Chi Square Test

CHI-SQUARE Prepared by: ANTONIO, RUSSEL N. BARADI, ARNEL F. LAZO, IVY U.

1

CHI-SQUARE

A chi-square (χ𝟐 ) statistic is a test that measures how expectations compare to actual observed data.

2

CHI-SQUARE TEST

It is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. 3

How to Calculate the Chi-Square Statistic (χ𝟐 )

1.

2. 3.

For each observed number in the table subtract the corresponding expected number 𝑂 − 𝐸. Square the difference 𝑂 − 𝐸 2 Divide the squares obtained for each cell in the table by the expected number for that cell

𝑂−𝐸 2 𝐸

4

CHI-SQUARE TESTS

GOODNESS OF FIT TEST

TEST FOR INDEPENDENCE

TEST FOR HOMOGENEITY

- used to test whether a frequency distribution obtained experimentally fits an “expected” frequency distribution that is based on the theoretical or previously known probability of each outcome.

- used to test whether or not two variables are independent.

- used to draw conclusion whether two populations have the same distribution

5

GOODNESS OF FIT TEST 6

GOODNESS OF FIT

• 𝐻𝑜 : The population frequencies are equal to the expected

• • •

frequencies. 𝐻𝑎 : The null hypothesis is false 𝛼: level of significance 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 𝑘 − 1 𝑶−𝑬 𝟐 𝑬

• =σ • From 𝛼 and 𝑘 − 1, a critical value is determined from the chi-square χ𝟐

table.

• Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test)

7

GOODNESS-OF-FIT TEST

1. Researchers have conducted a survey of 1600 coffee drinkers asking how much coffee they drink in order to confirm previous studies. Previous studies have indicated that 72% of Americans drink coffee. At 𝛼 = 0.05 , is there enough evidence to conclude that the distributions are the same?

Paytonbinnings.com 8

Researchers have conducted a survey of 1600 coffee drinkers asking how much coffee they drink in order to confirm previous studies. Previous studies have indicated that 72% of Americans drink coffee. At 𝛼 = 0.05, is there enough evidence to conclude that the distributions are the same?

GOODNESS-OF-FIT TEST

Response

Frequency

Response

Frequency

2 cups/week

15%

2 cups/week

206

1 cup/week

13%

1 cup/week

193

1 cup/day

27%

1 cup/day

462

2+cups/day

45%

2+cups/day

739

9

GOODNESS-OF-FIT TEST

Researchers have conducted a survey of 1600 coffee drinkers asking how much coffee they drink in order to confirm previous studies. Previous studies have indicated that 72% of Americans drink coffee. At 𝛼 = 0.05, is there enough evidence to conclude that the distributions are the same?

𝐻𝑜 :The population frequencies are equal to the expected frequencies. 𝐻𝑎 : The null is false. 𝛼 = 0.05 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 𝑘 − 1 = 4 − 1 = 3 10

Researchers have conducted a survey of 1600 coffee drinkers asking how much coffee they drink in order to confirm previous studies. Previous studies have indicated that 72% of Americans drink coffee. At 𝛼 = 0.05, is there enough evidence to conclude that the distributions are the same?

GOODNESS-OF-FIT TEST

Response 2 cups/week 1 cup/week 1 cup/day 2+ cups/day

%of Coffee Drinkers 15% 13% 27% 45% χ𝟐 ෍

E

O

O-E

𝑶−𝑬

240 208 432 720

206 193 462 739

-34 -15 30 19

1156 225 900 361

𝑶−𝑬 𝑬

𝟐

𝑶−𝑬 𝑬

𝟐

4.817 1.082 2.083 0.5014

𝟐

= 𝟖. 𝟒𝟖𝟑 11

χ𝟐 = 𝟖. 𝟒𝟖𝟑

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 12

Researchers have conducted a survey of 1600 coffee drinkers asking how much coffee they drink in order to confirm previous studies. Previous studies have indicated that 72% of Americans drink coffee. At 𝛼 = 0.05, is there enough evidence to conclude that the distributions are the same?

GOODNESS-OF-FIT TEST

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test) χ𝟐 (𝒕 − 𝒔𝒕𝒂𝒕)

𝟖. 𝟒𝟖𝟑

𝒕 − 𝒄𝒓𝒊𝒕

>

𝟕. 𝟖𝟏

There is enough statistical evidence to reject the null hypothesis and to believe that the old percentages no longer hold.

13

GOODNESS-OF-FIT TEST

2. A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly shoppers prefer are as follows. Is there enough evidence using 𝛼 = 0.05 to conclude that the proportions are really the same?

