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Stat wk12

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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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