8. In a study by Gonzaga et al. (2001), romantic couples answered questions about how much they loved their partner and also were videotaped while revealing something about themselves to their partner. The videotapes were later rated by trained judges for various signs of affiliation. Table 2–8 (reproduced from their Table 2) shows some of the results. Explain to a person who has never had a course in statistics the results for self-reported love for the partner and for the number of seconds “leaning toward the partner.” Table 2–8 Mean Levels of Emotions and Cue Display in Study 1 Women ( ) Men ( ) Indicator M SD M SD Emotion reports Self-reported love 5.02 2.16 5.11 2.08 Partner-estimated love 4.85 2.13 4.58 2.20 Affiliation-cue display Affirmative head nods 1.28 2.89 1.21 1.91 Duchenne smiles 4.45 5.24 5.78 5.59 Leaning toward partner 32.27 20.36 31.36 21.08 Gesticulation 0.13 0.40 0.25 0.77 Note: Emotions are rated on a scale of 0 (none) to 8 (extreme). Cue displays are shown as mean seconds displayed per 60 s. Source: Gonzaga, G. C., Keltner, D., Londahl, E. A., & Smith, M. D. (2001). Love and the commitment problem in romantic relationships and friendship. Journal of Personality and Social Psychology, 81, 247–262. Published by the American Psychological Association.Reprintedwithpermission.
Answer:
Self-reported love for the partner on average is good (around 5 on a cale from 0 to 8) of men and womed these numners are similar (5.02 and 5.11) sd for both groups is similar too
The number of seconds “leaning to the partner” on average is about 32 (32.27 for men and 31.36 for women) and the sd is similar too
11. For the following scores, find the
(a) mean,
(b) median,
(c) sum of squared deviations,
(d) variance, and
(e) standard deviation:
2, 2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0
a)Mean = (2+2+0+…..+1+0)/21 = 42/21= 2
Mean = 2
b)Ordered data is: 0,0,0,0,0,1,1,1,1,1,2,2,2,3,3,4,4,4,4,4,5
Median = 2
c)Sum of squared deviations =SS= (2-2)2+(2-2)2+…..+ (0-2)2=20+5+2+20+9=56
SS= 56
d)Variance = SS/N= 56/21=8/3
Variance = 8/3
e)Standard deviation = SD = 1.633
SD = 1.633
12. For the following scores, find the
(a) mean,
(b) median,
(c) sum of squared deviations,
(d) variance, and
(e) standard deviation:
1,112; 1,245; 1,361; 1,372; 1,472
a)Mean = (1,112+1,245+……..+1,472)/5 = 1,312.4
Mean = 1,312.4
b)Ordered data is: 1,112; 1,245; 1,361; 1,372; 1,472
Median = 1,361
c)Sum of squared deviations =SS= (1,112-1,312.4)2+…..+ (1,472-1,312.4)2=76,089.2
SS= 76,089.2
d)Variance = SS/N= 76,089.2/5=15,217.84
Variance = 15,217.84
e)Standard deviation = SD =
SD = 123.36
Chapt.3 14. On a standard measure of hearing ability, the mean is 300 and the standard deviation is 20. Give the Z scores for persons who score
(a) 340,
(b) 310, and
(c) 260.
a) z-score = (340-300)/20 = 2
b) z-score = (310-300)/20 = 0.5
c) z-score = (260-300)/20= -2
Give the raw scores for persons whose Z scores on this test are
(d) 2.4,
(e) 1.5,
(f) 0, and
(g)
d) raw score = 2.4(20)+300 = 348
e) raw score =1.5(20)+300 = 330
f) raw score = 0(20)+300 = 300
g) missing info
15. A person scores 81 on a test of verbal ability and 6.4 on a test of quantitative ability. For the verbal ability test, the mean for people in general is 50 and the standard deviation is 20. For the quantitative ability test, the mean for people in general is 0 and the standard deviation is 5. Which is this person’s stronger ability: verbal or quantitative? Explain your answer to a person who has never had a course in statistics.
z-score for verbal ability is (81-50)/20 = 1.55
z-score for quantitative ability is (6.4-0)/5=1.28
This person has a strong ability on verbal , this person is more above the average in verbal ability than the average in quantitative ability
25. You are conducting a survey at a college with 800 students, 50 faculty members, and 150 administrators. Each of these 1,000 individuals has a single listing in the campus phone directory. Suppose you were to cut up the directory and pull out one listing at random to contact. What is the probability it would be
(a) a student,
(b) a faculty member,
(c) an administrator,
(d) a faculty member or administrator, and
(e) anyone except an administrator?
(f) Explain your answers to someone who has never had a course in statistics.
a) P(student)= 800/1000= 0.8
Answer: 0.8
b) P(faculty member)= 50/1000= 0.05
Answer: 0.05
c)P(administrator)= 150/1000=0.15
Answer:0.15
d) P( faculty member or administrator)=P(faculty member)+P(administrator)= 0.05+0.15=0.2
Answer: 0.2
e) P(no administrator)=1-P(administrator)=1-0.15=0.85
Answer: 0.85
f) for all the answers above we computed the probability of A as #(A)/1000