Transcript Document

Statistics
Tests of
Goodness of Fit and Independence
1/71
Contents
STATISTICS in PRACTICE
 Goodness of Fit Test: A Multinomial
Population
 Test of Independence
 Goodness of Fit Test:

 Poisson and Normal Distributions
2/71
STATISTICS in PRACTICE

United Way of Greater
Rochester is a nonprofit
organization dedicated
to improving the quality
of life for people. Because of enormous
volunteer involvement, United Way of Greater
Rochester is able to hold its operating costs at
just eight cents of every dollar raised.
3/71
STATISTICS in PRACTICE


One of statistical tests was to determine whether
perceptions of administrative expenses were
independent of occupation.
In this chapter, you will learn how a statistical
test of independence.
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Hypothesis (Goodness of Fit)
Test for Proportions of a
Multinomial Population
 Consider the case that each element of a
population is assigned to one and only one
of several classes or categories.
 Such a population is a multinomial
population(多項母體).
 We want to test if the population follows a
multinominal distribution with specified
probabilities for each of the k categories.
5/71
Hypothesis Test for Proportions
of a Multinomial Population-Procedures
1. Set up the null and alternative hypotheses.
2. Select a random sample and record the
observed frequency, fi , for each of the k
categories.
3. Assuming H0 is true, compute the expected
frequency, ei , in each category by
multiplying the category probability by the
sample size.
6/71
Hypothesis Test for Proportions
of a Multinomial Population-Procedures
4. Compute the value of the test statistic.
( f i  ei ) 2
 
ei
i 1
2
k
where:
fi = observed frequency for category i
ei = expected frequency for category i
k = number of categories
Note: The test statistic has a chi-square distribution with (k – 1) df
provided that the expected frequencies are 5 or more for all
categories.
7/71
Hypothesis Test for Proportions
of a Multinomial Population-Procedures
5. Rejection rule:
p-value approach:
Reject H0 if p-value < 
Critical value approach:
Reject H0 if
 2  2
where  is the significance level and
there are k - 1 degrees of freedom
8/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population

Example:



The market share study being conducted by Scott
Marketing Research. Over the past year market shares
stabilized at 30% for company A, 50% for company B,
and 20% for company C.
Recently company C developed a “new and improved”
product to replace its current entry in the market.
Company C retained Scott Marketing Research to
determine whether the new product will alter market
shares.
9/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population



Scott Marketing Research conduct a sample survey
and compute the proportion preferring each
company’s product.
A hypothesis test will then be conducted to see
whether the new product caused a change in market
shares.
The null and alternative hypotheses are
10/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population


Perform a goodness of fit test that will
determine whether the sample of 200 customer
purchase preferences is consistent with the null
hypothesis.
Data
11/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population

Expected: the expected frequency for each
category is found by multiplying the sample
size of 200 by the hypothesized proportion for
the category.
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Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population

Computation of the Chi-square Test Statistic
for The Scott Marketing Research Market
Share Study
13/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population
p-Value Approach
The test statistic χ2 = 7.34 and 5.991 < 7.34 <
7.378. Thus, the corresponding upper tail area
or p-value must be between .05 and .025. with
p-value < .05, we reject H0.
Minitab or Excel can be used to show χ 2 = 7.34
provides a p-value = .0255.
14/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population

Critical Value Approach



With α = .05 and 2 degrees of freedom, the
critical value for the test statistic is
=
5.991. The upper tail rejection rule becomes
Reject H0 if χ 2 > 5.991
With 7.34 > 5.991, we reject H0.
Comparisons of the observed and expected
frequencies for the other two companies
indicate that company C’s gain in market share
will hurt company A more than company B. 15/71
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial
Population
Note:
The chi-square test is an approximate test
and the test result may not be valid when
the expected value for a category is less
than 5.
If one or more categories have expected
values less than 5, you can combine them
with adjacent categories to achieve the
minimum required expected value.
16/71
Multinomial Distribution
Goodness of Fit Test