Forbes.com 14

2. A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly shoppers prefer are as follows. Is there enough evidence using 𝛼 = 0.05 to conclude that the proportions are really the same?

GOODNESS-OF-FIT TEST

Store Number of Shoppers

A 262

B 234

C 204

D 190

E 210

15

GOODNESS-OF-FIT TEST

2. A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly shoppers prefer are as follows. Is there enough evidence using 𝛼 = 0.05 to conclude that the proportions are really the same?

𝐻𝑜 :The population frequencies are equal to the expected frequencies. 𝐻𝑎 : The null is false. 𝛼 = 0.05 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 𝑘 − 1 = 5 − 1 = 4 16

2. A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly shoppers prefer are as follows. Is there enough evidence using 𝛼 = 0.05 to conclude that the proportions are really the same?

GOODNESS-OF-FIT TEST

Preference

%of Shoppers

E

O

O-E

𝑶−𝑬

A B C D E

20% 20% 20% 20% 20%

220 220 220 220 220

262 234 204 190 210

42 14 -16 -30 -10

1764 196 256 900 100

χ𝟐 ෍

𝑶−𝑬 𝑬

𝟐

𝑶−𝑬 𝑬

𝟐

8.018 0.891 1.163 4.091 0.455

𝟐

= 𝟏𝟒. 𝟔𝟏𝟖

17

χ𝟐 = 𝟏𝟒. 𝟔𝟏𝟖

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 18

GOODNESS-OF-FIT TEST

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test) χ𝟐 (𝒕 − 𝒔𝒕𝒂𝒕)

14.618 >

2. A department store, A, has four competitors: B, C, D, and E. Store A hires a consultant to determine if the percentage of shoppers who prefer each of the five stores is the same. A survey of 1100 randomly shoppers prefer are as follows. Is there enough evidence using 𝛼 = 0.05 to conclude that the proportions are really the same?

𝒕 − 𝒄𝒓𝒊𝒕

9.49

There is enough statistical evidence to reject the null hypothesis and to believe that customers do not prefer each of the five stores equally.

19

TEST FOR INDEPENDENCE 20

TEST FOR INDEPENDENCE

• 𝐻𝑜 : The two variables are independent • 𝐻𝑎 : The two variables are not independent (dependent). • 𝛼: level of significance • The expected frequency 𝐸𝑟,𝑐 =

𝑆𝑢𝑚 𝑜𝑓 𝑟 (𝑠𝑢𝑚 𝑜𝑓 𝑐) 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒

• 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: (𝑟 − 1)(𝑐 − 1) •

χ𝟐

=σ

𝑶−𝑬 𝟐 𝑬

• From 𝛼 and 𝑘 − 1, a critical value is determined from the chisquare table. • Reject 𝐻𝑜 if χ 𝟐 is larger than the critical value (right-tailed test)

21

TEST FOR INDEPENDENCE

1. The result of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment and result are independent?

Dentagama.com 22

1. The result of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment and result are independent?

TEST FOR INDEPENDENCE

(r. 1) Significant Improvement

Acetaminophen (c. 1)

Ibuprofen (c. 2)

Codeine (c.3)

Total

58

81

61

200

66.7

66.7

(r. 2) Slight Improvement

42

19

33.3

33.3

Total

100

100

66.7

39

100

33.3

100

300 23

TEST FOR INDEPENDENCE

1. The result of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment and result are independent?

𝐻𝑜 :The population frequencies are equal to the expected frequencies. 𝐻𝑎 : The null is false. 𝛼 = 0.05 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 2 − 1 3 − 1 = 2 24

1. The result of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment and result are independent?

TEST FOR INDEPENDENCE

𝟐

Row, Column

E

O

O-E

𝑶−𝑬

1,1 1,2 1,3 2,1 2,2 2,3

66.7 66.7 66.7 33.3 33.3 33.3

58 81 61 42 19 39

-8.7 14.3 -5.7 8.7 -14.3 5.7

75.69 204.49 32.49 75.69 204.49 32.49

𝑶−𝑬 𝑬

1.135 3.067 0.487 2.271 6.135 0.975

𝟐

χ𝟐 ෍

𝑶−𝑬

𝟐

𝑬

= 𝟏𝟒. 𝟎𝟕

25

χ𝟐 = 𝟏𝟒. 𝟎𝟕

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 26

TEST FOR INDEPENDENCE

1. The result of a random sample of children with pain from musculoskeletal injuries treated with acetaminophen, ibuprofen, or codeine are shown in the table. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment and result are independent?

