Problem Set #5 Answer Key
1)a) p(C1) = 1 / 5 = .20
b) p(CCC) = (.2)3 = .008
c) p(ECE) = (.8)(.2)(.8) = .128
d) p(CEE or ECE or EEC) = (.128)(3) = .384
e) If the probability of getting none correct on trials 1-3 is .384 (see part d), the probability of get
Exam 2 Answer Key
1)a) The IV is choice, with two levels, limited and extensive. The DV is (in the analysis of interest) performance, measured by the
quality of papers. (Another DV in the experiment is what may be labeled motivation or effort, measured by
PSYC 5133
Fall '13
Exam 2
In completing this exam, you may use any resource except other people. Please adhere to the homework guidelines in deciding how
much of this to type, write out by hand, etc. Each problem's point value is listed before the problem
PSYC 5133
Fall '13
Exam 1
In completing this exam, you may use any resource except other people. Please adhere to the homework
guidelines in deciding how much of this to type, write out by hand, etc. Each problem's point value is listed
before the problem
Problem Set #14 Answer Key
1) For the simple-effects tests, I used the following syntax:
unianova
time by approach attract
/method = sstype(3)
/intercept = include
/emmeans = tables(approach*attract)compare(approach)adj(sidak)
/criteria = alpha(.05)
/desi
Problem Set 9 Answer Key
1) a) H0: FT C = 0; H1: FT C > 0
25
16
2
b) = 41 (7.99)2 + 41 (7.16)2 = 58.9
58.9
= 26 +
58.9
17
= 2.39
= .05, one-tailed, tcrit(41) = +1.684
tobs = (7.26 1.45) / 2.40 = 2.42
Reject H0, conclude that the girls in family therapy
Problem Set #10 Answer Key
1)a) Addition accuracy for sixth-grade females (M = 95.4, SD = 6.2) was slightly lower than for sixth-grade males (M = 98.7, SD =
3.2)1, but this difference was not significant, t(24) = 1.65, SE = 1.99, p = .11. Thus, we have no
Problem Set #11 Answer Key
1)a)
ANOVA
DV
Between Groups
Within Groups
Tot al
Sum of
Squares
78. 000
40. 000
118.000
df
2
6
8
Mean Square
39. 000
6. 667
F
5. 850
Sig.
.039
b) Yij = + j + ij
c) 1 = (3 6) = -3, 2 = (5 6) = -1, 3 = (10 6) = 4
d) Here are the
Problem Set #13 Answer Key
1)a)
b) The only significant effect is of major, so we may conclude that engineering students rate GWB more favorably than
do humanities students. Neither interviewer nor the interaction of interviewer and major had significant
Problem Set #12 Answer Key
D escriptives
1)a) For the sake of
thoroughness, I've
included some
descriptives .
notice the
heterogeneity of
variance issue (a
10:1 ratio of
highest-to-lowest
variances).
ERRORS
N
alway s
f requent
inf requent
nev er
Tot al
5
Problem Set #8 Answer Key
1) Exercise 5.2. I'm assuming normally-distributed populations here. Otherwise these questions are non-starters.
a) z = (170 200)/60 = -0.5, so p(Y > 170) = .6915
b) z = (170 200)/(60/9) = -1.5, so p( Y > 170) = .9332
c) d = 200
Problem Set #7 Answer Key
1) Exercise 5.1.
a) z = (115 100) / 15 = +1, so p(Y > 115) = .1587
b.i) z = (130 100) / 15 = +2, so p(Y > 130) = .0228
b.ii) z85 = (85 100) / 15 = -1; z145 = (145 100) / 15 = +3. For z = -1, the area in the body is 1 .1587 = .841
Problem Set #2 Answer Key
1) Here are two answers. I much prefer Version 1. Version 2, while creative and thoughtful, assumes strong non-independence of
samplingit's kind of reverse version of the gambler's fallacy (i.e., if you win once early, you're due
Problem Set #1 Answer Key
1)a) This is an experiment because subjects were randomly assigned to conditions; that is, the researcher was
able to manipulate a variable.
b) Because this sample is based on those who were interested in losing weight (presumabl
Problem Set 3 Answer Key
1)a) This hypothesis suggests a positive relationship
(high with high and low with low).
b) There appears to be a positive relationship in the
scatterplot at right.
c) Pearson's r for HRs and BBs is .48. This is a very
strong rela
Problem Set #4 Answer Key
1)a) p(yes) = 100/300 or .33
b) p(blue) = 90/300 or .3
c) p(yes or blue) = 130/300 or .43 [ = p(yes) + p(blue) p(yes and blue) = 100/300 + 90/300 60/300]
d) p(yes and blue) = 60/300 or .2
e) p(blue | yes) = 60/100 or .6
f) p(yes
Exam 1 Answer Key (corrected; corrections are in boldface type)
1)a) p(negative | dementia) = 1 p(positive | dementia) = 1 - .22 = .78.
b) p(negative | no dementia) = 1 p(positive | no dementia) = 1 - .01 = .99.
c) I prefer using frequencies, so if there