Download Chapter 9

Chapter 9
Fundamentals of
Hypothesis Testing:
One-Sample Tests
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 1
Learning Objectives
In this chapter, you learn:
n
The basic principles of hypothesis testing
n
How to use hypothesis testing to test a mean or
proportion
n
The assumptions of each hypothesis-testing
procedure, how to evaluate them, and the
consequences if they are seriously violated
n
Pitfalls & ethical issues involved in hypothesis testing
n
How to avoid the pitfalls involved in hypothesis testing
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 2
9.1 Fundamentals of Hypothesis –
DCOVA
Testing Methodology
What is a Hypothesis?
n
A hypothesis is a claim
(assertion) about a
population parameter:
n
population mean
Example: The mean monthly cell phone bill
in
this city is proportion
μ = $42
n population
Example: The proportion of adults in this
city with cell phones is π = 0.68
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 3
The Null Hypothesis, H0
n
DCOVA
States the claim or assertion to be tested
Example: The mean diameter of a manufactured
bolt is 30mm ( H 0 : µ = 30)
n
Is always about a population parameter,
not about a sample statistic
H0 : µ = 30
H0 : X = 30
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 4
The Null Hypothesis, H0
DCOVA
(continued)
n
n
n
Begin with the assumption that the null
hypothesis is true
Always contains “=“, or “≤”, or “≥” sign
May or may not be rejected
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 5
The Alternative Hypothesis, H1
DCOVA
n
Is the opposite of the null hypothesis
n
n
n
n
e.g., The average diameter of a manufactured
bolt is not equal to 30mm ( H1: μ ≠ 30 )
Never contains the “=“, or “≤”, or “≥” sign
May or may not be proven
Is generally the hypothesis that the
researcher is trying to prove
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 6
The Hypothesis Testing
Process
DCOVA
n
Claim: The population mean age is 50.
n
n
H0: μ = 50,
H1: μ ≠ 50
Sample the population and find the sample mean.
Population
Sample
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 7
The Hypothesis Testing
Process
(continued)
n
n
n
n
DCOVA
Suppose the sample mean age was X = 20.
This is significantly lower than the claimed mean
population age of 50.
If the null hypothesis were true, the probability of
getting such a different sample mean would be very
small, so you reject the null hypothesis .
In other words, getting a sample mean of 20 is so
unlikely if the population mean was 50, you conclude
that the population mean must not be 50.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 8
The Hypothesis Testing
Process
(continued)
DCOVA
Sampling
Distribution of X
20
If it is unlikely that you
would get a sample
mean of this value ...
µ = 50
If H0 is true
... When in fact this were
the population mean…
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
X
... then you reject
the null hypothesis
that µ = 50.
Chapter 9, Slide 9
The Test Statistic and
Critical Values
DCOVA
n
n
n
n
If the sample mean is close to the stated population
mean, the null hypothesis is not rejected.
If the sample mean is far from the stated population
mean, the null hypothesis is rejected.
How far is “far enough” to reject H0?
The critical value of a test statistic creates a “line in
the sand” for decision making -- it answers the
question of how far is far enough.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 10
The Test Statistic and
Critical Values
DCOVA
Sampling Distribution of the test statistic
Region of
Rejection
Region of
Non-Rejection
Region of
Rejection
Critical Values
“Too Far Away” From Mean of Sampling Distribution
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 11
Possible Errors in Hypothesis Test
Decision Making
DCOVA
n
n
Type I Error
n Reject a true null hypothesis
n Considered a serious type of error
n The probability of a Type I Error is a
n
Called level of significance of the test
n
Set by researcher in advance
Type II Error
n Failure to reject a false null hypothesis
n The probability of a Type II Error is β
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 12
Possible Errors in Hypothesis Test
DCOVA
Decision Making
(continued)
Possible Hypothesis Test Outcomes
Actual Situation
Decision
H0 True
H0 False
Do Not
Reject H0
No Error
Probability 1 - α
Type II Error
Probability β
Reject H0
Type I Error
Probability α
No Error
Power 1 - β
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 13
Possible Errors in Hypothesis Test
DCOVA
Decision Making
(continued)
n
n
n
The confidence coefficient (1-α) is the
probability of not rejecting H0 when it is true.
The confidence level of a hypothesis test is
(1-α)*100%.