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test) χ𝟐 (𝒕 − 𝒔𝒕𝒂𝒕)

14.07

𝒕 − 𝒄𝒓𝒊𝒕

>

4.61

There is enough statistical evidence to reject the null hypothesis and to believe that there is a relationship between the treatment and the response.

27

TEST FOR INDEPENDENCE

2. The side effects of a new drug being tested against a placebo. A simple random sample of 565 patients yields the results as follows. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment is independent of the side effect of nausea?

Nutritionaloutlook.com 28

2. The side effects of a new drug being tested against a placebo. A simple random sample of 565 patients yields the results as follows. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment is independent of the side effect of nausea?

TEST FOR INDEPENDENCE

Result

Drug (c. 1)

Nausea (r. 1)

36

No nausea (r. 2)

254

Total

290

Placebo (c. 2)

Total

(25.15)

13

(23.85)

49

(264.85)

262

(251.15)

516

275

565

29

TEST FOR INDEPENDENCE

2. The side effects of a new drug being tested against a placebo. A simple random sample of 565 patients yields the results as follows. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment is independent of the side effect of nausea?

𝐻𝑜 : the treatment and response are independent. 𝐻𝑎 : the treatment and response are dependent. 𝛼 = 0.10 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 2 − 1 2 − 1 = 1 30

2. The side effects of a new drug being tested against a placebo. A simple random sample of 565 patients yields the results as follows. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment is independent of the side effect of nausea?

TEST FOR INDEPENDENCE

Row, Column

E

O

O-E

1,1 1,2 2,1 2,2

25.15 23.85 264.85 251.15

36 13 254 262

10.85 -10.85 -10.85 10.85

χ𝟐 σ

𝑶−𝑬 𝟐 𝑬

(𝑶

117.72 117.72 117.72 117.72

= 𝟏𝟎. 𝟓𝟑

𝑶−𝑬 𝑬

𝟐

4.681 4.936 0.444 0.469

31

χ𝟐 = 𝟏𝟎. 𝟓𝟑

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 32

TEST FOR INDEPENDENCE

2. The side effects of a new drug being tested against a placebo. A simple random sample of 565 patients yields the results as follows. At 𝛼 = 0.10 , is there enough evidence to conclude that the treatment is independent of the side effect of nausea?

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test) χ𝟐 (𝒕 − 𝒔𝒕𝒂𝒕)

10.53

𝒕 − 𝒄𝒓𝒊𝒕

>

2.71

There is enough statistical evidence to reject the null hypothesis and to believe that there is a relationship between the treatment and response.

33

TEST FOR HOMOGENEITY 34

TEST FOR HOMOGENEITY

• 𝐻𝑜 : The distributions of the two populations are the same. • 𝐻𝑎 : The distributions of the two populations are not the same. • 𝛼: level of significance • The expected frequency 𝐸𝑟,𝑐 =

𝑆𝑢𝑚 𝑜𝑓 𝑟 (𝑠𝑢𝑚 𝑜𝑓 𝑐) 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒

• 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 𝑐 − 1 •

χ𝟐

=σ

𝑶−𝑬 𝟐 𝑬

• From 𝛼 and 𝑘 − 1, a critical value is determined from the chisquare table. • Reject 𝐻𝑜 if χ 𝟐 is larger than the critical value (right-tailed test)

35

TEST FOR HOMOGENEITY 1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

Wcu.edu 36

1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

TEST FOR HOMOGENEITY

Dormitory (c. 1)

(r. 1) Males

72

Apartment (c. 2)

84

(r. 2) Females Total

91

49 (77.27)

(74.09)

86 (88.91)

163

With Parents (c. 3) (62.27)

88 (92.73)

170

Other (c. 4)

Total

45

250

(36.36)

35

(74.73)

(43.64)

137

80

300 550 37

TEST FOR HOMOGENEITY

𝐻𝑜 :The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students. 𝐻𝑎 : The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students. 𝛼 = 0.05 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 4 − 1 = 3

1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

38

TEST FOR HOMOGENEITY

Row, Column

E

O

1,1 1,2 1,3

74.09 77.27 62.27

1,4 2,1 2,2 2,3 2,4

36.36 88.91 92.73 74.73 43.64

72 84 49 45 91 86 88 35

O-E

𝑶−𝑬

𝟐

4.372 -2.09 45.256 6.73 -13.27 176.165 74.587 8.64 4.372 2.09 45.256 -6.73 13.27 176.165 74.587 -8.64

𝑶−𝑬 𝑬

0.059 0.586 2.829 2.051 0.049 0.488 2.357 1.709

𝟐

1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

χ𝟐 ෍

𝑶−𝑬

𝟐

𝑬

= 𝟏𝟎. 𝟏𝟐𝟗

39

χ𝟐 = 𝟏𝟎. 𝟏𝟐𝟗

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 40

Chi-Square Distribution Calculator

1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

41

TEST FOR HOMOGENEITY

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test)

𝛼 = 0.05

1. Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are as follows. Do male and female college students have the same distribution of living arrangements?

