Soc. 3155 Answers to Homework 2

9.2       All N₁ + N₂ > 120, so use the Z distribution in all cases. The .05 critical value is + 1.96 for all tests.

  a)       H₀: μ_m = μ_f      H_a: μ_m ≠ μ_f      s_x1-x2 = √(.88²/122) + (.81²/251) = .095

            Z = (2.99 - 2.29)/.095 = 7.37   Reject H₀. Males are more tolerant of violence than females.

b)       H₀: μ_b = μ_w     H_a: μ_b ≠ μ_w     s_x1-x2 = √(.68² /43) + (.91² /304) = .116

            Z = (2.76 - 2.49)/.116 = 2.33   Reject H₀. Blacks are more tolerant of violence than whites.

c)       H₀: μ_w = μ_b     H_a: μ_w ≠ μ_b      s_x1-x2 = √(.91² /249) + (.87² /97) = .105

            Z = (2.46 - 2.67)/.105 = -1.99 Reject H₀. Blue collar workers are more tolerant of violence than white collar workers.

d)       While significant differences in tolerance for interpersonal violence were found in all of the comparisons done, the largest difference was found between males and females.

9.6 N₁ + N₂ > 120, so use the Z distribution. The .05 critical value is 1.65

           H₀: μ_r = μ_c       H_a: μ_r > μ_c      s_x1-x2 = √(2²/158) + (1.9²/173) = .215

            Z = (12.4 - 10.2)/.215 = 10.23             Reject H₀. Residential students devote more hours per week to extracurricular activities.

9.8 N₁ + N₂ < 120, so use the t distribution. There are 93 df; the 05 critical value is - 1.66

             H₀: μ_m = μ_n     H_a: μ_m < μ_n    s² = [(N₁ ^-1)s₁² + (N₂ -1)s₂²]/(N₁ + N₂ -2) =

                                                            [(39)(2.7²) + (54)(4.2²)]/(40 + 55 - 2) = 13.3

s_x1-x2 = √ s² (1/N₁ + 1/N₂) = √(13.3)(1/39 +1/54) = .77

            t = (10.3 - 12.3)/.77 = -2.60     Reject H₀. Members reported engaging in fewer acts of delinquency.

9.12 a. Let P_S = population proportion divorced in the special program, and P_R = population proportion divorced in the regular program. We use the Z distribution when working with proportions. The test is one-tailed because the problem asks if the program “worked.” The one-tailed .05 critical value is -1.65.

            H₀: P_S = P_R    H_a: P_S < P_R Common p = .54(78) + .59(82) = .56

                                                                                    78 + 82

s_p1-p2 = √(.56)(.44)[(1/78) + (1/82)] = .079

            Z = (.53 - .59)/.079    = -0.76 Retain H₀: There is no evidence that the program results in .079 lower divorce rates.

b. Let P_PC = population proportion “improved” in the peer counseling program, and P_NC = population proportion “improved” who had no peer counseling. We will use the Z distribution when working with proportions. The test is one-tailed because the problem asks if the program “worked.” The one-tailed .05 critical value is -1.65.

            H₀: P_PC = P_NC H_a: P_PC > P_NC Common p = .10(52) + .15(56) = .13

                                                                                      52 + 56

s_p1-p2 = √(.13)(.87[(1/52) + (1/56)] = .065

            Z = (.10 - .15)/.065    = -0.78 Retain H₀: There is no evidence that the program results in .065 more improvement in depression. In fact, the results were found to be in the opposite direction from what was predicted, indicating that the researchers were not justified in running a one-tailed test!

SPSS:

The questions asked were:

1. Are there differences between age groups in the use of cigarettes, alcohol, or marijuana?

2. Are there differences between males and females in the use of cigarettes, alcohol, or marijuana?

1. The relevant parts of the SPSS output for the tests of differences by age are:

Group Means

052(R) :AGE <>18	N	Mean
052B02 :# cigarettes last 30 days
< 18:(1)	659	1.42
18+:(2)	768	1.53
052B04C:# times drank alcohol last 30 days
< 18:(1)	639	1.91
18+:(2)	734	2.09
052B07C:# times marijuana last 30 days
< 18:(1)	648	1.51
18+:(2)	754	1.60

Independent Samples t- test, Equal Variances Assumed

	t	df	Sig. (2-tailed)
052B02 :# cigarettes last 30 days	-1.983	1425	.048
052B04C:# times drank alcohol last 30 days	-2.378	1371	.018
052B07C:# times marijuana last 30 days	-1.196	1400	.232

The group means show that the use of all three substances is slightly higher for those seniors who are over 18 years old. The t-tests of the null hypotheses of no difference between age groups are significant for cigarette use (t = -1.98, p = .048) and alcohol (t = -2.38, p = .018), but not for marijuana ( t = -1.12, p = .232). Students who are over 18 smoked and drank more in the last 30 days than those under 18. There is no significant difference between age groups in the amount of marijuana used.

2. The relevant parts of the SPSS output for the tests of difference by gender are:

Group Means

052C03 :R'S SEX	N	Mean
052B02 :# cigarettes last 30 days
MALE:(1)	683	1.52
FEMALE:(2)	720	1.44
052B04C:# times drank alcohol last 30 days
MALE:(1)	651	2.15
FEMALE:(2)	700	1.86
052B07C:# times marijuana last 30 days
MALE:(1)	666	1.72
FEMALE:(2)	715	1.41

Independent Samples t- test, Equal Variances Assumed

	t	df	Sig. (2-tailed)
052B02 :# cigarettes last 30 days	1.452	1401	.147
052B04C:# times drank alcohol last 30 days	3.844	1349	.000
052B07C:# times marijuana last 30 days	4.144	1379	.000

The group means show that the use of all three substances is higher for males than females. The t-tests of the null hypotheses of no difference between males and females are highly significant for alcohol use (t = 3.84, p < .000) and marijuana (t = 4.14, p < .000), but not for cigarettes ( t = 1.45, p = .147. Males drank and used marijuana more in the last 30 days than females. There is no significant difference between males and females in smoking.

It is not difficult to speculate about the reasons for these findings. Age may affect access to cigarettes and alcohol, because these substances are sold legally to those who are old enough. Students over 18 can buy cigarettes and are also more likely to have friends who are old enough to purchase alcohol. Marijuana cannot be purchased legally; therefore age may be less important than environmental factors in determining access. Gender roles may affect the propensity to take risks and also the positive or negative image associated with these behaviors. Males are more likely to engage in risky behaviors, and smoking and drinking have been associated with masculinity. There is no gender difference in cigarette use among this population (high school seniors). Historically, females have been much less likely than males to smoke. The gender gap has been diminishing in the past few decades, and recent generations of females have gained “equality” with males in smoking. One explanation that has been suggested is that smoking has become an increasingly popular method of weight control for girls.

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