CHAPTER 11:

CHI-SQUARE TESTS

2

THE CHI-SQUARE DISTRIBUTION

Definition The chi-square distribution has only one

parameter called the degrees of freedom. The shape of a chi-squared distribution curve is skewed to the right for small df and becomes symmetric for large df. The entire chi-square distribution curve lies to the right of the vertical axis. The chi-square distribution assumes nonnegative values only, and these are denoted by the symbol χ2 (read as “chi-square”).

3

Figure 11.1 Three chi-square distribution curves.

4

Example 11-1

Find the value of χ² for 7 degrees of freedom and an area of .10 in the right tail of the chi-square distribution curve.

5

Table 11.1 χ2 for df = 7 and .10 Area in the

Right Tail

Area in the Right Tail Under the Chi-Square Distribution Curve

df .995 … .100 … .005

12.7.

100

.000

.010…

.989…

67.328

………………

2.7064.605

…12.017

…118.498

………………

7.87910.597

…20.278

…140.169

Required value of χ²

6

Figure 11.2

df = 7

.10

12.017 χ² 0

7

Example 11-2

Find the value of χ² for 12 degrees of freedom and area of .05 in the left tail of the chi-square distribution curve.

8

Solution 11-2

Area in the right tail = 1 – Area in the left tail = 1 – .05 = .95

9

Table 11.2 χ2 for df = 12 and .95 Area in the Right Tail

Area in the Right Tail Under the Chi-Square Distribution Curve

df .995 … .950 … .005

12.

12.

100

.000

.010…

3.074…

67.328

………………

.004

.103…

5.226…

77.929

………………

7.87910.597

…28.300

…140.169

Required value of χ²

10

Figure 11.3

χ²

Shaded area = .95

df = 12

.05

5.226 0

11

A GOODNESS-OF-FIT TEST

Definition An experiment with the following

characteristics is called a multinomial experiment.

12

Multinomial Experiment cont.

1. It consists of n identical trials (repetitions).

2. Each trial results in one of k possible outcomes (or categories), where k > 2.

3. The trials are independent.4. The probabilities of the various

outcomes remain constant for each trial.

13

A GOODNESS-OF-FIT TEST cont.

Definition The frequencies obtained from the

performance of an experiment are called the observed frequencies and are denoted by O. The expected frequencies, denoted by E, are the frequencies that we expect to obtain if the null hypothesis is true. The expected frequency for a category is obtained as

E = np Where n is the sample size and p is the

probability that an element belongs to that category if the null hypothesis is true.

14

A GOODNESS-OF-FIT TEST cont.

Degrees of Freedom for a Goodness-of-Fit Test

In a goodness-of-fit test, the degrees of freedom are

df = k – 1 where k denotes the number of

possible outcomes (or categories) for the experiment.

15

Test Statistic for a Goodness-of-Fit Test

The test statistic for a goodness-of-fit test is χ2 and its value is calculated as

where O = observed frequency for a category E = expected frequency for a category = np

Remember that a chi-square goodness-of-fit test is always right-tailed.

E

EO 22 )(

16

Example 11-3 A bank has an ATM installed inside the bank, and

it is available to its customers only from 7 AM to 6 PM Monday through Friday. The manager of the bank wanted to investigate if the percentage of transactions made on this ATM is the same for each of the five days (Monday through Friday) of the week. She randomly selected one week and counted the number of transactions made on this ATM on each of the five days during this week. The information she obtained is given in the following table, where the number of users represents the number of transactions on this ATM on these days. For convenience, we will refer to these transactions as “people” or “users.”

17

Example 11-3

At the 1% level of significance, can we reject the null hypothesis that the proportion of people who use this ATM each of the five days of the week is the same? Assume that this week is typical of all weeks in regard to the use of this ATM.

Day Monday

Tuesday Wednesday

Thursday

Friday

Number of users

253 197 204 179 267

18

Solution 11-3

H0 : p1 = p2 = p3 = p4 = p5 = .20 H1 : At least two of the five

proportions are not equal to .20

19

Solution 11.3

There are five categories Five days on which the ATM is used Multinomial experiment

We use the chi-square distribution to make this test.

20

Solution 11-3

Area in the right tail = α = .01 k = number of categories = 5 df = k – 1 = 5 – 1 = 4 The critical value of χ2 = 13.277

21

Figure 11.4

Reject H0 Do not reject H0

α = .01

13.277

χ2

Critical value of χ2

22

Table 11.3

Category (Day)

Observed Frequenc

y

O p

Expected Frequency

E = np(O – E) (O – E)2

MondayTuesdayWednesdayThursdayFriday

253197204279267

.20

.20

.20

.20

.20

1200(.20) = 240

1200(.20) = 240

1200(.20) = 240

1200(.20) = 240

1200(.20) = 240

13-43-363927

169184912961521729

.7047.7045.4006.3383.038

n = 1200 Sum = 23.184

E

EO 2)(

23

Solution 11-3

All the required calculations to find the value of the test statistic χ2 are shown in Table 11.3.

