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Chapter
Thirteen
Hypothesis
Testing
Hypotheses Testing
• Oversimplified or incorrect assumptions must
be subjected to more formal hypothesis
testing
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13 | 2
Interesting Hypotheses
• Bankers assumed high-income earners are more
profitable than low-income earners
• Clients who carefully balance their checkbooks every
month and minimize fees due to overdrafts are
unprofitable checking account customers
• Old clients were more likely to diminish CD balances
by large amounts compared to younger clients
– This was nonintutive because conventional wisdom
suggested that older clients have a larger portfolio of
assets and seek less risky investments
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Data Analysis
• Descriptive
– Computing measures of central tendency and
dispersion,as well as constructing one-way
tables
• Inferential
– Data analysis aimed at testing specific
hypotheses is usually called inferential
analysis
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Null and Alternative Hypotheses
H0 -> Null Hypotheses
Ha -> Alternative Hypotheses
• Hypotheses always pertain to population
parameters or characteristics rather than to
sample characteristics. It is the population,
not the sample, that we want to make an
infernece about from limited data
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Steps in Conducting a Hypothesis Test
• Step 1. Set up H0 and Ha
• Step 2. Identify the nature of the sampling
distribution curve and specify the appropriate
test statistic
• Step 3. Determine whether the hypothesis
test is one-tailed or two-tailed
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Steps in Conducting a Hypothesis Test
(Cont’d)
• Step 4. Taking into account the specified significance
level, determine the critical value (two critical values
for a two-tailed test) for the test statistic from the
appropriate statistical table
• Step 5. State the decision rule for rejecting H0
• Step 6. Compute the value for the test statistic from
the sample data
• Step 7. Using the decision rule specified in step 5,
either reject H0 or reject Ha
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Launching a Product Line Into a
New Market Area
• Karen, product manager for a line of apparel, to
introduce the product line into a new market area
• Survey of a random sample of 400 households in that
market showed a mean income per household of
$30,000. Karen strongly believes the product line will
be adequately profitable only in markets where the
mean household income is greater than $29,000.
Should Karen introduce the product line into the new
market?
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Karen’s Criterion for Decision Making
• To reach a final decision, Karen has to make
a general inference (about the population)
from the sample data
• Criterion: mean income across across all
households in the market area under
consideration
• If the mean population household income is
greater than $29,000, then Karen should
introduce the product line into the new market
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Karen’s Hypothesis
• Karen’s decision making is equivalent to
either accepting or rejecting the hypothesis:
– The population mean household income in the
new market area is greater than $29,000
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One-Tailed Hypothesis Test
• The term one-tailed signifies that all - or zvalues that would cause Karen to reject H0,
are in just one tail of the sampling distribution
 -> Population Mean
H0:   $29,000
Ha:   $29,000
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Type I and Type II Errors
• Type I error occurs if the null hypothesis is
rejected when it is true
• Type II error occurs if the null hypothesis is
not rejected when it is false
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Significance Level
• -> Significance level—the upper-bound
probability of a Type I error
• 1 -  ->confidence level—the complement of
significance level
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Summary of Errors Involved in
Hypothesis Testing
Inference
Based on
Sample Data
H0 is True
H0 is False
Real State of Affairs
H0 is True
Correct decision
Confidence level
= 1- 
H0 is False
Type II error
P (Type II error) = 
Correct decision
Type I error
Significance level Power = 1-
= *
*Term  represents the maximum probability of
committing a Type I error
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Level of Risk
• Two firms considering introducing a new product that
radically differs from their current product line
– Firm ABC
• Well-established customer base, distinct reputation for its
existing product line
– Firm XYZ
• No loyal clientele, no distinct image for its present
products
• Which of these two firms should be more cautious in
making a decision to introduce the new product?
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Scenario - Firms ABC & XYZ
• Firm ABC
– ABC should be more cautious
• Firm XYZ
– XYZ should be less cautious
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Exhibit 13.1 Identifying the Critical Sample
Mean Value – Sampling Distribution
Sample mean (x) values greater than $29,000--that is x-values on the right-hand side
of the sampling distribution centered on µ = $29,000--suggest that H0 may be false.
More important the farther to the right x is , the stronger is the evidence against H0
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Karen’s Decision Rule for Rejecting
the Null Hypothesis
• Reject H0 if the sample mean exceeds xc
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Criterion Value
Every mean x has a corresponding equivalent
standard Normal Deviate:
The expression for z
x-
Z = --------sx
x =  + zsx
Substituting xc for x and zc for z
xc =  + zcsx where zc is standard normal deviate
corresponding to the critical sample mean, xc.