Example: Finger Lakes Homes (A)
Finger Lakes Homes manufactures four models of
prefabricated homes, a two-story colonial, a log
cabin, a split-level, and an A-frame. To help
in production planning, management would like to
determine if previous customer purchases indicate
that there is a preference in the style selected.
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Multinomial Distribution
Goodness of Fit Test

Example: Finger Lakes Homes (A)
The number of homes sold of each
model for 100 sales over the past two
years is shown below.
SplitAModel Colonial Log Level Frame
# Sold
30
20
35
15
18/71
Multinomial Distribution
Goodness of Fit Test
Hypotheses
H0: pC = pL = pS = pA = .25
Ha: The population proportions are not
pC = .25, pL = .25, pS = .25, and pA = .25
where:
pC = population proportion that purchase a colonial
pL = population proportion that purchase a log cabin
pS = population proportion that purchase a split-level
pA = population proportion that purchase an A-frame
19/71
Multinomial Distribution
Goodness of Fit Test
Rejection Rule
Reject H0 if p-value < .05 or χ2 > 7.815.
With  = .05 and
k-1=4-1=3
degrees of freedom
Do Not Reject H0
Reject H0
7.815
χ2
20/71
Multinomial Distribution
Goodness of Fit Test

Expected Frequencies
e1 = .25(100) = 25
e3 = .25(100) = 25

e2 = .25(100) = 25
e4 = .25(100) = 25
Test Statistic
( 30  25) 2 ( 20  25) 2 ( 35  25) 2 (15  25) 2
 



25
25
25
25
=1+1+4+4
= 10
2
21/71
Multinomial Distribution
Goodness of Fit Test

Conclusion Using the p-Value Approach
Area in Upper Tail
2 Value (df = 3)
.10
.05
.025
.01
.005
6.251 7.815 9.348 11.345 12.838
Because χ2 = 10 is between 9.348 and 11.345, the
area in the upper tail of the distribution is
between .025 and .01.
The p-value <  . We can reject the null hypothesis.
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Multinomial Distribution
Goodness of Fit Test

Conclusion Using the Critical Value
Approach
χ2 = 10 > 7.815
We reject, at the .05 level of significance, the
assumption that there is no home style preference.
23/71
Test of Independence:
Contingency Tables(列聯表）
 Another important application of the chisquare distribution involves using sample
data to test for the independence of two
variables.
 Contingency table is a table that lists all
possible combinations of the two variables.
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Test of Independence: Contingency
Tables

Example:


A test of independence addresses the question of
whether the beer preference (light, regular, or dark)
is independent of the gender of the beer drinker
(male, female).
If the independence assumption is valid, we argue
that the fraction of drinking light beer, regular beer
and dark beer must be applicable to both male and
female beer drinker.
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Test of Independence:
Contingency Tables
1. Set up the null and alternative hypotheses.
2. Select a random sample and record the
observed frequency, fij , for each cell of the
contingency table.
3. Compute the expected frequency, eij , for
each cell.
(Row i Total)(Column j Total)
eij 
Sample Size
26/71
Test of Independence:
Contingency Tables
4. Compute the test statistic.
2   
i
j
( f ij  eij ) 2
eij
5. Determine the rejection rule.
2
2
Reject H0 if p -value <  or   .
where  is the significance level and,
with n rows and m columns, there are
(n - 1)(m - 1) degrees of freedom.
27/71
Test of Independence: Contingency
Tables

Example:


A test of independence addresses the question of
whether the beer preference (light, regular, or dark)
is independent of the gender of the beer drinker
(male, female).
The hypotheses for this test of independence are:
H0: Beer preference is independent of the gender
of the beer drinker
Ha: Beer preference is not independent of the
28/71
gender of the beer drinker
Test of Independence:
Contingency Tables

Sample Results
1/3

7/15
1/5
1
Expected Frequencies
80*1/3=26.67
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Test of Independence:
Contingency Tables
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Test of Independence:
Contingency Tables


