Statistics for business economics 7th by paul newbold chapter 14

Statistics for
Business and Economics
7th Edition

Chapter 14
Analysis of Categorical Data

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Ch. 14-1


Chapter Goals
After completing this chapter, you should be able to:


Use the chi-square goodness-of-fit test to determine
whether data fits specified probabilities



Perform tests for the Poisson and Normal distributions



Set up a contingency analysis table and perform a chisquare test of association

 Use the sign test for paired or matched samples
 Recognize when and how to use the Wilcoxon signed
rank test for paired or matched samples
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Ch. 14-2


Chapter Goals
(continued)

After completing this chapter, you should be able to:

 Use a sign test for a single population median
 Apply a normal approximation for the Wilcoxon signed
rank test

 Know when and how to perform a Mann-Whitney U-test
 Explain Spearman rank correlation and perform a test
for association

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Ch. 14-3


Nonparametric Statistics


Nonparametric Statistics


Fewer restrictive assumptions about data
levels and underlying probability distributions




Population distributions may be skewed
The level of data measurement may only be
ordinal or nominal

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Ch. 14-4


Goodness-of-Fit Tests

14.1



Does sample data conform to a hypothesized
distribution?
 Examples:






Do sample results conform to specified expected
probabilities?
Are technical support calls equal across all days of

the week? (i.e., do calls follow a uniform
distribution?)
Do measurements from a production process
follow a normal distribution?

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Ch. 14-5


Chi-Square Goodness-of-Fit Test
(continued)


Are technical support calls equal across all days of the
week? (i.e., do calls follow a uniform distribution?)
 Sample data for 10 days per day of week:
Sum of calls for this day:
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday

290
250
238
257

265
230
192

 = 1722
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Ch. 14-6


Logic of Goodness-of-Fit Test


If calls are uniformly distributed, the 1722 calls
would be expected to be equally divided across
the 7 days:

1722
246 expected calls per day if uniform
7


Chi-Square Goodness-of-Fit Test: test to see if
the sample results are consistent with the
expected results

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Ch. 14-7

Observed vs. Expected
Frequencies
Observed
Oi

Expected
Ei

Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday

290
250
238
257
265
230
192

246
246
246
246
246

246
246

TOTAL

1722

1722

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Ch. 14-8


Chi-Square Test Statistic
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform


The test statistic is
2
(O

E
)
i
 2  i
Ei
i1

K

(where d.f. K  1)

where:
K = number of categories
Oi = observed frequency for category i
Ei = expected frequency for category i
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Ch. 14-9


The Rejection Region
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform
2
(O

E
)
i
 2  i
Ei
i 1
K



Reject H0 if

2

 

2
α

(with k – 1 degrees
of freedom)
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

0

2
Do not
reject H0



2

Reject H0

Ch. 14-10

Chi-Square Test Statistic
H0: The distribution of calls is uniform
over days of the week
H1: The distribution of calls is not uniform
2
2
2
(290

246)
(250

246)
(192

246)
2 

 ... 
23.05
246
246
246

k – 1 = 6 (7 days of the week) so
use 6 degrees of freedom:

2.05 = 12.5916
Conclusion:
2 = 23.05 > 2 = 12.5916 so

reject H0 and conclude that the
distribution is not uniform
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 = .05

0

Do not
reject H0

Reject H0

2.05 = 12.5916

2

Ch. 14-11


14.2

Goodness-of-Fit Tests, Population
Parameters Unknown

Idea:
 Test whether data follow a specified distribution
(such as binomial, Poisson, or normal) . . .


. . . without assuming the parameters of the
distribution are known



Use sample data to estimate the unknown
population parameters

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Ch. 14-12


Goodness-of-Fit Tests, Population
Parameters Unknown
(continued)


Suppose that a null hypothesis specifies category
probabilities that depend on the estimation (from the
data) of m unknown population parameters



The appropriate goodness-of-fit test is the same as in
the previously section . . .
(Oi  Ei )2
 
Ei
i1

2

K



. . . except that the number of degrees of freedom for
the chi-square random variable is
Degrees of Freedom (K  m  1)



Where K is the number of categories

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Ch. 14-13


14.3

Test of Normality



The assumption that data follow a normal
distribution is common in statistics



Normality was assessed in prior chapters (for
example, with Normal probability plots in
Chapter 5)