184.23)( 2

2

E

EO

24

Solution 11.3

The value of the test statistic χ2 = 23.184 is larger than the critical value of χ2 = 13.277 It falls in the rejection region

Hence, we reject the null hypothesis

25

Example 11-4 In a National Public Transportation survey

conducted in 1995 on the modes of transportation used to commute to work, 79.6% of the respondents said that they drive alone, 11.1% car pool, 5.1% use public transit, and 4.2% depend on other modes of transportation (USA TODAY, April 14, 1999). Assume that these percentages hold true for the 1995 population of all commuting workers. Recently 1000 randomly selected workers were asked what mode of transportation they use to commute to work. The following table lists the results of this survey.

26

Example 11-4

Mode of transportation

Drive alone Carpool Public transit Other

Number of workers 812 102 57 29

Test at the 2.5% significance level whether the current pattern of use of transportation modes is different from that for 1995.

27

Solution 11-4

H0: The current percentage distribution of the use of transportation modes is the same as that for 1995.

H1: The current percentage distribution of the use of transportation modes is different from that for 1995.

28

Solution 11-4 There are four categories

Drive alone, carpool, public transit, and other

Multinomial experiment We use the chi-square distribution to

make the test.

29

Solution 11-4

Area in the right tail = α = .025 k = number of categories = 4 df = k – 1 = 4 – 1 = 3 The critical value of χ2 = 9.348

30

Figure 11.5

Reject H0 Do not reject H0

α = .025

9.348 χ2

Critical value of χ2

31

Table 11.4

Category

Observed Frequenc

y

O p

Expected Frequency

E = np(O – E) (O – E)2

Drive aloneCar poolPublic transitOther

8121025729

.796

.111

.051

.042

1000(.796) = 796

1000(.111) = 111

1000(.051) = 51

1000(.042) = 42

16-96

-13

2568136169

.322

.730

.7064.024

n = 1000 Sum = 5.782

E

EO 2)(

32

Solution 11-4

All the required calculations to find the value of the test statistic χ2 are shown in Table 11.4.

782.5)( 2

2

E

EO

33

Solution 11-4

The value of the test statistic χ2 = 5.782 is less than the critical value of χ2 = 9.348 It falls in the nonrejection region

Hence, we fail to reject the null hypothesis.

34

CONTINGENCY TABLES

Full-Time Part-Time

Male 6768 2615

Female 7658 3717

Table 11.5 Total 2002 Enrollment at a University

Students who are male and enrolled part-time

35

A TEST OF INDEPENDENCE OR HOMOGENEITY

A Test of Independence A Test of Homogeneity

36

A Test of Independence Definition A test of independence involves a test of

the null hypothesis that two attributes of a population are not related. The degrees of freedom for a test of independence are

df = (R – 1)(C – 1) Where R and C are the number of rows

and the number of columns, respectively, in the given contingency table.

37

Test Statistic for a Test of Independence The value of the test statistic χ2 for a

test of independence is calculated as

where O and E are the observed and expected frequencies, respectively, for a cell.

A Test of Independence cont.

E

EO 22 )(

38

Example 11-5 Violence and lack of discipline have

become major problems in schools in the United States. A random sample of 300 adults was selected, and they were asked if they favor giving more freedom to schoolteachers to punish students for violence and lack of discipline. The two-way classification of the responses of these adults is represented in the following table.

39

Calculate the expected frequencies for this table assuming that the two attributes, gender and opinions on the issue, are independent.

Example 11-5

In Favor

(F)

Against(A)

No Opinions(N)

Men (M)Women (W)

9387

7032

126

40

Table 11.6

Solution 11-5

In Favor(F)

Against(A)

No Opinion(N)

Row Total

s

Men (M) 93 70 12 175

Women (W) 87 32 6 125

Column Totals

180 102 18 300

41

Expected Frequencies for a Test of Independence

The expected frequency E for a cell is calculated as

size sample

)lumn total total)(CoRow(E

42

Table 11.7

Solution 11-5

In Favor(F)

Against(A)

No Opinion(O)

Row Total

s

Men (M)93

(105.00)

70(59.50)

12(10.50)

175

Women (W)

87(75.00)

32(42.50)

6(7.50)

125

Column Totals

180 102 18 300

43

Example 11-6 Reconsider the two-way classification table

given in Example 11-5. In that example, a random sample of 300 adults was selected, and they were asked if they favor giving more freedom to schoolteachers to punish students for violence and lack of discipline. Based on the results of the survey, a two-way classification table was prepared and presented in Example 11-5. Does the sample provide sufficient information to conclude that the two attributes, gender and opinions of adults, are dependent? Use a 1% significance level.