Computation of the Chi-square Test
Statistic
p-value = .0468. At the .05 level of
significance, p-value < α= .05. We
reject the null hypothesis
31/71
Test of Independence:
Contingency Tables

Example: Finger Lakes Homes (B)
Each home sold by Finger Lakes Homes can be
classified according to price and to style. Finger
Lakes’manager would like to determine if the
price of the home and the style of the home are
independent variables.
32/71
Test of Independence:
Contingency Tables

Example: Finger Lakes Homes (B)
The number of homes sold for each model
and price for the past two years is shown below.
For convenience, the price of the home is listed
as either $99,000 or less or more than $99,000.
Price Colonial
< $99,000
18
> $99,000
12
Log
6
14
Split-Level A-Frame
19
12
16
3
33/71
Test of Independence:
Contingency Tables

Hypotheses
H0: Price of the home is independent of the
style of the home that is purchased
Ha: Price of the home is not independent of the
style of the home that is purchased
34/71
Test of Independence:
Contingency Tables
Expected Frequencies
Price
Colonial
< $99K
18
> $99K
Total
Log
Split-Level
A-Frame
Total
6
19
12
55
12
14
16
3
45
30
20
35
15
100
35/71
Contingency Table
(Independence) Test

Rejection Rule
2

With α = .05 and (2 - 1)(4 - 1) = 3 d.f., .05 =7.815
Reject H0 if p-value < .05 or 2 > 7.815

Test Statistic
2
2
2
(
18

16
.
5
)
(
6

11
)
(
3

6
.
75
)
2 


16.5
11
6.75
= .1364 + 2.2727 + . . . + 2.0833 = 9.149
36/71
Contingency Table
(Independence) Test

Conclusion Using the p-Value
Approach
Area in Upper Tail
2 Value (df = 3)
.10
.05
.025
.01
.005
6.251 7.815 9.348 11.345 12.838
Because χ2 = 9.145 is between 7.815 and 9.348,
the area in the upper tail of the distribution is
between .05 and .025.
The p-value <  . We can reject the null
hypothesis.
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Contingency Table
(Independence) Test
Conclusion Using the Critical Value
Approach
χ2 = 9.145 > 7.815
We reject, at the .05 level of significance,
the assumption that the price of the home is
independent of the style of home that is
purchased.
38/71
Goodness of Fit Test: Poisson
Distribution
 In general, the goodness of fit test can be used
with any hypothesized probability distribution.
Here we will demonstrate the cases of Poisson
distribution and Normal distribution.
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Goodness of Fit Test: Poisson
Distribution
1. Set up the null and alternative hypotheses.
H0: Population has a Poisson probability
distribution
Ha: Population does not have a Poisson
distribution
2. Select a random sample and
a. Record the observed frequency fi for each
value of the Poisson random variable.
b. Compute the mean number of occurrences40/71
.
Goodness of Fit Test: Poisson
Distribution
3. Compute the expected frequency of occurrences ei
for each value of the Poisson random variable.
4. Compute the value of the test statistic.
where:
fi = observed frequency for category i
ei = expected frequency for category i
k = number of categories
41/71
Goodness of Fit Test: Poisson
Distribution
5. Rejection rule:
p-value approach:
Reject H0 if p-value < 
Critical value approach:
Reject H0 if
 2  2
where  is the significance level and
there are k - 2 degrees of freedom
42/71
Goodness of Fit Test: Poisson
Distribution—Example


Consider the arrival of customers at
Dubek’s Food Market in Tallahassee,
Florida.
A statistical test conducted to see
whether an assumption of a Poisson
distribution for arrivals is reasonable.
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Goodness of Fit Test: Poisson
Distribution—Example

Hypotheses are
H0: The number of customers entering
the store during 5-minute intervals
has a Poisson probability distribution
Ha: The number of customers entering the
store during 5-minute intervals does
not have a Poisson distribution
44/71
Goodness of Fit Test: Poisson
Distribution—Example