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Computing the Criterion Value
Standard deviation for the sample of 400 households is
$8,000. The standard error of the mean (sx ) is given by
S
s x = ---- = $400
n
Critical mean household income xc through the
following two steps:
1. Determine the critical z-value, zc. For  =.05, From
Appendix 1, zc = 1.645.
2. Substitute the values of zc, s, and  (under the assumption
that H0 is "just" true ), xc =  + zc s = $29,658.
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Karen’s Decision Rule
• If the sample mean household income is
greater than $29,658, reject the null
hypothesis and introduce the product line into
the new market area.
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Test Statistic
The value of the test statistic is simply the zvalue corresponding to = $30,000.
x-
Z = ------ = 2.5
s
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Exhibit 13.2 Critical Value for Rejecting
the Null Hypothesis
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P - Value – Actual Significance Level
• The probability of obtaining an x-value as high as
$30,000 or more when  is only $29,000 = .0062
• This value is sometimes called the actual significance
level, or the p-value
• The actual significance level of .0062 in this case
means the odds are less than 62 out of 10,000 that
the sample mean income of $30,000 would have
occurred entirely due to chance (when the population
mean income is $29,000 or less)
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T-test
Conduct T-Test when sample is small
Let the sample size, n = 25
X = $30,000 , s = $8,000
From the t-table in Appendix 3, tc = 1.71 for  = .05
and d.f. = 24.
Decision rule: “Reject H0 if t  1.7l.”
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T-test (Cont’d)
The value of t from the sample data:
S = 8000/25 = $1,600
x-
t = ------ = 0.625
sx
The computed value of t is less than 1.71, H0 cannot
be rejected.
Karen should not introduce the product line into the
new market area.
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Two-Tailed Hypothesis Test
• Two-tailed test is one in which values of the
test statistic leading to rejectioin of the null
hypothesis fall in both tails of the sampling
distribution curve
H0 :  = $29,000
Ha :   $29,000
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Test of Two Means
• A health service agency has designed a public
service campaign to promote physical fitness and the
importance of regular exercise. Since the campaign
is a major one, the agency wants to make sure of its
potential effectiveness before running it on a national
scale
– To conduct a controlled test of the campaign’s
effectiveness, the agency needs two similar cities
– The agency identified two similar cities
• city 1 will serve as the test city
• city 2 will serve as a control city
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Test of Two Means (Cont’d)
• Random survey of was conducted to measure the
average time per day a typical adult in each city
spent on some form of exercise
– 300 adults in city 1,
– 200 adults in city 2
• Results of the survey :
– average was 30 minutes per day (with a standard
deviation of 22 minutes) in city 1
– Average was 35 minutes per day (with a standard
deviation of 25 minutes) in city 2
• Question
– From these results, can the agency conclude
confidently that the two cities are well matched for the
controlled test?
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Basic Statistics and Hypotheses
City 1: n1 = 300
x1 = 30 s1 = 22
City 2: n2 = 200
x2 = 35 s2 = 25
The hypotheses are
H0: 1 =2
or
1 -2 = 0
Ha: 1  2
or
1 -2  0
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Test Statistic
Test statistic is the z-statistic, given by
z
=
(x1 - x 2) - (1 - 2 )
------------------------------ s12/n1 + s22/n2
n1 and n2 are greater than 30.
The z-statistic can therefore be used as the
test statistic.
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Decision – Two-Tailed Test
• For Two-Tailed tests
– Identify two critical values of z, one for each tail of the
sampling distribution
– The probability corresponding to each tail is .025, since
 = .05
– From the Normal Table, the z-value, for /2 =.025 is
1.96
• Decision rule : “Reject H0 if z  -1.96 or if z  1.96.”
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Computing Z-value – Two-Tailed Test
Computing the value of z from the survey results
and under the customary assumption that the null
hypothesis is true (i.e., 1 - 2 = 0):
z=
(30 - 35) - (0)
--------------------------------- = -2.29
(22)2/300 + (25)2/200
Since z  -1.96, we should reject H0.
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Exhibit 13.5 Hypothesis Test Related to
Mean Exercising in Two Cities
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T- Test for Independent Samples
Test statistic
(x1 - x2) - (1 - 2 )
t = ------------------------s* (  1/n1 + 1/n2 )
with d.f. = n1 + n2 - 2. In this expression, s* is the pooled
standard deviation, given by
(n1 – 1)s12 + (n2 – 1)s22
s* = --------------------------------n1 + n2 - 2
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T- Test for Independent Samples Two Cities
n1 = 20
n2 = 10
x1 = 30
x2 = 35
s1 = 22
s2 = 25
The degrees of freedom for the t-statistic are
d.f. = 28
Critical value of t with 28 d.f for a tail probability
of .025 is 2.05.
