Download Summer 2004 Paper - Goldsmiths, University of London

GOLDSMITHS
University of London
PSYCHOLOGY DEPARTMENT
MSc in RESEARCH METHODS IN PSYCHOLOGY 2004
PS71020A STATISTICAL METHODS
3 HOURS
Answer THREE questions.
ONE from Section A, ONE from Section B and ONE from Section C. Each question is worth
50 marks and the marks for each part of each question are indicated where appropriate. The
mark for the whole paper will be converted to a mark on a 0-100 scale.
Note that there are attached SPSS printouts for questions 1, 2, 3, 4, 6, 8, & 9
You may use a calculator.
SECTION A
1. A psychologist was hired by a sports centre to investigate the factors associated with
customers’ satisfaction in relation to the aerobics classes running at the centre. Over a
1-month period, the psychologist distributed a questionnaire which recorded
customers’ satisfaction ratings on a 0-100 scale (0 completely dissatisfied; 100
completely satisfied; SPSS variable= satis). The psychologist planned to carry out
multiple regression analyses to predict the satisfaction ratings from other variables
recorded on the questionnaire. These included: the participants' estimates of the size
of the class (SPSS variable sizeclas); how difficult they found the class (on a 0-100
scale, with higher scores indicating higher difficulty; SPSS variable difficul); and how
long in months they had been doing aerobics (SPSS variable howlong).
Complete data were returned by 160 female aerobics participants. Ratings (of the class
just completed) were taken at the end of 30 different classes, each run by one of the
centre’s 3 aerobics instructors. From each class, there were between 3 and 12
respondents.
The first predictor explored was the teacher. The researcher coded the teacher
variable in two ways, known as dummy coding and effect coding. The coding scheme
is shown in the table below.
1
PTO
SPSS Dummy Coded Variable
tdumcod1
tdumcod2
SPSS Effect Coded
Variable
teffcod1
teffcod2
1
Teacher
2
3
1
0
0
1
0
0
1
0
0
1
-1
-1
(i) Explain what these two coding schemes for the teacher variable are able to tell the
researcher about the effect of the class teacher on participant satisfaction. (10
marks)
The psychologist carried out one regression predicting satisfaction from the pair of
dummy coded teacher variables and another predicting satisfaction from the effect
coded teacher variables. In each case, the overall model R 2 was 0.053, which
reflected a statistically significant effect of teacher on satisfaction (F[1,157]=4.4,
p<0.02).
(ii) Inspect the printouts of the regression coefficients for the teacher variables and
say, with reference to the answer given for part (i) above, what the researcher
might infer from them. (5 marks)
The psychologist then carried out a regression predicting satisfaction from 5 predictors
simultaneously (difficulty, length of past aerobics experience, estimated number of
class participants, plus the dummy coded pair of teacher variables). First he calculated
basic descriptive variables and bivariate correlations for the dependent variable plus
the non-teacher predictor variables (see printout). After univariate screening, he
attempted to check for multivariate outliers by computing a Mahalanobis distance (MD)
for each participant. To determine which participants might be multivariate outliers, the
researcher had to compare the MD values with the critical value from a well-known
statistical distribution.
(iii) What distribution should the researcher look up? What probability level and
number of degrees of freedom should he use to obtain the critical value? (5 marks)
There were no multivariate outliers in the dataset. The researcher carried out the
regression and the results are shown in the attached printout.
(iv) Inspect the regression output and earlier printouts, and describe in detail what
these results would allow the researcher to report to the sports centre. Comment
on any aspect(s) of the actual statistical findings which might cause the researcher
to be concerned about the validity of his analysis, and suggest how he might
modify his regressions to deal with his concerns. (25 marks)
(v) Identify a particular aspect of the design and execution of the study that is likely to
mean that the data do not satisfy an important assumption of multiple regression
analysis. (5 marks)
2
PTO
2. A researcher wanted to investigate the factors which are related to whether an
individual can “see” the 3-D images hidden in “Magic Eye” stereograms. He coded his
120 participants into 3 groups based on his findings (1= saw the correct image in all
the pictures; 2= saw some of the images; 3 = did not see any of the images; SPSS
variable name = cansee). The researcher fitted a series of multinomial logistic
regression models with cansee as the dependent variable. The predictor variables he
was interested in included the following: the participant’s age (younger vs. older),
gender and extraversion score (1-4 based on the quartiles within the sample; SPSS
variable = extra4). The output produced by the final model fitted is attached.
