© 2005 thomson/south-western slide 1 · 2 © 2005 thomson/south-western slide 4 hypothesis...

1

1Slide© 2005 Thomson/South-Western

Slides Prepared byJOHN S. LOUCKS

St. Edward’s University


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


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.

2



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:



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


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.

3


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.


Example: Finger Lakes Homes (A)


Hypotheses


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


Rejection Rule

χ2

7.815

Do Not Reject H0 Reject H0


With α = .05 andk - 1 = 4 - 1 = 3

degrees of freedom

Reject H0 if p-value < .05 or χ2 > 7.815.

4


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=

−+

−+

−+

−( ) ( ) ( ) ( )


e1 = .25(100) = 25 e2 = .25(100) = 25e3 = .25(100) = 25 e4 = .25(100) = 25

= 1 + 1 + 4 + 4 = 10



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


Conclusion Using the Critical Value Approach


We reject, at the .05 level of significance,the assumption that there is no home stylepreference.

χ2 = 10 > 7.815

5


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.


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.


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)

6


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


Example: Finger Lakes Homes (B)


Hypotheses


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




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

7


Rejection Rule


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



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



Conclusion Using the Critical Value Approach


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

8


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.



χ22

1=

−∑=

( )f ee

i i

ii

kχ2

2

1=

−∑=

( )f ee

i i

ii

k


fi = observed frequency for category iei = expected frequency for category ik = number of categories

where:


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


9


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.



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.


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


Hypotheses


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

10


Estimate of Poisson Probability Function

f x ex

x( )

!=

−6 6f x e

x

x( )

!=

−6 6


Τotal Arrivals = 0(0) + 1(1) + 2(4) + . . . + 12(1) = 600

Hence,

Estimate of µ = 600/100 = 6Total Time Periods = 100




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 )


Observed and Expected Frequencies


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

11


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


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 =


Rejection Rule



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: 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

12



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≥


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.


A simple random sample of 30 ofthe salespeople was taken and theirnumbers of units sold are below.


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

13


Hypotheses


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.


Interval Definition


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.


Interval Definition

Areas= 1.00/6= .1667

7153.0271 − .43(18.54) = 63.03 78.97

88.98 = 71 + .97(18.54)


14


Observed and Expected Frequencies


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


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 =


Rejection Rule





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

15


End of Chapter 12

© 2005 thomson/south-western slide 1 · 2 © 2005 thomson/south-western slide 4 hypothesis...

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