
zα=1.28 0
Reject H0
Reject H0 if z > 1.28
Example: Find Rejection Region (continued)
9-40
Review: Finding Critical Value - One Tail
Z .07 .09
1.1 .3790 .3810 .3830
1.2 .3980 .4015
1.3 .4147 .4162 .4177 z 0 1.28
.08
Standard Normal
Distribution Table (Portion) What is z given = 0.10?
= 0.10
Critical Value
= 1.28
0.90
.3997
0.10
0.40 0.50
9-41
Obtain sample evidence and compute the test
statistic
Suppose a sample is taken with the following
results: n = 64, x = 53.1 (=10 was assumed known)
Then the test statistic is:
0.88
64
10
5253.1
n
σ
μx z
Example: Test Statistic (continued)
9-42
Reject H0 Do not reject H0
Example: Decision
= 0.10
1.28 0
Reject H0
Do not reject H0 since z = 0.88 ≤ 1.28
i.e.: there is not sufficient evidence that the mean bill is over $52
z = 0.88
Reach a decision and interpret the result:
(continued)
9-43
.18940
.31060.500.88)P(z
64 10
52.053.1 zP
52.0)μ|53.1xP(
Reject H0
= 0.10
Do not reject H0 1.28
0
Reject H0
z = 0.88
Calculate the p-value and compare to
(continued)
p-value =
0.1894
p -Value Solution
Do not reject H0 since p-value = 0.1894 > = 0.10
9-44
Critical Value Approach to Testing
When σ is unknown, convert sample statistic ( )
to a t test statistic
x
Known Unknown
Hypothesis
Tests for
The test statistic is:
n
s
μx t
1n
(The population must be
approximately normal)
9-45
Hypothesis Tests for μ, σ Unknown
1. Specify the population value of interest
2. Formulate the appropriate null and alternative hypotheses
3. Specify the desired level of significance
4. Determine the rejection region (critical values are from the t-distribution with n-1 d.f.)
5. Obtain sample evidence and compute the test statistic
6. Reach a decision and interpret the result
9-46
Example: Two-Tail Test ( Unknown)
The average cost of a
hotel room in New York
is said to be $168 per
night. A random sample
of 25 hotels resulted in
x = $172.50 and
s = $15.40. Test at the
= 0.05 level (Assume the population distribution is normal)
H0: μ = 168
HA: μ 168
9-47
= 0.05
n = 25
Critical Values:
t24 = ± 2.0639
is unknown, so
use a t statistic
Example Solution: Two-Tail Test
Do not reject H0: not sufficient evidence that
true mean cost is different than $168
Reject H0 Reject H0
/2=0.025
-tα/2 Do not reject H0
0 tα/2
/2=0.025
-2.0639 2.0639
1.46
25
15.40
168172.50
n
s
μx t
1n
1.46
H0: μ = 168
HA: μ 168
9-48
Hypothesis Tests for Proportions
Involves categorical values
Two possible outcomes
“Success” (possesses a certain characteristic)
“Failure” (does not possesses that characteristic)
Fraction or proportion of population in the
“success” category is denoted by π
9-49
Proportions
The sample proportion of successes is denoted
by p :
When both nπ and n(1- π) are at least 5, p
is approximately normally distributed with mean
and standard deviation
sizesample
sampleinsuccessesofnumber
n
x p
πμ p
n
π)π(1 σ
p
9-50
The sampling
distribution of p is
normal, so the test
statistic is a z
value:
Hypothesis Tests for Proportions
n
π)π(1
πp z
nπ 5
and
n(1-π) 5
Hypothesis
Tests for π
nπ < 5
or
n(1-π) < 5
Not discussed
in this chapter
9-51
Example: z Test for Proportion
A marketing company
claims that it receives
8% responses from its
mailing. To test this
claim, a random sample
of 500 were surveyed
with 25 responses. Test
at the = 0.05
significance level.
Check:
n π = (500)(0.08) = 40
n(1-π) = (500)(0.92) = 460
Both > 5, so assume normal
9-52
Z Test for Proportion: Solution
= 0.05
n = 500, p = 0.05
Reject H0 at = 0.05
H0: π = 0.08
HA: π 0.08
Critical Values: ± 1.96
Test Statistic:
Decision:
Conclusion:
z 0
Reject Reject
0.025 0.025
1.96
-2.47
There is sufficient
evidence to reject the
company’s claim of 8%
response rate.
-1.96
2.47
500
.08)00.08(1
.0800.05
n
π)π(1
πp z
9-53
Do not reject H0 Reject H0 Reject H0
/2 = 0.025
1.96 0
z = -2.47
Calculate the p-value and compare to (For a two tailed test the p-value is always two tailed)
0.01362(0.0068)
.4932)02(0.5
2.47)P(x2.47)P(z
p-value = .0136:
p -Value Solution
Reject H0 since p-value = 0.0136 < = 0.05
z = 2.47
-1.96
/2 = 0.025
0.0068 0.0068
9-54
Reject
H0: μ 52
Do not reject H0 : μ 52
Type II Error
Type II error is the probability of
failing to reject a false H0
52 50
Suppose we fail to reject H0: μ 52
when in fact the true mean is μ = 50
9-55
Reject
H0: 52
Do not reject H0 : 52
Type II Error
Suppose we do not reject H0: 52 when in fact
the true mean is = 50
52 50
This is the true
distribution of x if = 50
This is the range of x where
H0 is not rejected
(continued)
9-56
Chapter Summary
Addressed hypothesis testing methodology
Performed z Test for the mean (σ known)
Discussed p–value approach to
hypothesis testing
Performed one-tail and two-tail tests
9-57
Chapter Summary
Performed t test for the mean (σ
unknown)
Performed z test for the proportion
(continued)