The power of a statistical test (1-β) is the
probability of rejecting H0 when it is false.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 14
Type I & II Error Relationship
DCOVA
§ Type I and Type II errors cannot happen at
the same time
§A
Type I error can only occur if H0 is true
§A
Type II error can only occur if H0 is false
If Type I error probability (a)
, then
Type II error probability (β)
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 15
Factors Affecting Type II Error
DCOVA
n
All else equal,
n
β
when the difference between
hypothesized parameter and its true value
n
β
when
a
n
β
when
σ
n
β
when
n
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 16
Level of Significance
and the Rejection Region
H0: μ = 30
H1: μ ≠ 30
DCOVA
Level of significance = α
α /2
α /2
30
Critical values
Rejection Region
This is a two-tail test because there is a rejection region in both tails
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 17
Hypothesis Tests for the Mean
DCOVA
Hypothesis
Tests for m
sKnown
(Z test)
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
sUnknown
(t test)
Chapter 9, Slide 18
Z Test of Hypothesis for the
Mean (σ Known)
n
DCOVA
Convert sample statistic ( X ) to a ZSTAT test statistic
Hypothesis
Tests for m
sKnown
σ Known
(Z test)
sUnknown
σ Unknown
(t test)
The test statistic is:
ZSTAT =
X −µ
σ
n
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 19
Critical Value
Approach to Testing
n
n
n
n
DCOVA
For a two-tail test for the mean, σ known:
Convert sample statistic ( X ) to test statistic
(ZSTAT)
Determine the critical Z values for a specified
level of significance a from a table or
computer
Decision Rule: If the test statistic falls in the
rejection region, reject H0 ; otherwise do not
reject H0
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 20
Two-Tail Tests
DCOVA
n
There are two
cutoff values
(critical values),
defining the
regions of
rejection
H0: μ = 30
H1: μ ¹ 30
a/2
a/2
X
30
Reject H0
Do not reject H0
-Zα/2
Lower
critical
value
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
0
Reject H0
+Zα/2
Z
Upper
critical
value
Chapter 9, Slide 21
6 Steps in
Hypothesis Testing
DCOVA
1. State the null hypothesis, H0 and the
alternative hypothesis, H1
2. Choose the level of significance, a, and the
sample size, n
3. Determine the appropriate test statistic and
sampling distribution
4. Determine the critical values that divide the
rejection and nonrejection regions
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 22
6 Steps in
Hypothesis Testing
5.
6.
DCOVA
(continued)
Collect data and compute the value of the test
statistic
Make the statistical decision and state the
managerial conclusion. If the test statistic falls
into the nonrejection region, do not reject the
null hypothesis H0. If the test statistic falls into
the rejection region, reject the null hypothesis.
Express the managerial conclusion in the
context of the problem
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 23
Hypothesis Testing Example
DCOVA
Test the claim that the true mean diameter
of a manufactured bolt is 30mm.
(Assume σ = 0.8)
1. State the appropriate null and alternative
hypotheses
n H0: μ = 30
H1: μ ≠ 30 (This is a two-tail test)
2. Specify the desired level of significance and the
sample size
n Suppose that a= 0.05 and n = 100 are chosen
for this test
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 24
Hypothesis Testing Example
(continued)
DCOVA
3. Determine the appropriate technique
n σ is assumed known so this is a Z test.
4. Determine the critical values
n For a= 0.05 the critical Z values are ±1.96
5. Collect the data and compute the test statistic
n Suppose the sample results are
n = 100, X = 29.84 (σ = 0.8 is assumed known)
So the test statistic is:
Z STAT
X−µ
29.84 − 30
− 0.16
=
=
=
= −2.0
σ
0.8
0.08
n
100
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 25
Hypothesis Testing Example (continued)
DCOVA
n
6. Is the test statistic in the rejection region?
a/2 = 0.025
Reject H0 if
ZSTAT < -1.96 or
ZSTAT > 1.96;
otherwise do
not reject H0
Reject H0
-Zα/2 = -1.96
a/2 = 0.025
Do not reject H0
0
Reject H0
+Zα/2 = +1.96
Here, ZSTAT = -2.0 < -1.96, so the
test statistic is in the rejection
region
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 26
Hypothesis Testing Example
(continued)
DCOVA
6 (continued). Reach a decision and interpret the result
a= 0.05/2
Reject H0
-Zα/2 = -1.96
a= 0.05/2
Do not reject H0
0
Reject H0
+Zα/2= +1.96
-2.0
Since ZSTAT = -2.0 < -1.96, reject the null hypothesis
and conclude there is sufficient evidence that the mean
diameter of a manufactured bolt is not equal to 30
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 27
p-Value Approach to Testing
DCOVA
n
p-value: Probability of obtaining a test
statistic equal to or more extreme than the
observed sample value given H0 is true
n
n
The p-value is also called the observed level of
significance
It is the smallest value of a for which H0 can be
rejected
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 28
p-Value Approach to Testing:
Interpreting the p-value
DCOVA
n
n
Compare the p-value with a
n
If p-value < a, reject H0
n
If p-value ³a, do not reject H0
Remember
n
If the p-value is low then H0 must go
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 29
The 5 Step p-value approach to
Hypothesis Testing
DCOVA
1.