44

Solution 11-6

H0: Gender and opinions of adults are independent

H1: Gender and opinions of adults are dependent

45

Solution 11-6

α = .01 df = (R – 1)(C – 1) = (2 – 1)(3 – 1) = 2 The critical value of χ2 = 9.210

46

Figure 11.6

Reject H0 Do not reject H0

α = .01

9.210 χ2

Critical value of χ2

47

Table 11.8

In Favor(F)

Against

(A)

No Opinion(N)

Row Totals

Men (M)

93(105.00)

70(59.50)

12(10.50)

175

Women (W)

87(75.00)

32(42.50)

6(7.50)

125

Column

Totals180 102 18 300

48

Solution 11-6

252.8300.594.2920.1214.853.1371.1 50.7

50.76

50.42

50.4232

00.75

00.7587

50.10

50.1012

50.59

50.5970

00.105

00.10593

)(

222

222

22

E

EO

49

Solution 11-6

The value of the test statistic χ2 = 8.252 It is less than the critical value of χ2

It falls in the nonrejection region Hence, we fail to reject the null

hypothesis

50

Example 11-7

A researcher wanted to study the relationship between gender and owning cell phones. She took a sample of 2000 adults and obtained the information given in the following table.

51

Example 11-7

At the 5% level of significance, can you conclude that gender and owning cell phones are related for all adults?

Own Cell Phones

Do Not Own Cell Phones

Men Women

640440

450470

52

Solution 11-7

H0: Gender and owning a cell phone are not related

H1: Gender and owning a cell phone are related

53

Solution 11-7

We are performing a test of independence

We use the chi-square distribution α = .05. df = (R – 1)(C – 1) = (2 – 1)(2 – 1) = 1 The critical value of χ2 = 3.841

54

Figure 11.7

Reject H0 Do not reject H0

α = .05

3.841 χ2

Critical value of χ2

55

Table 11.9

Own Cell Phones (Y)

Do Not Own Cell Phones

(N)

Row Total

s

Men (M)

640(588.60)

450(501.40)

1090

Women (W)

440(491.40)

470(418.60)

910

Column Totals

1080 920 2000

56

Solution 11-7

445.21311.6376.5269.5489.4 60.481

60.418470

40.491

40.491440

40.501

40.501450

60.588

60.588640

)(

22

22

22

E

EO

57

Solution 11-7

The value of the test statistic χ2 = 21.445 It is larger than the critical value of χ2 It falls in the rejection region

Hence, we reject the null hypothesis

58

A Test of Homogeneity

Definition A test of homogeneity involves

testing the null hypothesis that the proportions of elements with certain characteristics in two or more different populations are the same against the alternative hypothesis that these proportions are not the same.

59

Example 11-8

Consider the data on income distributions for households in California and Wisconsin given in following table:Californi

aWisconsi

nRow

Totals

High Income 70 34 104

Medium Income

80 40 120

Low Income 100 76 176

Column Totals 250 150 400

60

Example 11-8

Using the 2.5% significance level, test the null hypothesis that the distribution of households with regard to income levels is similar (homogeneous) for the two states.

61

Solution 11-8

H0: The proportions of households that belong to different income

groups are the same in both states

H1: The proportions of households that belong to different income

groups are not the same in both states

62

Solution 11-8

α = .025 df = (R – 1)(C – 1) = (3 – 1)(2 – 1) = 2 The critical value of χ2 = 7.378

63

Figure 11.7

Reject H0 Do not reject H0

α = .025

7.378 χ2

Critical value of χ2

64

Table 11.11

California Wisconsin Row Totals

High income 70

(65)34

(39)104

Medium income80

(75)40

(45)120

Low income100

(110)76

(66)176

Column Totals 250 150 400

65

Solution 11-8

339.4515.1909.566.333.641.385. 66

6676

110

110100

45

4540

75

7580

39

3934

65

6570

)(

222

222

22

E

EO

66

Solution 11-8

The value of the test statistic χ2 = 4.339 It is less than the critical value of χ2

It falls in the nonrejection region Hence, we fail to reject the null

hypothesis

67

INFERENCES ABOUT THE POPULATION VARIANCE

Estimation of the Population Variance Hypothesis Tests About the

Population Variance

68

INFERENCES ABOUT THE POPULATION VARIANCE cont.

Sampling Distribution of (n – 1)s2 / σ2

If the population from which the sample is selected is (approximately) normally distributed, then

has a chi-square distribution with n – 1 degrees of freedom.

