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1Slide© 2005 Thomson/South-Western
Slides Prepared byJOHN S. LOUCKS
St. Edward’s University
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Chapter 12Tests of Goodness of Fit and Independence
Goodness of Fit Test: A Multinomial Population
Goodness of Fit Test: Poissonand Normal Distributions
Test of Independence
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Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population
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 expectedfrequency, ei , in each category by multiplying thecategory probability by the sample size.
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Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population
χ22
1=
−∑=
( )f ee
i i
ii
kχ2
2
1=
−∑=
( )f ee
i i
ii
k
4. Compute the value of the test statistic.
Note: The test statistic has a chi-square distributionwith k – 1 df provided that the expected frequenciesare 5 or more for all categories.
fi = observed frequency for category iei = expected frequency for category ik = number of categories
where:
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Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population
where α is the significance level andthere are k - 1 degrees of freedom
p-value approach:
Critical value approach:
Reject H0 if p-value < α
5. Rejection rule:
2 2αχ χ≥2 2αχ χ≥Reject H0 if
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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, asplit-level, and an A-frame. To helpin production planning, managementwould like to determine if previous customer purchases indicate that thereis a preference in the style selected.
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Split- A-Model Colonial Log Level Frame# Sold 30 20 35 15
The number of homes sold of eachmodel for 100 sales over the past twoyears is shown below.
Multinomial Distribution Goodness of Fit Test
Example: Finger Lakes Homes (A)
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Hypotheses
Multinomial Distribution Goodness of Fit Test
where:pC = population proportion that purchase a colonialpL = population proportion that purchase a log cabinpS = population proportion that purchase a split-levelpA = population proportion that purchase an A-frame
H0: pC = pL = pS = pA = .25Ha: The population proportions are not
pC = .25, pL = .25, pS = .25, and pA = .25
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Rejection Rule
χ2
7.815
Do Not Reject H0 Reject H0
Multinomial Distribution Goodness of Fit Test
With α = .05 andk - 1 = 4 - 1 = 3
degrees of freedom
Reject H0 if p-value < .05 or χ2 > 7.815.
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Expected Frequencies
Test Statistic
χ22 2 2 230 25
2520 25
2535 25
2515 25
25=
−+
−+
−+
−( ) ( ) ( ) ( )χ22 2 2 230 25
2520 25
2535 25
2515 25
25=
−+
−+
−+
−( ) ( ) ( ) ( )
Multinomial Distribution Goodness of Fit Test
e1 = .25(100) = 25 e2 = .25(100) = 25e3 = .25(100) = 25 e4 = .25(100) = 25
= 1 + 1 + 4 + 4 = 10
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Multinomial Distribution Goodness of Fit Test
Conclusion Using the p-Value Approach
The p-value < α . We can reject the null hypothesis.
Because χ2 = 10 is between 9.348 and 11.345, thearea in the upper tail of the distribution is between.025 and .01.
Area in Upper Tail .10 .05 .025 .01 .005χ2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838
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Conclusion Using the Critical Value Approach
Multinomial Distribution Goodness of Fit Test
We reject, at the .05 level of significance,the assumption that there is no home stylepreference.
χ2 = 10 > 7.815
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Test of Independence: Contingency Tables
e i jij =
(Row Total)(Column Total) Sample Size
e i jij =
(Row Total)(Column Total) Sample Size
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.
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Test of Independence: Contingency Tables
χ22
=−
∑∑( )f e
eij ij
ijjiχ2
2
=−
∑∑( )f e
eij ij
ijji
5. Determine the rejection rule.
Reject H0 if p -value < α or . 2 2αχ χ≥2 2αχ χ≥
4. Compute the test statistic.
where α is the significance level and,with n rows and m columns, there are(n - 1)(m - 1) degrees of freedom.
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Each home sold by Finger LakesHomes can be classified according toprice and to style. Finger Lakes’manager would like to determine ifthe price of the home and the style ofthe home are independent variables.
Contingency Table (Independence) Test
Example: Finger Lakes Homes (B)
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Price Colonial Log Split-Level A-Frame
The number of homes sold foreach model and price for the past twoyears is shown below. For convenience,the price of the home is listed as either$99,000 or less or more than $99,000.