(i)
Explain in detail, step-by-step, how the researcher would go about conducting
the logistic regression analyses. Pay particular emphasis to the sequence of
models that he would have fitted, ending up at the model which produced the
attached printout. (20 marks)
(ii)
Explain in detail what conclusions the researcher could draw from the each
element of the output from the final regression model, and the associated
crosstabulations. (20 marks)
(iii)
The researcher felt that his choice of coding for the cansee variable obscured
some findings of interest in his sample, so he recoded an alternative variable
cansee2. A participant was given a particular value of cansee2 depending on their
value for cansee. The mappings were as detailed in the following table. Explain
why the researcher did this and, by looking at the regression parameter estimates
for the analysis with cansee2 as the dependent variable, outline the additional
conclusions that this allowed him to draw. (10 marks)
Value of cansee2
1 (always sees)
2 (not used)
3 (never sees)
4 (sees sometimes)
Equivalent value of cansee
1
not applicable
3
2
3. A researcher administered a short scale containing 9 trait adjectives (angry; stressed;
happy etc) to 100 students. The participants were asked to rate themselves in terms of
how much they possessed the trait in question, relative to the general population. A
rating scale of 50-150 was to be used, with higher numbers indicating a greater level
of the trait. A rating of 100 was to be used if they felt they were completely average in
terms of that trait. The researcher carried out a factor analysis using unweighted least
squares factoring with a varimax rotation. On the basis of an initial scree plot, 3 factors
were retained. The results of his analysis are shown in the attached printout. Using
this analysis as an example when answering parts (i) and (ii):
(i) describe the complete sequence of steps that researchers go through when
carrying out a factor analysis; (30 marks)
3
PTO
(ii) comment on the conclusions that researchers can draw from factor analyses and
outline the features which indicate that the analysis is likely to be reliable and valid.
(20 marks)
SECTION B
4. In parts (i)-(vi) below illustrate your answers, where appropriate, with hypothetical
examples of psychological research designs.
(i) Outline the main similarities and highlight the key differences between a two-way
univariate analysis of variance (ANOVA) with between-groups factors, and its
multivariate equivalent (“classical” MANOVA). (5 marks)
(ii) Outline the main similarities and highlight the key differences between classical
MANOVA and repeated measures MANOVA. (5 marks)
(iii) Outline the main similarities and highlight the key differences between repeated
measures analysis of variance (ANOVA) and repeated measures MANOVA. (5
marks)
(iv) Outline the main similarities and highlight the key differences between an analysis
of covariance (ANCOVA) and its multivariate equivalent (MANCOVA). (5 marks)
(v) Describe what is meant by a doubly multivariate MANOVA, highlighting its
relationship with classical MANOVA and repeated-measures MANOVA. (5 marks)
(vi) What are trend contrasts and when is it appropriate to analyse them? If a factor
has 5 levels how many trend components can be analysed? How many trend
components would one usually be interested in? (5 marks)
A psychologist studied football fans from 3 leading teams (Arsenal; Man Utd;
Chelsea), and investigated their mood at 3 phases of the season (1=early; 2=mid;
3=late). At each time-point in the season she asked each fan to rate 4 positive mood
states (happiness, excitedness, hopefulness, and confidence) using a self-rating
questionnaire which measured their feelings over the preceding week. The attached
printout shows the results of an analysis which the researcher conducted on the data.
Graphs of the mean mood profiles over time, for each mood and fan-type separately,
are also attached.
(vii) Comment on the type of analysis which the researcher has carried out. Go
through the printout and describe what conclusions the researcher can draw from
the findings. Suggest other analyses that would be useful in interpreting the data
in more detail. (20 marks)
5. “Analysis of Covariance (ANCOVA) is one of the most misused statistical techniques
in psychology.“ Discuss this statement in detail, and use as many examples as
feasible to illustrate situations in which ANCOVA might be misused by psychologists.
