Chapter 17: Chi-Square Tests

 Дата 22.04.2016 Памер 60.99 Kb. #17737

Chapter 17: Chi-Square Tests

1. Nonparametric tests make few if any assumptions about the populations from which the data are obtained. For example, the populations do not need to form normal distributions, nor is it required that different populations in the same study have equal variances (homogeneity of variance assumption). Parametric tests require data measured on an interval or ratio scale. For nonparametric tests, any scale of measurement is acceptable.

2. a. The null hypothesis states that the gender distribution for theater goers is not different from

the distribution for the general population of students. For a sample of 600 students, the

expected frequencies are 330 females (55%) and 270 males (45%), and chi-square = 20.37.

With df = 1, the critical value is 3.84. Reject H0 and conclude that the gender distribution

for theater goers is significantly different from the distribution for the population of

students.

b. The null hypothesis states that the gender distribution for basketball fans is not different

from the distribution for the general population of students. For a sample of 180 students,

the expected frequencies are 99 females (55%) and 81 males (45%), and chi-square = 5.75.

With df = 1, the critical value is 3.84. Reject H0 and conclude that the gender distribution

for basketball fans is significantly different from the distribution for the population of

students.

3. a. The null hypothesis states that there is no preference among the four colors; p = 1/4 for all

categories. The expected frequencies are fe = 15 for all categories, and chi square = 4.53.

With df = 3, the critical value is 7.81. Fail to reject H0 and conclude that there are no

significant preferences.

b. The results indicate that there are no significant preferences among the four colors, χ2(3, N

= 60) = 4.53, p > .05.

4. a. The null hypothesis states that the age distribution for people who get speeding tickets is

not different from the distribution for the population of licensed drivers. With df = 1, the

critical value is 3.84. The expected frequencies are 48 over age 25 and 12 under age 25,

and chi-square = 10.42. Reject the null hypothesis and conclude that the age distribution

for people who receive speeding tickets is significantly different from the distribution for

the population of drivers.

b. The null hypothesis states that the age distribution for people who get parking tickets is

not different from the distribution for the population of licensed drivers. With df = 1, the

critical value is 3.84. The expected frequencies are 48 over age 25 and 12 under age 25,

and chi-square = 2.60. Fail to reject the null hypothesis and conclude that the age

distribution for people who receive speeding tickets is not significantly different from the

distribution for the population of drivers.

5. The null hypothesis states that wins and loses are equally likely. With 64 games, the expected frequencies are 32 wins and 32 losses. With df = 1 the critical value is 3.84, and the data produce a chi-square of 6.25. Reject the null hypothesis and conclude that home team wins are significantly more common that would be expected by chance.
6. The null hypothesis states that couples with the same initial do not occur more often than would be expected by chance. For a sample of 200, the expected frequencies are 13 with the same initial and 187 with different initials. With df = 1 the critical value is 3.84, and the data produce a chi-square of 2.96. Fail to reject the null hypothesis.
7. a. The null hypothesis states that couples with the same initial do not occur more often than

would be expected by chance. For a sample of 400, the expected frequencies are 26 with

the same initial and 374 with different initials. With df = 1 the critical value is 3.84, and

the data produce a chi-square of 5.92. Reject the null hypothesis.

b. A larger sample should be more representative of the population. If the sample continues

to be different from the hypothesis as the sample size increases, eventually the difference will be significant.

8. The null hypothesis states that the grade distribution for last semester has the same proportions as it did in 1985. For a sample of n = 200, the expected frequencies are 28, 52, 62, 38, and 20 for grades of A, B, C, D, and F, respectively. With df = 4, the critical value for chi-square is 9.49. For these data, the chi-square statistic is 6.68. Fail to reject H0 and conclude that there is no evidence that the distribution has changed.
9. a. H0 states that the distribution of automobile accidents is the same as the distribution of

registered drivers: 16% under age 20, 28% age 20 to 29, and 56% age 30 or older.

With df = 2, the critical value is 5.99. The expected frequencies for these three categories

are 48, 84, and 168. Chi square = 13.76. Reject H0 and conclude that the distribution of

automobile accidents is not identical to the distribution of registered drivers.

b. The chi-square test shows that the age distribution for people in automobile accidents is

significantly different from the age distribution of licensed drivers, χ2(3, N= 180) =

13.76, p < .05.

10. a. The null hypothesis states that there is no advantage (no preference) for red or blue. With

df = 1, the critical value is 3.84. The expected frequency is 25 wins for each color, and

chi-square = 2.88. Fail to reject H0 and conclude that there is no significant advantage for

one color over the other.

b. The null hypothesis states that there is no advantage (no preference) for red or blue. With

df = 1, the critical value is 3.84. The expected frequency is 50 wins for each color, and

chi-square = 5.76. Reject H0 and conclude that there is a significant advantage for

the color red.

c. Although the proportions are identical for the two samples, the sample in part b is twice

as big as the sample in part a. The larger sample provides more convincing evidence of

an advantage for red than does the smaller sample.

11. The null hypothesis states that there are no preferences among the three designs; p = 1/3 for all categories. With df = 2, the critical value is 5.99. The expected frequencies are fe = 40 for all categories, and chi square = 8.60. Reject H0 and conclude that there are significant preferences.
12. The null hypothesis states that the distribution of preferences is the same for both groups (same proportions). With df = 2, the critical value is 5.99. The expected frequencies are:

Design 1 Design 2 Design 3

 Students 24 27 9 Older Adults 24 27 9

Chi-square = 7.94. Reject H0.
13. The null hypothesis states that there is no relationship between the type of music and whether the women give their phone numbers. With df = 1, the critical value is 3.84. The expected frequencies are:

Phone Number No Number

 Romantic Music 15 25 40 Neutral Music 15 25 40

30 50

Chi-square = 7.68. Reject H0.