State the null hypothesis, H0 and the alternative hypothesis,
H1
2.
Choose the level of significance, a, and the sample size, n
3.
Determine the appropriate test statistic and sampling
distribution
4.
Collect data and compute the value of the test statistic and
the p-value
5.
Make the statistical decision and state the managerial
conclusion. If the p-value is < α then reject H0, otherwise do
not reject H0. State the managerial conclusion in the context
of the problem
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 30
p-value Hypothesis Testing
Example
DCOVA
Test the claim that the true mean
diameter of a manufactured bolt is 30mm.
(Assume σ = 0.8)
1. State the appropriate null and alternative
hypotheses
n H0: μ = 30
H1: μ ≠ 30 (This is a two-tail test)
2. Specify the desired level of significance and the
sample size
n Suppose that a= 0.05 and n = 100 are chosen
for this test
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 31
p-value Hypothesis Testing
Example
(continued)
DCOVA
3. Determine the appropriate technique
n σ is assumed known so this is a Z test.
4. Collect the data, compute the test statistic and the
p-value
n Suppose the sample results are
n = 100, X = 29.84 (σ = 0.8 is assumed known)
So the test statistic is:
Z STAT =
X − µ
29.84 − 30
− 0.16
=
=
= −2.0
σ
0.8
0.08
n
100
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 32
p-Value Hypothesis Testing Example:
(continued)
Calculating the p-value
DCOVA
4. (continued) Calculate the p-value.
n
How likely is it to get a ZSTAT of -2 (or something further from the
mean (0), in either direction) if H0 is true?
P(Z > 2.0) = 0.0228
P(Z < -2.0) = 0.0228
Z
0
-2.0
2.0
p-value = 0.0228 + 0.0228 = 0.0456
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 33
p-value Hypothesis Testing
Example
(continued)
DCOVA
n
5. Is the p-value < α?
n
n
Since p-value = 0.0456 < α = 0.05 Reject H0
5. (continued) State the managerial conclusion
in the context of the situation.
n
There is sufficient evidence to conclude the average diameter of
a manufactured bolt is not equal to 30mm.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 34
Connection Between Two Tail Tests
and Confidence Intervals
n
DCOVA
For X = 29.84, σ = 0.8 and n = 100, the 95%
confidence interval is:
0.8
0.8
29.84 - (1.96)
to 29.84 + (1.96)
100
100
29.6832 ≤ μ ≤ 29.9968
n
Since this interval does not contain the hypothesized
mean (30), we reject the null hypothesis at a= 0.05
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 35
Do You Ever Truly Know σ?
DCOVA
n
n
n
n
Probably not!
In virtually all real world business situations, σ is not
known.
If there is a situation where σ is known then µ is also
known (since to calculate σ you need to know µ.)
If you truly know µ there would be no need to gather a
sample to estimate it.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 36
Hypothesis Testing:
σ Unknown
DCOVA
n
n
n
n
If the population standard deviation is unknown, you
instead use the sample standard deviation S.
Because of this change, you use the t distribution instead
of the Z distribution to test the null hypothesis about the
mean.
When using the t distribution you must assume the
population you are sampling from follows a normal
distribution.
All other steps, concepts, and conclusions are the same.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 37
9.2 t Test of Hypothesis for the Mean
(σ Unknown)
DCOVA
n
Convert sample statistic ( X ) to a tSTAT test statistic
Hypothesis
Tests for m
sKnown
σ Known
(Z test)
sUnknown
σ Unknown
(t test)
The test statistic is:
t STAT
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
X −µ
=
S
n
Chapter 9, Slide 38
Example: Two-Tail Test
(sUnknown)
The average cost of a hotel
room in New York is said to
be $168 per night. To
determine if this is true, a
random sample of 25 hotels
is taken and resulted in an X
of $172.50 and an S of
$15.40. Test the appropriate
hypotheses at a= 0.05.
