chi-square distribution

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Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved.

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Understand the nature and role of chi-square distribution

Identify a wide variety of uses of the chi-square distribution

Conduct a test of hypothesis comparing an observed frequency distribution to an

expected frequency distribution

When you have completed this chapter, you will be able to:

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Conduct a hypothesis test to determine whether two attributes are

independent

Conduct a test of hypothesis for normality using the chi-square distribution

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Characteristics of the Chi-Square Distribution


… it is positively skewed

… it is non-negative

… it is based on degrees of freedom

…when the degrees of freedom change a new distribution is created

…e.g.

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df = 3

df = 5

df = 10



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Goodness-of-Fit Test: Equal Expected

Frequencies


Frequencies

Let f0 and fe be the observed and expected frequencies respectively

H0: There is no difference between the observed and expected frequencies

H1: There is a difference between the observed and the expected frequencies

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…the critical value is a chi-square value with (k-1) degrees of freedom,

where k is the number of categories


Frequencies


Frequencies

e

eo

f

ff 22

… the test statistic is:

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The following information shows the number of employees absent by day of the week

at a large a manufacturing plant.

Day Frequency

Monday 120Tuesday 45Wednesday 60Thursday 90Friday 130

Total 445

Day Frequency

Monday 120Tuesday 45Wednesday 60Thursday 90Friday 130

Total 445


Frequencies


Frequencies

At the .05 level of significance, is there a difference in the absence rate by day of the week?

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Hypothesis Test Hypothesis Test

Step 1Step 1

Step 2Step 2

Step 3Step 3

Step 4Step 4

H0: There is no difference in absence rate by day of the week…

H1: Absence rates by day are not all equal

= 0.05

Use Chi-Square test

Reject H0 if 2 >

(5-1) = 4Degrees of freedom

9.488. (see Appendix I)

(120+45+60+90+130)/5 = 89


Frequencies


Frequencies

Chi-Square

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Right-Tail Area

= 0.05

Degrees of Freedom 5 – 1

= 4

Reject H0 if 2 > 9.488

Using the Table…Using the Table…

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= 1.98 = 1.98

Day Frequency Expected (fo – fe)2/fe

Monday 120 89 10.80Tuesday 45 89 21.75Wednesday 60 89 9.45Thursday 90 89 0.01Friday 130 89

18.89Total 445 445 60.90

Day Frequency Expected (fo – fe)2/fe

Monday 120 89 10.80Tuesday 45 89 21.75Wednesday 60 89 9.45Thursday 90 89 0.01Friday 130 89

18.89Total 445 445 60.90

e

eo

f

ff 22

e

eo

f

ff 22

2

Reject the null hypothesis. Absentee rates are not the same for each day

of the week.

(120-89)2/89Step 5Step 5

Test Statistics

Reject H0 if 2 > 9.488

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At the .05 significance level, can we conclude that the Philadelphia area is different from the U.S. as a whole?

Married Widowed Divorced Single

63.9% 7.7% 6.9% 21.5%

A U.S. Bureau of the Census indicated that…

Not re-married Never married

A sample of 500 adults from the Philadelphia area showed:

310 40 30 120

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Married 310 *319.5 ** .2825

Widowed 40 38.5 .0584

Divorced 30 34.5 .5870

Single 120 107.5 1.4535

Total 500 2.3814

… continued

22.3814

* Census figures would predict: i.e. 639*500 = 319.5

** Our sample: (310-319.5)2/319.5 = .2825

(fo – fe)2/feExpected

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Step 1Step 1

Step 2Step 2

Step 3Step 3

Step 4Step 4

H0: The distribution has not changed

… continued

H1: The distribution has changed.

H0 is rejected if

2 >7.815, df = 3

= 0.05

2 = 2.3814

Reject the null hypothesis.The distribution regarding marital status in Philadelphia is different from the rest of the United States.

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Goodness-of-Fit Test: Normality


… the test investigates if the observed frequencies in a frequency distribution

match the theoretical normal distribution

- Compute the z-value for the lower class limit and the upper class limit for each

class

- Determine fe for each category- Use the chi-square goodness-of-fit test to

determine if fo coincides with fe

…to determine the mean and standard deviation of the frequency distribution

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A sample of 500 donations to the Arthritis Foundation is reported in the

following frequency distribution

Is it reasonable to conclude that the distribution is normally distributed with a mean of $10 and a

standard deviation of $2?

Use the .05 significance level



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Amount Spent

fo Area

fe

(fo- fe )2/fe

<$6 20

$6 up to $8 60

$8 up to $10 140

$10 up to $12 120

$12 up to $14 90

>$14 70

Total 500

… continued

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To compute fe for the first class, first determine the z - value

To compute fe for the first class, first determine the z - value

… continued

Now… find the probability of a z - value less than –2.00

Now… find the probability of a z - value less than –2.00

00.22

106

Xz

.02284772.5000.0)00.2( zP

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Amount Spent

fo Area

fe

(fo- fe )2/fe

<$6 20 .02

$6 up to $8 60 .14

$8 up to $10 140 .34

$10 up to $12 120 .34

$12 up to $14 90 .14

>$14 70 .02

Total 500

… continued

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The expected frequency is the probability of a z-value less than –2.00 times the sample size

… continued

The other expected frequencies are computed similarly

40.11)500)(0228(. ef

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Amount Spent

fo Area

fe

(fo- fe )2/fe

<$6 20 .02 11.40 6.49

$6 up to $8 60 .14 67.95 .93

$8 up to $10 140 .34 170.65 5.50

$10 up to $12 120 .34 170.65 15.03

$12 up to $14 90 .14 67.95 7.16

>$14 70 .02 11.40 301.22

Total 500 500 336.33

… continued

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… continued

Step 1Step 1

Step 2Step 2

Step 3Step 3

Step 4Step 4

H0: The observations follow the normal distribution

H0 is rejected if 2 >7.815, df = 6

= 0.05

2 = 336.33

H0: is rejected.

The observations do NOT follow the normal distribution

H0: The observations do NOT follow the normal distribution

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A contingency table is used to investigate whether two traits or characteristics

are related

A contingency table is used to investigate whether two traits or characteristics

are related

… the expected frequency is computed as: Expected Frequency = (row total)(column total)/grand

total

… each observation is classified according to two criteria

…the usual hypothesis testing procedure is used

… the degrees of freedom is equal to: (number of rows -1)(number of columns -1)

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Is there a relationship between the location of an accident and the gender of the person involved in the accident?

A sample of 150 accidents reported to the police were classified by type and gender.

At the .05 level of significance, can we conclude that gender and the location of

the accident are related?

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… continued

SexWork Home Other

Total

Male 60 20 10 90

Female 20 30 10 60

Total 80 50 20 150

Location

The expected frequency for the work-male intersection is computed as (90)(80)/150 =48

Similarly, you can compute the expected frequencies for the other cells

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Step 1Step 1

Step 2Step 2

Step 3Step 3

Step 4Step 4

H0: The Gender and Location are NOT related

H0 is rejected if 2 >5.991, df = 2

= 0.05

H0: is rejected.Gender and Location are related!

H0: The Gender and Location are related

(…there are (3- 1)(2-1) = 2 degrees of freedom)

Find the value of 2

8

810...

48

4860 222

… continued

667.16

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This completes Chapter 15This completes Chapter 15

chi-square distribution

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