Decision rule : “Reject H0 if t  -2.05 or if t 
2.05." The pooled standard deviation is
s* =  529 (approximately) = 23
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T- Test for Independent Samples
The test statistic is
t = -.56
Since t is neither less than -2.05 nor greater than 2.05,
we cannot reject H0
The sample evidence is not strong enough to conclude
that the two cities differ in terms of levels of
exercising activity of their residents.
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National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
• Test of Two Means Using the SPSS T-TEST
Program
– On the 10-point scale,
• males gave a mean rating of approximately
7.87
• females gave a mean rating of approximately
7.83.
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National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
• In SPSS,
1. Select ANALYZE from the menu,
2. Click COMPARE MEANS
3. Select INDEPENDENT-SAMPLES T -TEST
4. Move “OQ – Over all Service Quality” to the “TEST
VARIABLES(S)” box
5. Move “gender” to “GROUPING VARIABLE” box
6. DEFINE GROUPS (SEX = 1 for male and 2 for female)
7. Click OK.
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National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
OQ – Overall Perceived Service Quality
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Gender – Sex = 1 for male
Sex = 2 for female
13 | 40
National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
Group Statistics
OQ
gender
male
female
N
137
126
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Mean
7.87
7.83
Std. Deviation
2.26
2.31
Std. Error
Mean
.19
.21
13 | 41
National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
P-value >  = 0.05 -- Do
not Reject, Equal variance
assumed is correct
Use this
row
when the
null
hypothesi
s of
equality
of
variance
is
rejected
F-Test--to see if the variance of the
2 groups are assumed to be equal
p-value = .210 --> null hypothesis
cannot be rejected at  = 0.05
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National Insurance Company Study – Perceived Service
Quality Differences Between Males and Females
P-value=.88 is
greater than the  =
of 0.05.
Do
not reject Ho.
The p-value implies that the odds are 88 to 100
that a difference of magnitude .04 (i.e., 7.87 7.83) could have occurred from chance. The null
hypothesis cannot be rejected at the customary
significance level of .05.
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Test of Two Means When
Samples Are Dependent
• The need to check for significant differences
between two mean values when the samples
are not independent
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Test of Two Means When
Samples Are Dependent (Cont’d)
• A retail chain ran a special promotion in a
representative sample of 10 of its stores to
boost sales
• Weekly sales per store before and after the
introduction of the special promotion are
shown
• Did the special promotion lead to a significant
increase in sales?
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Sales Per Store Before and After a
Promotional Campaign
Sales per Store (In Thousands)
Store
Before
Number (i) Promotion
(xbi )
After
Promotion
(xai )
Change in
Sales (In
Thousands)
xdi = xai - xbi
1
250
260
10
2
3
4
5
235
150
145
120
240
151
140
124
5
1
-5
4
6
98
100
2
75
85
180
212
70
95
200
220
-5
10
20
8
50
7
8
9
10
Total
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Test of Two Means When
Samples Are Dependent (Cont’d)
One-Tailed Hypothesis Test:
H0: d  0; Ha: d  0.
The sample estimate of d is xd, given
by
n
Xdi
i=1
xd = ----n
where n is the sample size.
xd = 50/10 = 5
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Test of Two Means When
Samples Are Dependent (Cont’d)
Test statistic is
t =
xd - 
----------- = 2.10
s/n
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Test of Two Means When
Samples Are Dependent (Cont’d)
Standard deviation (s) = 7.53,  = 0.05,
tc for 9 d.f = 1.83 from the Appendix 3
Decision rule: “Reject H0 if t  1.83.”
Test Statistic, t  1.83, we reject H0 and
conclude that the mean change in sales per
store was significantly greater than zero.
The special promotion was indeed effective.
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Exhibit 13.6 Hypothesis Test Related to
Change in Weekly Sales Per Store
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Test for a Single Proportion
• Ms.Jones wants to substantially increase her firm's
advertising budget.
• The firm sells a variety of personal computer
accessories
• Random sample : 20/100 (20%) know the brand
name
• True awareness rate for the brand name across all
personal computer owners is less than .3
• Should Ms. Jones increase the advertising budget on
the basis of survey results?