2

2)1(

sn

69

Estimation of the Population Variance

Assuming that the population from which the sample is selected is (approximately) normally distributed, the (1 – α)100% confidence interval for the population variance σ2 is

2

2

22/

2

2/1

)1( to

)1(

snsn

70

Example 11-9 One type of cookie manufactured by

Haddad Food Company is Cocoa Cookies. The machine that fills packages of these cookies is set up in such a way that the average net weight of these packages is 32 ounces with a variance of .015 square ounce.

71

Example 11-9 From time to time the quality control

inspector at the company selects a sample of a few such packages, calculates the variance of the net weights of these packages, and construct a 95% confidence interval for the population variance. If either both or one of the two limits of this confidence interval is not the interval .008 to .030, the machine is stopped and adjusted.

72

Example 11-9 A recently taken random sample of 25

packages from the production line gave a sample variance of .029 square ounce. Based on this sample information, do you think the machine needs an adjustment? Assume that the net weights of cookies in all packages are normally distributed.

73

Solution 11-9 n = 25 s2 = .029 α = 1 - .95 = .05 α / 2 = .05 / 2 = .025 1 – α / 2 = 1 – .025 = .975 df = n – 1 = 25 – 1 = 24 χ2 for 24 df and .025 area in the right tail =

39.364 χ2 for 24 df and .975 area in the right tail =

12.401

74

Figure 11.9

df = 24

2

= .025

39.364

Value of 2

2/χ2

75

Figure 11.9

df = 24

21

= .025

12.401

Value of 221

χ2

76

Solution 11-9

.0561 to.0177 12.401

)029)(.125( to

364.39

)029)(.125(

)1( to

)1(

2

2

22/

2

2/1

snsn

77

Solution 11-9

Thus, with 95% confidence, we can state that the variance for all packages of Cocoa Cookies lies between .0177 and .0561 square ounce.

78

Hypothesis Tests About the Population Variance

The value of the test statistic χ2 is calculated as

where s2 is the sample variance, σ2 is the hypothesized value of the population variance, and n – 1 represents the degrees of freedom. The population from which the sample is selected is assumed to be (approximately) normally distributed.

2

22 )1(

sn

79

Example 11-10 One type of cookie manufactured by Haddad

Food Company is Cocoa Cookies. The machine that fills packages of these cookies is set up in such a way that the average net weight of these packages is 32 ounces with a variance of .015 square ounce. From time to time the quality control inspector at the company selects a sample of a few such packages, calculates the variance of the net weights of these packages, and makes a test of hypothesis about the population variance.

80

Example 11-10

She always uses α = .01. The acceptable value of the population variance is .015 square ounce or less. If the conclusion from the test of hypothesis is that the population variance is not within the acceptable limit, the machine is stopped and adjusted.

81

Example 11-10 A recently taken random sample of

25 packages from the production line gave a sample variance of .029 square ounce. Based on this sample information, do you think the machine needs an adjustment? Assume that the net weights of cookies in all packages are normally distributed.

82

Solution 11-10

H0 :σ2 ≤ .015 The population variance is within the

acceptable limit H1: σ2 >.015

The population variance exceeds the acceptable limit

83

Solution 11-10

α = .01 df = n – 1 = 25 – 1 = 24 The critical value of χ2 = 42.980

84

Figure 11.10

Reject H0 Do not reject H0

α = .01

42.980

χ2

Critical value of χ2

85

Solution 11-10

400.46015.

)029)(.125()1(2

22

sn

From H0

86

Solution 11-10 The value of the test statistic χ2 = 46.400

It is greater than the critical value of χ 2

It falls in the rejection region Hence, we reject the null hypothesis H0

We conclude that the population variance is not within the acceptable limit The machine should be stopped and adjusted

87

Example 11-11 The variance of scores on a standardized

mathematics test for all high school seniors was 150 in 2002. A sample of scores for 20 high school seniors who took this test this year gave a variance of 170. Test at the 5% significance level if the variance of current scores of all high school seniors on this test is different from 150. Assume that the scores of all high school seniors on this test are (approximately) normally distributed.

88

Solution 11-11

H0: σ2 = 150 The population variance is not different

from 150 H1: σ2 ≠ 150

The population variance is different from 150

89

Solution 11-11

α = .05 Area in the each tail = .025 df = n – 1 = 20 – 1 = 19 The critical values of χ2 32.852 and

8.907

90

Figure 11.11

Reject H0Do not reject H0

Reje

ct H

0

α /2 = .025α /2 = .025

8.907 32.852

Two critical values of χ2

91

Solution 11-11

533.21150

)170)(120()1(2

22

sn

From H0

92

Solution 11-11

The value of the test statistic χ2 = 21.533 It is between the two critical values of χ2

It falls in the nonrejection region Consequently, we fail to reject H0.