> $99,000 12 14 16 3< $99,000 18 6 19 12
Contingency Table (Independence) Test
Example: Finger Lakes Homes (B)
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Hypotheses
Contingency Table (Independence) Test
H0: Price of the home is independent of thestyle of the home that is purchased
Ha: Price of the home is not independent of thestyle of the home that is purchased
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Expected Frequencies
Contingency Table (Independence) Test
Price Colonial Log Split-Level A-Frame Total< $99K> $99K
Total 30 20 35 15 100
12 14 16 3 4518 6 19 12 55
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Rejection Rule
Contingency Table (Independence) Test
2.05 7.815χ =2.05 7.815χ =With α = .05 and (2 - 1)(4 - 1) = 3 d.f.,
Reject H0 if p-value < .05 or χ2 > 7.815
χ22 2 218 16 5
16 56 11
113 6 75
6 75=
−+
−+ +
−( . ).
( ) . . ( . ).
. χ22 2 218 16 5
16 56 11
113 6 75
6 75=
−+
−+ +
−( . ).
( ) . . ( . ).
.
= .1364 + 2.2727 + . . . + 2.0833 = 9.149
Test Statistic
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Conclusion Using the p-Value Approach
The p-value < α . We can reject the null hypothesis.
Because χ2 = 9.145 is between 7.815 and 9.348, thearea in the upper tail of the distribution is between.05 and .025.
Area in Upper Tail .10 .05 .025 .01 .005χ2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838
Contingency Table (Independence) Test
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Conclusion Using the Critical Value Approach
Contingency Table (Independence) Test
We reject, at the .05 level of significance,the assumption that the price of the home isindependent of the style of home that ispurchased.
χ2 = 9.145 > 7.815
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Goodness of Fit Test: Poisson Distribution
1. Set up the null and alternative hypotheses.H0: Population has a Poisson probability distributionHa: Population does not have a Poisson distribution
3. Compute the expected frequency of occurrences eifor each value of the Poisson random variable.
2. Select a random sample anda. Record the observed frequency fi for each value of
the Poisson random variable.b. Compute the mean number of occurrences µ.
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Goodness of Fit Test: Poisson Distribution
χ22
1=
−∑=
( )f ee
i i
ii
kχ2
2
1=
−∑=
( )f ee
i i
ii
k
4. Compute the value of the test statistic.
fi = observed frequency for category iei = expected frequency for category ik = number of categories
where:
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where α is the significance level andthere are k - 2 degrees of freedom
p-value approach:
Critical value approach:
Reject H0 if p-value < α
5. Rejection rule:
2 2αχ χ≥2 2αχ χ≥Reject H0 if
Goodness of Fit Test: Poisson Distribution
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Example: Troy Parking GarageIn studying the need for an
additional entrance to a city parking garage, a consultant has recommended an analysisapproach that is applicable only in situations where the number of carsentering during a specified time period follows aPoisson distribution.
Goodness of Fit Test: Poisson Distribution
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A random sample of 100 one-minute time intervals resultedin the customer arrivals listedbelow. A statistical test mustbe conducted to see if theassumption of a Poisson distribution is reasonable.
Goodness of Fit Test: Poisson Distribution
Example: Troy Parking Garage
# Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1
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Hypotheses
Goodness of Fit Test: Poisson Distribution
Ha: Number of cars entering the garage during aone-minute interval is not Poisson distributed
H0: Number of cars entering the garage duringa one-minute interval is Poisson distributed
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Estimate of Poisson Probability Function
f x ex
x( )
!=
−6 6f x e
x
x( )
!=
−6 6
Goodness of Fit Test: Poisson Distribution
Τotal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600
Hence,
Estimate of µ = 600/100 = 6Total Time Periods = 100
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Expected Frequencies
Goodness of Fit Test: Poisson Distribution
x f (x ) nf (x )
0123456
13.7710.33
6.884.132.252.01
100.00
.1377
.1033
.0688
.0413
.0225
.02011.0000
789
101112+
Total
.0025
.0149
.0446
.0892
.1339
.1606
.1606
.251.494.468.92
13.3916.0616.06
x f (x ) nf (x )
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Observed and Expected Frequencies
Goodness of Fit Test: Poisson Distribution
i fi ei fi - ei
-1.201.080.613.94
-4.06-1.77-1.331.121.61
6.208.92
13.3916.0616.0613.7710.33
6.888.39
51014201212
98
10
0 or 1 or 23456789
10 or more
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Test Statistic
χ −= + + + =
2 2 22 ( 1.20) (1.08) (1.61) . . . 3.268
6.20 8.92 8.39χ −
= + + + =2 2 2
2 ( 1.20) (1.08) (1.61) . . . 3.2686.20 8.92 8.39
Goodness of Fit Test: Poisson Distribution
With α = .05 and k - p - 1 = 9 - 1 - 1 = 7 d.f.(where k = number of categories and p = numberof population parameters estimated), 2
.05 14.067χ =2.05 14.067χ =
Reject H0 if p-value < .05 or χ2 > 14.067.