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Test for a Single Proportion (Cont’d)
• Need to test the population proportion of
personal computer owners who are aware of
the brand:
H0:   .3
Ha:   .3
( is the symbol for population proportion)
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Test for a Single Proportion (Cont’d)
The test statistic:
p-
Z = -------------------- (1- )/n
where p is the sample proportion.
From the Normal Table, zc, = -1.645 for  = .05.
Decision rule here is: “Reject Ho if z  - 1.645.”
p = .2,  = .3, and n = 100, z = -2.174
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Test for a Single Proportion
Since -2.174  -1.645, we reject H0;
The sample awareness rate of .2 is too low to support the
hypothesis that the population awareness rate is .3 or more.
The actual significance level (p-value) corresponding to
z = -2.174 is approximately .015 (from Appendix 1).
Level of significance implies that the odds are lower than
15 in 1,000 that the sample awareness rate of .2 would have
occurred entirely by chance (that is, when the population
awareness rate is .3 or higher).
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Exhibit 13.4 Hypothesis Test Related to
Proportion of Personal Computer Owners
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Test of Two Proportions: Choosing Between
Commercial X & Commercial Y For a New Product
• Tom, advertising manager for a frozen-foods,
company, is in the process of deciding between two
TV commercials (X and Y) for a new frozen food to
be introduced
– Commercial X
• Runs for 20 seconds
• Random sample: 20 % awareness out of 200
respondents
– Commercial Y
• Runs for 30 seconds
• Random sample: 25 % awareness out of 200
respondents
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Test of Two Proportions (Cont’d)
• Question
– Can Tom conclude that commercial Y will be
more effective in the total market for the new
product?
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Criterion for Decision Making
• To reach a final decision, Tom has to make a
general inference (about the population) from
the sample data
• Criterion: relative degrees of awareness likely
to be created by the 2 commercials in the
population of all adult consumers
• Tom should conclude that commercial Y is
more effective than commercial X only if the
anticipated population awareness rate for
commercial Y is greater than that for X
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Hypothesis
• Tom’s decision-making is equvalent to either
accepting or rejecting the hypothesis
– The potential awareness rate that commercial
Y can generate among the population of
consumers is greater than that which
commercial X can generate
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13 | 59
Null and Alternative Hypotheses
Sample sizes:
Sample proportions:
Commercial
X
n1 = 200
p1 = .25
Commercial
Y
n2 = 200
p2 = .20
The hypotheses are:
H0:
1  2
or
1 - 2
 0
Ha:
1  2
or
1 - 2
 0
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Test of Two Proportions –
Sample Standard Error
(p1 – p2) - (1 - 2)
z = -----------------------p1 - p2 -- is estimated by the sample
standard error formula
Sample Standard Error
sp1 - p2 = PQ ( 1/n1 + 1/n2)
n1p1 + n2p2
P = ------------------n1 + n2
Q = 1-P
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Test of Two Proportions
For  = .05, the critical value of z (from Appendix 1)
is 1.645.
Decision rule: “Reject H0 if z  1.645.”
First compute P and Q, then sp1
P=
- p2
and z:
200(.25) + 200(.2)
----------------------- = .225
200 + 200
Q = 1 - .225
= .775
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Test of Two Proportions
sp1
- p2
= (.225)(.775) (1/200 + 1/200)
=0.042
z=
(.25 - .20) - (0)
---------------------- = 1.19
.042
Since z  1.645, we cannot reject H0.
The sample evidence is not strong enough to suggest that
commercial Y will be more effective than commercial X.
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Hypothesis Test Related to Awareness
Generated by Two Commercials
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Cross-Tabulations:
Chi-square Contingency Test
• Technique used for determining whether
there is a statistically significant relationship
between two categorical (nominal or ordinal)
variables
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Telecommunications Company
• Marketing manager of a telecommunications
company is reviewing the results of a study of
potential users of a new cell phone
– Random sample of 200 respondents
• A cross-tabulation of data on whether target consumers
would buy the phone (Yes or No) and whether the cell
phone had Bluetooth wireless technology (Yes or No)
• Question
– Can the marketing manager infer that an association
exists between Bluetooth technology and buying the
cell phone?
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13 | 66
Table 13.3 Two-Way Tabulation of Bluetooth Technology and
Whether Customers Would Buy Cell Phone
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Cross Tabulations - Hypotheses
H0: There is no association between wireless
technology and buying the cell phone (the two
variables are independent of each other).
Ha: There is some association between the Bluetooth
feature and buying the cell phone (the two
variables are not independent of each other).