The Poisson probability function,
f ( x) 
x 
 e
x!
where
--μ represents the expected number of customers
arriving per 5-minute period,
--x is the random variable indicating the number of
customers arriving during a 5-minute period, and
--f (x) is the probability that x customers will arrive in
a 5-minute interval.
45/71
Goodness of Fit Test: Poisson
Distribution—Example


Sample Data
μ known, an estimate
of μ is 640/128 = 5
customers
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Goodness of Fit Test: Poisson
Distribution—Example

Expected frequency
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Goodness of Fit Test: Poisson
Distribution—Example


Computation of the Chi-square Test
Statistic
p-value = .1403. With p-value > α
= .05, we cannot reject H0.
48/71
Goodness of Fit Test:
Poisson Distribution

Example: Troy Parking Garage
In studying the need for an additional
entrance to a city parking garage, a
consultant has recommended an analysis
approach that is applicable only in situations
where the number of cars entering during a
specified time period follows a Poisson
distribution.
49/71
Goodness of Fit Test:
Poisson Distribution
A random sample of 100 one-minute time
intervals resulted in the customer arrivals
listed below. A statistical test must be
conducted to see if the assumption of a
Poisson distribution is reasonable.
# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12
Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1
50/71
Goodness of Fit Test:
Poisson Distribution

Hypotheses
H0: Number of cars entering the garage during
a one-minute interval is Poisson distributed
Ha: Number of cars entering the garage during a
one-minute interval is not Poisson distributed
51/71
Goodness of Fit Test:
Poisson Distribution

Estimate of Poisson Probability
Function
otal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600
Estimate of  = 600/100 = 6
Total Time Periods = 100
Hence,
x
6
6 e
f ( x) 
x!
52/71
Goodness of Fit Test:
Poisson Distribution

Expected Frequencies
x
f (x )
nf (x )
x
f (x )
nf (x )
0
1
2
3
4
5
6
.0025
.0149
.0446
.0892
.1339
.1606
.1606
.25
1.49
4.46
8.92
13.39
16.06
16.06
7
8
9
10
11
12+
Total
.1377
.1033
.0688
.0413
.0225
.0201
1.0000
13.77
10.33
6.88
4.13
2.25
2.01
100.00
53/71
Goodness of Fit Test:
Poisson Distribution

Observed and Expected Frequencies
i
fi
ei
f i - ei
0 or 1 or 2
3
4
5
6
7
8
9
10 or more
5
10
14
20
12
12
9
8
10
6.20
8.92
13.39
16.06
16.06
13.77
10.33
6.88
8.39
-1.20
1.08
0.61
3.94
-4.06
-1.77
-1.33
1.12
1.61
54/71
Goodness of Fit Test:
Poisson Distribution

Rejection Rule
2

With  = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. .05  14.067
(where k = number of categories and p = number
of population parameters estimated),
Reject H0 if p-value < .05 or 2 > 14.067.

Test Statistic
2
2
2
(

1
.
20
)
(
1
.
08
)
(
1
.
61
)
2 


 3.268
6.20
8.92
8.39
55/71
Goodness of Fit Test:
Poisson Distribution

Conclusion Using the p-Value Approach
Area in Upper Tail
2 Value (df = 7)
.90
.10
.05
.025
.01
2.833 12.017 14.067 16.013 18.475
Because χ2 = 3.268 is between 2.833 and 12.017 in the
Chi-Square Distribution Table, the area in the upper tail
of the distribution is between .90 and .10.
The p-value >  . We cannot reject the null hypothesis.
There is no reason to doubt the assumption of a Poisson
distribution.
56/71
Goodness of Fit Test: Normal
Distribution
1. Set up the null and alternative hypotheses.
2. Select a random sample and
a. Compute the mean and standard deviation.
b. Define intervals of values so that the expected
frequency is at least 5 for each interval.
c. For each interval record the observed frequencies
3. Compute the expected frequency, ei , for each interval.
57/71
Goodness of Fit Test: Normal
Distribution
4. Compute the value of the test statistic.
5. Reject H0 if  2   2 (where  is the significance
level and there are k - 3 degrees of freedom).
58/71
Goodness of Fit Test: Normal
Distribution--Example



