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- Published: October 27, 2022
- Updated: October 27, 2022
- Level: Masters
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Assignment Problem Set 3: Chapter 18, problems 2, 6, 20 #2a The null and alternate hypotheses are H0: The gender distribution for theatergoers is not different from the distribution for the general college.

H1: The gender distribution for theatergoers is different from the distribution for the general college.

The selected level of significance, α is . 05.

The selected test is Chi-Square Goodness-Of-Fit Test.

The degrees of freedom for the Chi-Square Goodness-Of-Fit Test are,

df = k – 1 = 2 – 1 = 1

At the level of significance of . 05 with 1 degree of freedom, the critical vale of χ2 is 3. 84. Therefore, decision rule will be

Reject H0 if χ2 > 3. 84. Otherwise, do not reject H0.

Gender

Observed Frequency

O

Expected Frequency

E

Difference

O – E

Square of Difference

(O – E)2

Chi-square subtotal

(O – E)2/E

Female

385

330

55

3025

9. 17

Male

215

270

-55

3025

11. 20

600

600

0

χ2 = 20. 37

The test statistics is

χ2 = 20. 37

The chi-square value of 20. 37 is greater than critical value of 3. 84. Therefore, we can reject the null hypothesis H0. The gender distribution for theatergoers is significantly different from the distribution for the general college, χ2(1, n = 600) = 20. 37, p < . 05.

#2b

The null and alternate hypotheses are

H0: The gender distribution for basketball fans is not different from the distribution for the general college.

H1: The gender distribution for basketball fans is different from the distribution for the general college.

The selected level of significance, α is . 05.

The selected test is Chi-Square Goodness-Of-Fit Test.

The degrees of freedom for the Chi-Square Goodness-Of-Fit Test are,

df = k – 1 = 2 – 1 = 1

At the level of significance of . 05 with 1 degree of freedom, the critical vale of χ2 is 3. 84. Therefore, decision rule will be

Reject H0 if χ2 > 3. 84. Otherwise, do not reject H0.

Gender

Observed Frequency

O

Expected Frequency

E

Difference

O – E

Square of Difference

(O – E)2

Chi-square subtotal

(O – E)2/E

Female

83

99

-16

256

2. 59

Male

97

81

16

256

3. 16

180

180

0

χ2 = 5. 75

The test statistics is

χ2 = 5. 75

The chi-square value of 5. 75 is greater than critical value of 3. 84. Therefore, we can reject the null hypothesis H0. The gender distribution for basketball fans is significantly different from the distribution for the general college, χ2(1, n = 180) = 5. 75, p < . 05.

#6

The null and alternate hypotheses are

H0: The self-hypnosis has no effect on increasing or decreasing allergic reactions.

H1: The self-hypnosis has a effect on increasing or decreasing allergic reactions.

The selected level of significance, α is . 05.

The selected test is Chi-Square Goodness-Of-Fit Test.

The degrees of freedom for the Chi-Square Goodness-Of-Fit Test are,

df = k – 1 = 2 – 1 = 1

At the level of significance of . 05 with 1 degree of freedom, the critical vale of χ2 is 3. 84. Therefore, decision rule will be

Reject H0 if χ2 > 3. 84. Otherwise, do not reject H0.

Allergic Reactions

Observed Frequency

O

Expected Frequency

E

Difference

O – E

Square of Difference

(O – E)2

Chi-square subtotal

(O – E)2/E

Reduced

47

32

15

225

7. 03

Increased

17

32

-15

225

7. 03

64

64

0

χ2 = 14. 06

The test statistics is

χ2 = 14. 06

The chi-square value of 14. 06 is greater than critical value of 3. 84. Therefore, we can reject the null hypothesis H0. The self-hypnosis has a effect on increasing or decreasing allergic reactions, χ2(1, n = 64) = 14. 06, p < . 05.

#20

The null and alternate hypotheses are

H0: There is no relationship between gender and the aggression content of dreams.

H1: There is a relationship between gender and the aggression content of dreams.

The selected level of significance, α is . 01.

The selected test is Chi-Square Independence Test.

The degrees of freedom for the Chi-Square test of Independence are,

df = (c – 1)(r – 1) = (3 – 1)(2 – 1) = 2

At the level of significance of . 01 with 2 degree of freedom, the critical vale of χ2 is 9. 21. Therefore, decision rule will be

Reject H0 if χ2 > 9. 21. Otherwise, do not reject H0.

Gender

Aggression Content

Low

Medium

High

Total

Female

Observed

18

4

2

24

Expected

8. 80

8. 40

6. 80

24. 00

O – E

9. 20

-4. 40

-4. 80

0. 00

(O – E)² / E

9. 62

2. 30

3. 39

15. 31

Male

Observed

4

17

15

36

Expected

13. 20

12. 60

10. 20

36. 00

O – E

-9. 20

4. 40

4. 80

0. 00

(O – E)² / E

6. 41

1. 54

2. 26

10. 21

Total

Observed

22

21

17

60

Expected

22. 00

21. 00

17. 00

60. 00

O – E

0. 00

0. 00

0. 00

0. 00

(O – E)² / E

16. 03

3. 84

5. 65

25. 52

25. 52

chi-square

2

df

0. 0000

p-value

The test statistics is

χ2 = 25. 52

The chi-square value of 25. 52 is greater than critical value of 9. 21. Therefore, we can reject the null hypothesis H0. There is a significant relationship between gender and the aggression content of dreams, χ2(2, n = 60) = 25. 52, p < . 01.

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