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13 | 68
Conducting the Test
• Test involves comparing the actual, or
observed, cell frequencies in the crosstabulation with a corresponding set of
expected cell frequencies (Eij)
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Expected Values
Eij =
n inj
----n
Where ni and nj are the marginal frequencies,
that is, the total number of sample units in
category i of the row variable and category j
of the column variable, respectively
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Computing Expected Values
The expected frequency for the first-row, first-column
cell is given by
100  100
E11 =
------------ = 50
200
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Table 13.4 Observed and
Expected Cell Frequencies
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Chi-square Test Statistic
c
(Oij - Eij)2

-----------------
i=1 j=1
Eij
r
2 = 
= 72.00
Where r and c are the number of rows and columns,
respectively, in the contingency table. The number of
degrees of freedom associated with this chi-square
statistic are given by the product (r - 1)(c - 1).
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Chi-square Test Statistic
in a Contingency Test
For d.f. = 1, Assuming  =.05, from Appendix 2, the
critical chi-square value (2c) = 3.84.
Decision rule is: “Reject H0 if 2  3.84.”
Computed 2 = 72.00
Since the computed Chi-square value is greater than
the critical value of 3.84, reject H0.
The apparent relationship between “Bluetooth
technology"and "would buy the cellular phone"
revealed by the sample data is unlikely to have
occurred because of chance
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13 | 74
Interpretation
• The actual significance level associated with
a chi-square value of 72 is less than .001
(from Appendix 2). Thus, the chances of
getting a chi-square value as high as 72
when there is no relationship between
Bluetooth technology and purchase of cell
phones are less than 1 in 1,000.
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13 | 75
Cross-Tabulation Using SPSS
for National Insurance Company
• One crucial issue in the customer survey of
National Insurance Company was how a
customer's education was associated with
whether or not she or he would recommend
National to a friend.
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13 | 76
Need to Conduct Chi-square
Test to Reach a Conclusion
• The hypotheses are
– H0:There is no association between
educational level and willingness to
recommend National to a friend (the two
variables are independent of each other)
– Ha:There is some association between
educational level and willingness to
recommend National to a friend (the two
variables are not independent of each other)
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Association Between Education and Customer’s
Willingness to recommend National to a Friend
For two-way tabulation:
1. Select ANALYZE on the SPSS menu,
2. Click on DESCRIPTIVE STATISTICS,
3. Select CROSS-TABS.
4. Move the “highest level of schooling” to ROW(S) box,
5. Move “rec” variable to “COLUMN(S) box.
6. Click on CELLS,
7. Select OBSERVED, and ROW PERCENTAGES.
8. Click CONTINUE and
9. Click OK.
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Association Between Education and Customer’s Willingness
to recommend National to a Friend
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Association Between Education and Customer’s
Willingness to recommend National to a Friend
COUNT
represents the
actual number
of customers
in each cell.
The
percentages
are based on
the
corresponding
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Association Between Education and Customer’s
Willingness to recommend National to a Friend
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National Insurance Company Study Chi-Square Test
For Chi-Square Assessment:
1.
2.
3.
4.
5.
6.
7.
Select ANALYZE
Click on DESCRIPTIVE STATISTICS
Select CROSS-TABS
Move the variable “highest level of schooling” to ROW(s) box
Move “rec” to COLUMN(s) box;
Click on “STATISTICS”
Select CHI-SQUARE, CONTINGENCY COEFFICIENT, and
CRAMER’S V
8. Click on CELLS,
9. Select OBSERVED and EXPECTED FREQUENCIES
10.Click CONTINUE
11.Click OK.
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National Insurance Company Study Chi-Square Test (Cont’d)
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National Insurance Company Study –
Expected Frequency Table
Interpret
the Table
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National Insurance Company Study
Computed
Chi-square
value
P-value
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National Insurance Company Study –
P-Value Significance
• The actual significance level (p-value) =
0.019
• The chances of getting a chi-square value as
high as 10.007 when there is no relationship
between education and recommendation are
less than 19 in 1000
• The apparent relationship between education
and recommendation revealed by the sample
data is unlikely to have occurred because of
chance
• Jill and Tom can safely reject null hypothesis
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Precautions in Interpreting
Cross Tabulation Results
• Two-way tables cannot show conclusive
evidence of a causal relationship
• Watch out for small cell sizes
• Increases the risk of drawing erroneous
inferences when more than two variables are
involved
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Two-way Table Based on a Survey of 200
Hospital Patients:
Patients who Patients who
jog
do not jog
Patients with
heart disease
20
40
Patients
without heart
disease
80
60
100
100
Is there a causal relationship between patients
who jog and patients with hearth disease?
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