Business statistics, 6e, 2005, groebner CH12

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Prentice-Chapter Goals

After completing this chapter, you should be

able to:

 Use the chi-square goodness-of-fit test to

determine whether data fits a specified distribution

 Set up a contingency analysis table and perform a chi-square test of independence

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 Does sample data conform to a hypothesized

distribution?

 Examples:

days of the week? (i.e., do calls follow a uniform distribution?)

process follow a normal distribution?

Chi-Square Goodness-of-Fit

Test

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Prentice- 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:

 = 1722

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Logic of Goodness-of-Fit Test

would be expected to be equally divided across the 7 days:

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

expected results

uniform if

day per

calls expected

246 7

1722

=

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Tuesday Wednesday Thursday Friday Saturday Sunday

290 250 238 257 265 230 192

246 246 246 246 246 246 246

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Chi-Square Test Statistic

 The test statistic is

1) k

df (where

e

) e (o

i

2 i i

where:

k = number of categories

o i = observed cell frequency for category i

e i = expected cell frequency for category i

H 0 : The distribution of calls is uniform over days of the week

H A : The distribution of calls is not uniform

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Prentice-The Rejection Region

2

e

) e o

2 > c c

(with k – 1 degrees of

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23.05 246

246)

(192

246

246)

(250 246

246)

2 = - + - + + - =

c

Chi-Square Test Statistic

H0: The distribution of calls is uniform over days of the week

HA: The distribution of calls is not uniform

0

 = 05

Reject H0

Do not reject H0

 2

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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= 15?

 Process:

 Get sample data

 Group sample results into classes (cells) (Expected cell frequency must be at least

5 for each cell)

 Compare actual cell frequencies with expected cell frequencies

Normal Distribution Example

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Normal Distribution Example

150 Sample Measurements

80 65 36 66 50 38 57 77 59

 Sample data and values grouped into classes:

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Prentice- What are the expected frequencies for these classes for

a normal distribution with μ = 50 and σ = 15?

(continue d)

• Class • Frequency

• Expected Frequency

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μ=50, σ=15 Example:

.0912

1.3333) P(z

15

50 30

z P 30) P(x

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Prentice-The Test Statistic

• Class

• Frequency

• Expected Frequency, ei

e

) e o

2

2 á

2 > c c

The test statistic is

(with k – 1 degrees of freedom)

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The Rejection Region

097

12 57

0

) 57 0 2

(

68 13

) 68 13 10

( e

) e o

i

2 i i

 2

8 classes so use 7 d.f.:

 2 05 = 14.0671

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Prentice-Contingency Tables

Contingency Tables

proportions

to two or more characteristics

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Contingency Table Example

H 0 : Hand preference is independent of gender

H A : Hand preference is not independent of gender

Left-Handed vs Gender

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Prentice-Contingency Table Example

Sample results organized in a contingency table:

(continue d)

Gender

Hand Preference Left Right Female 12 108 120

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Logic of the Test

 If H 0 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

H 0 : Hand preference is independent of gender

H A : Hand preference is not independent of gender

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Prentice-Finding Expected

Frequencies

Overall:

P(Left Handed) = 36/300 = 12

120 Females, 12 were left handed

180 Males, 24 were left handed

If independent, then

P(Left Handed | Female) = P(Left Handed | Male) = 12

So we would expect 12% of the 120 females and 12% of the 180

males to be left handed…

i.e., we would expect (120)(.12) = 14.4 females to be left handed

(180)(.12) = 21.6 males to be left handed

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Expected Cell Frequencies

 Expected cell frequencies:

(continue d)

size sample

Total

total) Column

j total)(

Row

i(

e

th th

4

14 300

) 36 )(

120

(

Example:

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The Chi-Square Test Statistic

 where:

o ij = observed frequency in cell (i, j)

e ij = expected frequency in cell (i, j)

c

1

2 ij ij

2

e

) e o

(

The Chi-square contingency test statistic is:

) 1 c )(

1 r ( d

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0 4

158

) 4 158 156

( 6

21

) 6 21 24

( 6

105

) 6 105 108

( 4

14

) 4 14 12

c

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Contingency Analysis

 

 2 05 = 3.841

Reject H 0

 = 0.05

Decision Rule:

If  2 > 3.841, reject H 0 , otherwise, do not reject H 0

1 (1)(1)

1) - 1)(c -

(r d.f.

with 6848

0

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Prentice-Chapter Summary

 Used the chi-square goodness-of-fit test to

determine whether data fits a specified distribution

 Example of a discrete distribution (uniform)

 Example of a continuous distribution (normal)

 Used contingency tables to perform a chi-square test of independence

 Compared observed cell frequencies to expected cell frequencies

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