DCOVA
H0: μ = 168
H1: μ ¹ 168
(Assume the population distribution is normal)
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 39
Example Solution:
Two-Tail t Test
H0: μ = 168
H1: μ ¹ 168
n α = 0.05
α/2=.025
α/2=.025
Reject H0
n n = 25, df = 25-1=24
n sis unknown, so
use a t statistic
n Critical Value:
DCOVA
-t 24,0.025
-2.0639
t STAT =
Do not reject H0
Reject H0
0
1.46
t 24,0.025
2.0639
X−µ
172.50 − 168
=
= 1.46
S
15.40
n
25
±t24,0.025 = ± 2.0639 Do not reject H0: insufficient evidence that true
mean cost is different from $168
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 40
To Use the t-test Must Assume
DCOVA
the Population Is Normal
n
n
As long as the sample size is not very small and
the population is not very skewed, the t-test can
be used.
To evaluate the normality assumption can use:
n
n
n
How closely sample statistics match the normal
distribution’s theoretical properties.
A histogram or stem-and-leaf display or boxplot or a
normal probability plot.
Section 6.3 has more details on evaluating normality.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 41
Example Two-Tail t Test Using A
p-value from Excel
DCOVA
n
n
Since this is a t-test we cannot calculate the p-value
without some calculation aid.
The Excel output below does this:
t Test for the Hypothesis of the Mean
Data
Null Hypothesis
µ=
Level of Significance
Sample Size
Sample Mean
Sample Standard Deviation
$
$
$
Intermediate Calculations
Standard Error of the Mean
$
Degrees of Freedom
t test statistic
p-value > α
So do not reject H0
168.00
0.05
25
172.50
15.40
3.08 =B8/SQRT(B6)
24 =B6-1
1.46 =(B7-B4)/B11
Two-Tail Test
Lower Critical Value
-2.0639 =-TINV(B5,B12)
Upper Critical Value
2.0639 =TINV(B5,B12)
p-value
0.157 =TDIST(ABS(B13),B12,2)
Do Not Reject Null Hypothesis
=IF(B18<B5, "Reject null hypothesis",
"Do not reject null hypothesis")
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 42
Example Two-Tail t Test Using A
p-value from Minitab
DCOVA
1
2
3
p-value > α
So do not reject H0
One-Sample T
4
Test of mu = 168 vs not = 168
N Mean StDev SE Mean
95% CI
T
P
25 172.50 15.40 3.08 (166.14, 178.86) 1.46 0.157
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 43
Connection of Two Tail Tests to
Confidence Intervals DCOVA
n
For X = 172.5, S = 15.40 and n = 25, the 95%
confidence interval for µ is:
172.5 - (2.0639) 15.4/ 25
to 172.5 + (2.0639) 15.4/ 25
166.14 ≤ μ ≤ 178.86
n
Since this interval contains the Hypothesized mean (168),
we do not reject the null hypothesis at a= 0.05
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 44
9.3 One-Tail Tests
n
DCOVA
In many cases, the alternative hypothesis
focuses on a particular direction
H0: μ ≥ 3
H1: μ < 3
H0: μ ≤ 3
H1: μ > 3
This is a lower-tail test since the
alternative hypothesis is focused on
the lower tail below the mean of 3
This is an upper-tail test since the
alternative hypothesis is focused on
the upper tail above the mean of 3
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 45
Lower-Tail Tests
H0: μ ≥ 3
n
There is only one
critical value, since
the rejection area is
in only one tail
DCOVA
H1: μ < 3
α
Reject H0
-Zα or -tα
Do not reject H0
0
μ
Z or t
X
Critical value
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 46
Upper-Tail Tests
DCOVA
n
There is only one
critical value, since
the rejection area is
in only one tail
Z or t
_
X
H0: μ ≤ 3
H1: μ > 3
α
Do not reject H0
0
Reject H0
Zα or tα
μ
Critical value
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 47
Example: Upper-Tail t Test
for Mean (sunknown)
DCOVA
A phone industry manager thinks that
customer monthly cell phone bills have
increased, and now average over $52 per
month. The company wishes to test this
claim. (Assume a normal population)
Form hypothesis test:
H0: μ ≤ 52 the average is not over $52 per month
H1: μ > 52
the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 48
Example: Find Rejection Region
(continued)
DCOVA
n Suppose that a= 0.10 is chosen for this test and
n = 25.