14. The null hypothesis states that the distribution of satisfaction scores is the same for both groups. With df = 1, the critical value is 3.84. The expected frequencies are:

Satisfied Not satisfied

 Less reimbursement 55 45 100 Same or more reimbursement 33 27 60

88 72

Chi-square = 8.73. Reject H0.

15. a. The null hypothesis states that the distribution of opinions is the same for those who live

in the city and those who live in the suburbs. For df = 1 and α = .05, the critical value for

chi-square is 3.84. The expected frequencies are:

Favor Oppose

 City 30 20 Suburb 60 40

For these data, chi-square = 3.12 Fail to reject H0 and conclude that opinions in the city are not different from those in the suburbs.

b. The phi coefficient is 0.144.

16. a. The null hypothesis states that the distribution of opinions is the same for those who

live in the city and those who live in the suburbs. For df = 1 and α = .05, the critical

value for chi-square is 3.84. The expected frequencies are:

Favor Oppose

 City 60 40 Suburb 120 80

For these data, chi-square = 6.25. Reject H0 and conclude that opinions in the city are different from those in the suburbs. The larger sample produces a significant relationship.

b. The phi coefficient is still 0.144. The sample size has no effect on the strength of the relationship.

17. a. The null hypothesis states that the proportion who falsely recall seeing broken glass

should be the same for all three groups. The expected frequency of saying yes is 9.67 for

all groups, and the expected frequency for saying no is 40.33 for all groups. With df = 2,

the critical value is 5.99. For these data, chi-square = 7.78. Reject the null hypothesis

and conclude that the likelihood of recalling broken glass is depends on the question that

b. Cramérs V = 0.228.

c. Participants who were asked abou the speed with the cars “smashed into” each other,

were more than two times more likely to falsely recall seeing broken glass.

d. The results of the chi-square test indicate that the phrasing of the question had a

significant effect on the participants’ recall of the accident, χ2(2, N = 150) = 7.78, p < .05,

V = 0.228.

18. a. The null hypothesis states that the distribution of weights for men is the same as the

distribution for women. The expected frequencies are 81.6 desirable and 38.4

overweight for men, and 54.4 desirable and 25.6 overweight for women. With df = 1, the

critical value is 3.84. For these data, chi-square = 5.53. Reject the null hypothesis.

b. The phi-coefficient is 0.166.

c. The chi-square test shows that the proportion of men who are overweight is significantly

greater than the proportion of women, χ2(1, N = 200) = 5.53, p < .05, φ = 0.166.
19. The null hypothesis states that IQ and gender are independent. The distribution of IQ scores for boys should be the same as the distribution for girls. With df = 2 and and α = .05, the critical value is 5.99. The expected frequencies are 15 low IQ, 48 medium, and 17 high for both boys and girls. For these data, chi-square is 3.76. Fail to reject the null hypothesis. These data do not provide evidence for a significant relationship between IQ and gender.

20. The null hypothesis states that there is no relationship between dream content and gender; the distribution of aggression content should be the same for males and females. The critical value is 9.21. The expected frequencies are:

Low Medium High

 Female 8.8 8.4 6.8 Male 13.2 12.6 10.2

The chi-square statistic is 25.52. Reject H0 with α = .01 and df = 2.
21. The null hypothesis states that there is no difference between the distribution of preferences predicted by women and the actual distribution for men. With df = 3 and  = .05, the critical value is 7.81. The expected frequencies are:
somewhat slightly slightly somewhat

thin thin heavy heavy

 Women 22.9 22.9 22.9 11.4 Men 17.1 17.1 17.1 8.6

Chi square = 9.13. Reject H0 and conclude that there is a significant difference in the preferences predicted by women and the actual preferences expressed by men.

22. a. The null hypothesis states that there is no relationship between the personalities of the participants and the personalities of the avatars they create. With df = 1 and α = .05, the critical value is 3.84. The expected frequencies are:

Participant Personality

Introverted Extroverted

 Introverted Avatar 17.1 27.9 45 Extroverted Avatar 20.9 34.1 55

38 62

The chi-square statistic is 4.12. Reject H0.

b. The phi-coefficient is 0.203.
23. a. The null hypothesis states that there is no relationship between IQ and volunteering. With

df = 2 and α = .05, the critical value is 5.99. The expected frequencies are:

IQ

High Medium Low

 volunteer 37.5 75 37.5 not volunteer 12.5 25 12.5

The chi-square statistic is 4.75. Fail to reject H0 with α = .05 and df = 2.
24. a. The null hypothesis states that littering is independent of the amount of litter already on

the ground. With df = 2, the critical value is 5.99. The expected frequencies are:

Amount of Litter

None Small Large

 Litter 31.33 31.33 31.33 Not Litter 58.67 58.67 58.67

Chi-square = 25.88. Reject H0.

b. V = 0.310 (a medium effect)

25. The null hypothesis states that there is no relationship between the season of birth and schizophrenia. With df = 3 and  = .05, the critical value is 7.81. The expected frequencies are:

Summer Fall Winter Spring

 No Disorder 23.33 23.33 26.67 26.67 Schizophrenia 11.67 11.67 13.33 13.33

Chi square = 3.62. Fail to reject H0 and conclude that these data do not provide enough evidence to conclude that there is a significant relationship between the season of birth and schizophrenia.

Solutions - Chapter 17 - page

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