Chemline Employee
Aptitude Scores
To test the null
hypothesis that the
population of test
scores has a normal
distribution.
Sample Data
59/71
Goodness of Fit Test: Normal
Distribution—Example


Estimates of μ and  are x= 68.42, s = 10.41.
Hypotheses
H0: The population of test scores has a
normal distribution with mean
68.42 and standard deviation 10.41.
Ha: The population of test scores does not
have a normal distribution with mean
68.42 and standard deviation 10.41
60/71
Goodness of Fit Test: Normal
Distribution—Example



Expected Frequency
Normal Distribution with 10 EqualProbability Intervals (The Chemiline
Example)
Note: 55.10 = 68.42 - 1.28(10.41)
61/71
Goodness of Fit Test: Normal
Distribution—Example
62/71
Goodness of Fit Test: Normal
Distribution—Example

Computation of the Chi-square Test
Statistic

p-value = .4084 >α = .10. Cannot reject
H0.
63/71
Normal Distribution
Goodness of Fit Test

IQ
Example: IQ Computers
IQ Computers (one better than
HP?) manufactures and sells a
general purpose microcomputer. As part of
a study to evaluate sales personnel, management
wants to determine, at a .05 significance level, if the
annual sales volume (number of units sold by a
salesperson) follows a normal probability
64/71
distribution.
Normal Distribution
Goodness of Fit Test

IQ
Example: IQ Computers
A simple random sample of 30 of the
salespeople was taken and their
numbers of units sold are below.
33
64
83
43
65
84
44
66
85
45
68
86
52
70
91
52
72
92
56
73
94
58 63 64
73 74 75
98 102 105
(mean = 71, standard deviation = 18.54)
65/71
Normal Distribution
Goodness of Fit Test

Hypotheses
H0: The population of number of units sold
has a normal distribution with mean 71
and standard deviation 18.54.
Ha: The population of number of units sold
does not have a normal distribution with
mean 71 and standard deviation 18.54.
66/71
Normal Distribution
Goodness of Fit Test

Interval Definition
To satisfy the requirement of an expected
frequency of at least 5 in each interval we
will divide the normal distribution into
30/5 = 6 equal probability intervals.
67/71
Normal Distribution
Goodness of Fit Test

Interval Definition
Areas
= 1.00/6
= .1667
53.02
71
88.98 = 71 + .97(18.54)
71  .43(18.54) = 63.03 78.97
68/71
Normal Distribution
Goodness of Fit Test

Observed and Expected Frequencies
i
fi
ei
f i - ei
Less than 53.02
53.02 to 63.03
63.03 to 71.00
71.00 to 78.97
78.97 to 88.98
More than 88.98
6
3
6
5
4
6
30
5
5
5
5
5
5
30
1
-2
1
0
-1
1
Total
69/71
Normal Distribution
Goodness of Fit Test

Rejection Rule
With  = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f.
(where k = number of categories and p = number
of population parameters estimated),
Reject H0 if p-value < .05 or 2 > 7.815.

Test Statistic
2
2
2
2
2
2
(1)
(

2)
(1)
(0)
(

1)
(1)
2 





 1.600
5
5
5
5
5
5
70/71
Normal Distribution
Goodness of Fit Test

Conclusion Using the p-Value Approach
Area in Upper Tail
.90
.10
.05
2 Value (df = 3)
.584 6.251 7.815
.025
.01
9.348 11.345
 Because χ2 = 1.600 is between .584 and 6.251 in the ChiSquare Distribution Table, the area in the upper tail of the
distribution is between .90 and .10. The p-value >  . We
cannot reject the null hypothesis.
There is little evidence to support rejecting the
assumption the population is normally distributed with  =
71/71
71 and  = 18.54.