𝑷 − 𝒗𝒂𝒍𝒖𝒆

Type 0.05 > equation The distribution of living arrangements for male here. college students is not the same as the distribution of living arrangements for female college students.

42

TEST FOR HOMOGENEITY

2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

punjabiCenter.com 43

2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

TEST FOR HOMOGENEITY

Lone Ranger (c. 1)

(r. 1) Boys

50

(r. 2) Girls

50

Total

Sesame Street (c. 2)

30 (33.33)

20 (36.67)

80 (66.67)

100

The Simpsons (c. 3)

(73.33)

110

Total

100

(30)

70

200

(60)

90

300 44

TEST FOR HOMOGENEITY

2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

𝐻 of boys who prefer the Lone 𝐻𝑜𝑜 ::The 𝑃𝑏𝑜𝑦𝑠proportion 𝑙𝑖𝑘𝑒 𝐿𝑜𝑛𝑒 𝑅𝑎𝑛𝑔𝑒𝑟 = 𝐻𝑔𝑖𝑟𝑙𝑠 𝑙𝑖𝑘𝑒 𝐿𝑜𝑛𝑒 𝑅𝑎𝑛𝑔𝑒𝑟 Ranger is identical to the proportion of girls. 𝐻𝑜 : 𝑃𝑏𝑜𝑦𝑠 𝑙𝑖𝑘𝑒 𝑆𝑒𝑠𝑎𝑚𝑒 𝑆𝑡𝑟𝑒𝑒𝑡 = 𝐻𝑔𝑖𝑟𝑙𝑠 𝑙𝑖𝑘𝑒 𝑆𝑒𝑠𝑎𝑚𝑒 𝑆𝑡𝑟𝑒𝑒𝑡 Similarly, for the other programs. 𝐻𝑜 : 𝑃𝑏𝑜𝑦𝑠 𝑙𝑖𝑘𝑒 𝑡ℎ𝑒 𝑆𝑖𝑚𝑝𝑠𝑜𝑛𝑠 = 𝐻𝑔𝑖𝑟𝑙𝑠 𝑙𝑖𝑘𝑒𝑡ℎ𝑒 𝑆𝑖𝑚𝑝𝑠𝑜𝑛𝑠 𝐻𝑎 : At least one of the null hypothesis statement is false. 𝛼 = 0.05 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚: 3 − 1 2 − 1 = 2

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2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

TEST FOR HOMOGENEITY

𝟐

Row, Column

E

O

O-E

𝑶−𝑬

1,1 1,2 1,3 2,1 2,2 2,3

33.33 36.67 30 66.67 73.33 60

72 30 20 50 80 70

16.67 -6.67 -10 -16.67 6.67 10

277.78 44.44 100 277.78 44.44 100

𝑶−𝑬 𝑬

8.33 1.21 3.33 4.17 0.61 1.67

𝟐

χ𝟐 ෍

𝑶−𝑬

𝟐

𝑬

= 𝟏𝟗. 𝟑𝟏𝟖

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χ𝟐 = 𝟏𝟎. 𝟏𝟐𝟗

Percentage Points of the Chi-Square Distribution

0.50

0.25

0.10

0.05

0.01

1

0.455

1.32

2.71

3.84

6.63

2

1.386

2.77

4.61

5.99

9.21

3

2.366

4.11

6.25

7.81

11.34

4

3.357

5.39

7.78

9.49

13.28

5

4.351

6.63

9.24

11.07

15.09

6

5.348

7.84

10.64

12.59

16.81

7

6.346

9.04

12.02

14.07

18.48 47

Chi-Square Distribution Calculator

2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

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TEST FOR HOMOGENEITY

Reject 𝐻𝑜 if χ𝟐 is larger than the critical value (right-tailed test)

𝛼 = 0.05

𝑷 − 𝒗𝒂𝒍𝒖𝒆

0.05

> 0.00005

There is enough statistical evidence to reject the null hypothesis and to believe that the proportion of boys who like Lone Ranger/Sesame Street/The Simpsons is not identical to girls who like the same TV programs, respectively.

2. In a study of the television viewing habits of children, a developmental psychologist selects a random sample of 300 first graders – 100 boys and 200 girls. Each child is asked which of the following TV programs they like best. The Long Ranger, Sesame Street, or The Simpsons. Results are shown in the contingency as follows.

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