Reject H0
Find the rejection region:
a= 0.10
Do not reject H0
0
1.318
Reject H0
Reject H0 if tSTAT > 1.318
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 49
Example: Test Statistic
(continued)
DCOVA
Obtain sample and compute the test statistic
Suppose a sample is taken with the following
results: n = 25, X = 53.1, and S = 10
n
Then the test statistic is:
t STAT
X−µ
53.1 − 52
=
=
= 0.55
S
10
n
25
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 50
Example: Decision
(continued)
DCOVA
Reach a decision and interpret the result:
Reject H0
a= 0.10
Do not reject H0
0
tSTAT
1.318
= 0.55
Reject H0
Do not reject H0 since tSTAT = 0.55 ≤ 1.318
there is not sufficient evidence that the
mean bill is over $52
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 51
Example: Utilizing The p-value
for The Test
DCOVA
n
Calculate the p-value and compare to a(p-value below
calculated using excel spreadsheet on next page)
p-value = .2937
Reject H0
a= .10
0
Do not reject
H0
1.318
Reject
H0
tSTAT = .55
Do not reject H0 since p-value = .2937 > a= .10
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 52
Excel Spreadsheet Calculating The
p-value for The Upper Tail t Test
DCOVA
t Test for the Hypothesis of the Mean
Data
Null Hypothesis
µ=
Level of Significance
Sample Size
Sample Mean
Sample Standard Deviation
Intermediate Calculations
Standard Error of the Mean
Degrees of Freedom
t test statistic
52.00
0.1
25
53.10
10.00
2.00 =B8/SQRT(B6)
24 =B6-1
0.55 =(B7-B4)/B11
Upper Tail Test
Upper Critical Value
1.318 =TINV(2*B5,B12)
p-value
0.2937 =TDIST(ABS(B13),B12,1)
Do Not Reject Null Hypothesis
=IF(B18<B5, "Reject null hypothesis",
"Do not reject null hypothesis")
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 53
Using Minitab to calculate The p-value
for The Upper Tail t Test
DCOVA
1
2
3
p-value > α
So do not reject H0
One-Sample T
4
Test of mu = 52 vs > 52
N Mean StDev SE Mean
25 53.10 10.00 2.00
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
95% Lower
Bound T
P
49.68 0.55 0.294
Chapter 9, Slide 54
Hypothesis Tests for Proportions
DCOVA
n
Involves categorical variables
n
Two possible outcomes
n
n
Possesses characteristic of interest
n
Does not possess characteristic of interest
Fraction or proportion of the population in the
category of interest is denoted by π
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 55
Proportions
n
DCOVA
Sample proportion in the category of interest is
denoted by p
n
n
(continued)
X number in category of interest in sample
p=
=
n
sample size
When both nπ and n(1-π) are at least 5, p can
be approximated by a normal distribution with
mean and standard deviation
n
μp = π
σp =
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
π (1− π )
n
Chapter 9, Slide 56
9.4 Z Test of Hypothesis Tests for
Proportions
DCOVA
n
The sampling
distribution of p is
approximately
normal, so the test
statistic is a ZSTAT
value:
ZSTAT =
p−π
π (1 − π )
n
Hypothesis
Tests for p
nπ ³5
and
n(1-π) ³5
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
nπ < 5
or
n(1-π) < 5
Not discussed
in this chapter
Chapter 9, Slide 57
Z Test for Proportion in Terms of
Number in Category of Interest
n
An equivalent form
to the last slide,
but in terms of the
number in the
category of
interest, X:
ZSTAT
DCOVA
Hypothesis
Tests for X
X ³5
and
n-X ³5
X − nπ
=
nπ (1 − π )
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
X<5
or
n-X < 5
Not discussed
in this chapter
Chapter 9, Slide 58
Example: Z Test for Proportion
DCOVA
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 a= 0.05
significance level.
Check:
nπ = (500)(.08) = 40
n(1-π) = (500)(.92) = 460
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
ü
Chapter 9, Slide 59
Z Test for Proportion: Solution
DCOVA
Test Statistic:
H0: π = 0.08
H1: π ¹ 0.08
ZSTAT =
α = 0.05
n = 500, p = 0.05
Reject
.025
.025
-1.96
-2.47
0
1.96
π (1 − π )
n
=
.05 − .08
.08(1 − .08)
500
= −2.47
Decision:
Critical Values: ± 1.96
Reject
p −π
z
Reject H0 at a= 0.05
Conclusion:
There is sufficient
evidence to reject the
company’s claim of 8%
response rate.
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 60
p-Value Solution
(continued)
DCOVA
Calculate the p-value and compare to a
(For a two-tail test the p-value is always two-tail)
Do not reject H0
Reject H0
a/2 = .025
Reject H0
p-value = 0.0136:
a/2 = .025
0.0068
0.0068
-1.96
Z = -2.47
0
P(Z ≤ −2.47) + P(Z ≥ 2.47)
= 2(0.0068) = 0.0136
1.96
Z = 2.47
Reject H0 since p-value = 0.0136 < a= 0.05
Copyright © 2015, 2012, 2009 Pearson Education, Inc.
Chapter 9, Slide 61