Here, a chi-square test is developed

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Ch. 14-14


Test of Normality
(continued)


Two population parameters can be estimated using
sample data:
n

3
(x

x
)
 i

Skewness  i 1

ns3

n

4
(x

x
)
 i

Kurtosis  i 1


ns 4

For a normal distribution,
Skewness = 0
Kurtosis = 3

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Ch. 14-15


Jarque-Bera
Test for Normality


Consider the null hypothesis that the population

distribution is normal



The Jarque-Bera Test for Normality is based on the closeness the
sample skewness to 0 and the sample kurtosis to 3



The test statistic is

 (Skewness)2 (Kurtosis  3)2 
JB n


6
24




as the number of sample observations becomes very large, this
statistic has a chi-square distribution with 2 degrees of freedom



The null hypothesis is rejected for large values of the test statistic

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Ch. 14-16


Jarque-Bera
Test for Normality
(continued)


The chi-square approximation is close only for very
large sample sizes



If the sample size is not very large, the BowmanShelton test statistic is compared to significance points
from text Table 14.9
Sample
size N

10%
point

5% point

Sample
size N

10%
point

5% point

20
30
40
50
75
100
125
150

2.13
2.49
2.70
2.90
3.09
3.14
3.31
3.43

3.26
3.71
3.99
4.26
4.27
4.29
4.34
4.39

200
250

300
400
500
800
∞

3.48
3.54
3.68
3.76
3.91
4.32
4.61

4.43
4.61
4.60
4.74
4.82
5.46
5.99

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Ch. 14-17


Example: Jarque-Bera
Test for Normality


The average daily temperature has been recorded for
200 randomly selected days, with sample skewness
0.232 and kurtosis 3.319



Test the null hypothesis that the true distribution is
normal

 (Skewness)2 (Kurtosis 3) 2 
 (0.232)2 (3.319  3) 2 
JB n


 200
 2.642
6
24
6
24






From Table 14.9 the 10% critical value for n = 200 is
3.48, so there is not sufficient evidence to reject that the
population is normal

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Ch. 14-18


14.3

Contingency Tables

Contingency Tables


Used to classify sample observations according
to a pair of attributes



Also called a cross-classification or crosstabulation table



Assume r categories for attribute A and c
categories for attribute B


Then there are (r x c) possible cross-classifications

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Ch. 14-19


r x c Contingency Table
Attribute B
Attribute A

1

2

...

C

Totals

1
2
.
.
.
r
Totals

O11

O12

O1c

R1

O21

O22

O2c

R2

.
.
.
Or1

.
.
.
Or2

.
.
.
Orc

.
.
.
Rr

C1

C2

…
…
…
…
…
…
…

Cc

n

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Ch. 14-20


Test for Association


Consider n observations tabulated in an r x c
contingency table



Denote by Oij the number of observations in
the cell that is in the ith row and the jth column



The null hypothesis is

H0 : No association exists
between the two attributes in the population


The appropriate test is a chi-square test with
(r-1)(c-1) degrees of freedom

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Ch. 14-21


Test for Association
(continued)



Let Ri and Cj be the row and column totals
The expected number of observations in cell row i and
column j, given that H0 is true, is

Eij 


R iC j
n

A test of association at a significance level  is based
on the chi-square distribution and the following decision
rule
r

c

Reject H0 if χ 2 
i1 j1

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(Oij  Eij )2
Eij

 χ (r2  1)c  1),α
Ch. 14-22


Contingency Table Example
Left-Handed vs. Gender
 Dominant Hand: Left vs. Right
 Gender: Male vs. Female
H0: There is no association between
hand preference and gender
H1: Hand preference is not independent of gender

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Ch. 14-23


Contingency Table Example
(continued)

Sample results organized in a contingency table:
Hand Preference
sample size = n = 300:
120 Females, 12
were left handed
180 Males, 24 were
left handed

Gender

Left

Right

Female

12

108

120

Male

24

156

180

36

264

300

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Ch. 14-24


Logic of the Test
H0: There is no association between
hand preference and gender
H1: Hand preference is not independent of gender


If H0 is true, then the proportion of left-handed females
should be the same as the proportion of left-handed
males



The two proportions above should be the same as the
proportion of left-handed people overall

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Ch. 14-25