Rejection Rule
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Conclusion Using the p-Value Approach
The p-value > α . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution.
Because χ2 = 3.268 is between 2.833 and 12.017 in the Chi-Square Distribution Table, the area in the upper tailof the distribution is between .90 and .10.
Area in Upper Tail .90 .10 .05 .025 .01 χ2 Value (df = 7) 2.833 12.017 14.067 16.013 18.475
Goodness of Fit Test: Poisson Distribution
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Goodness of Fit Test: Normal Distribution
1. Set up the null and alternative hypotheses.
3. Compute the expected frequency, ei , for each interval.
2. Select a random sample anda. 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
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4. Compute the value of the test statistic.
Goodness of Fit Test: Normal Distribution
χ22
1=
−∑=
( )f ee
i i
ii
kχ2
2
1=
−∑=
( )f ee
i i
ii
k
5. Reject H0 if (where α is the significance leveland there are k - 3 degrees of freedom).
2 2αχ χ≥2 2αχ χ≥
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Normal Distribution Goodness of Fit Test
Example: IQ Computers
IQIQ Computers (one better than HP?)
manufactures and sells a generalpurpose microcomputer. As part ofa study to evaluate sales personnel, managementwants to determine, at a .05 significance level, if theannual sales volume (number of units sold by asalesperson) follows a normal probability distribution.
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A simple random sample of 30 ofthe salespeople was taken and theirnumbers of units sold are below.
Normal Distribution Goodness of Fit Test
Example: IQ Computers
(mean = 71, standard deviation = 18.54)
33 43 44 45 52 52 56 58 63 6464 65 66 68 70 72 73 73 74 7583 84 85 86 91 92 94 98 102 105
IQ
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Hypotheses
Normal Distribution Goodness of Fit Test
Ha: The population of number of units solddoes not have a normal distribution withmean 71 and standard deviation 18.54.
H0: The population of number of units soldhas a normal distribution with mean 71and standard deviation 18.54.
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Interval Definition
Normal Distribution Goodness of Fit Test
To satisfy the requirement of an expectedfrequency of at least 5 in each interval we willdivide the normal distribution into 30/5 = 6equal probability intervals.
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Interval Definition
Areas= 1.00/6= .1667
7153.0271 − .43(18.54) = 63.03 78.97
88.98 = 71 + .97(18.54)
Normal Distribution Goodness of Fit Test
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Observed and Expected Frequencies
Normal Distribution Goodness of Fit Test
1-210
-11
555555
30
636546
30
Less than 53.0253.02 to 63.0363.03 to 71.0071.00 to 78.9778.97 to 88.98
More than 88.98
i fi ei fi - ei
Total
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2 2 2 2 2 22 (1) ( 2) (1) (0) ( 1) (1) 1.600
5 5 5 5 5 5χ − −
= + + + + + =2 2 2 2 2 2
2 (1) ( 2) (1) (0) ( 1) (1) 1.6005 5 5 5 5 5
χ − −= + + + + + =
Test Statistic
With α = .05 and k - p - 1 = 6 - 2 - 1 = 3 d.f.(where k = number of categories and p = numberof population parameters estimated), 2
.05 7.815χ =2.05 7.815χ =
Reject H0 if p-value < .05 or χ2 > 7.815.
Rejection Rule
Normal Distribution Goodness of Fit Test
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Normal Distribution Goodness of Fit Test
Conclusion Using the p-Value Approach
The p-value > α . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with µ = 71 and σ = 18.54.
Because χ2 = 1.600 is between .584 and 6.251 in the Chi-Square Distribution Table, the area in the upper tailof the distribution is between .90 and .10.
Area in Upper Tail .90 .10 .05 .025 .01 χ2 Value (df = 3) .584 6.251 7.815 9.348 11.345
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End of Chapter 12
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