In answering the question make clear differentiations between the various uses of
4
PTO
ANCOVA and, for each use of ANCOVA, make recommendations about when the
technique is appropriate or inappropriate.
6. A personality researcher was interested in the effects of background noise on
performance in relation to extraversion. In his first study, he tested a group of
extraverts and introverts on a reaction time task while playing continuous white noise
to the participants. Approximately half the extraverts and half the introverts were
randomly allocated to be tested under a low noise (50 db), with the remainder of each
group being tested under high noise conditions (90 db). He recorded a mean reaction
time for each participant (SPSS variable = rtmean), and analysed the data using a 2x2
ANOVA. He predicted an interaction between extraversion group and noise condition.
The output from this analysis of study 1 is attached, and this shows a marginal
interaction (F[1,25]= 4.2, p=0.05). He decided before the experiment that he would
explore the predicted interaction with planned contrasts to compare low vs. high noise
conditions in extraverts and introverts separately.
(i) The psychologist used the following syntax commands to compute the contrasts.
Explain how the LMATRIX subcommands are constructed, bearing in mind that
the extraversion group variable (extgp) is coded 1=extravert, 2=introvert. The
noise condition variable is coded 1=low, 2=high. (10 marks)
SPSS syntax for Planned Contrast:GLM
rtmean BY extgp noise
/LMATRIX = "simple noise effect for introverts" noise 1 -1 extgp*noise 0 0 1 -1
/LMATRIX = "simple noise effect for extraverts" noise 1 -1 extgp*noise 1 -1 0 0
/DESIGN = extgp noise extgp*noise .
The contrast for the simple main effect of noise in extraverts yielded t[25]=2.75,
p=0.011 (two-tailed); the contrast for the simple main effect of noise in introverts
yielded t[25]=-0.27, p>0.5 (two-tailed).
(ii) What did the researcher conclude based on these contrast findings? Explain your
reasoning carefully. (5 marks)
(iii) Explain how this method of exploring simple main effects differs from simply
computing a planned t-test based on the data from extraverts in low noise and
extraverts in high noise (or an equivalent t-test based on the introverts’ data). (5
marks)
(iv) Outline the reasons why the planned contrasts will generally be more powerful
than the planned t-tests described in section (iii). (10 marks)
The researcher carried out a second study which followed the same design as the first
study, except that he added another between-groups factor of reinforcement condition
(SPSS variable = reinf). Approximately half the participants in each combination of
noise and extraversion group were randomly allocated to receive monetary rewards for
each correct response during the task; the remaining participants received monetary
punishments (loss of money) for every error during the task. It was predicted that this
factor might interact with extraversion group in determining reaction time, and there
5
PTO
might be a 3-way interaction (extgp X noise X reinf). The researcher planned to use
contrasts to explore the interaction between extraversion and reinforcement. The
contrasts looked at the simple main effect of reinforcement condition for extraverts and
introverts separately, plus the simple main effect of extraversion group for the reward
and punishment conditions separately.
The only significant effect from the 2x2x2 ANOVA carried out on study 2 was the
interaction between reinforcement condition and extraversion group (F[1,71]=4.3,
p<0.05). The relevant means are shown in the following table:
ex traversi on group * reinforceme nt condition
Dependent Variable: RTMEAN
ex traversion group
ex travert
int rovert
reinforc ement c ondition
reward
punishment
reward
punishment
Mean
564.550
533.225
543.985
575.900
St d. Error
15.180
15.180
15.596
15.180
95% Confidenc e
Int erval
Lower
Upper
Bound
Bound
534.282
594.818
502.957
563.493
512.888
575.083
545.632
606.168
The planned contrasts were as shown in the following table:
Planned Contrast
Reward vs. punishment for extraverts
Reward vs. punishment for introverts
Extraverts vs. introverts for reward
Extraverts vs. introverts for punishment
t-value
1.46
-1.47
0.945
-1.99
df
71
71
71
71
p (2-tailed)
0.15
0.15
0.35
0.05
(v) How should the researcher interpret the findings from study 2? Explain your
arguments carefully. (10 marks)
(vi) Describe what is meant by a set of orthogonal contrasts. Illustrate your answer
with reference to a one-way design with 3 levels of the factor. Describe one set of
contrasts for the factor which are orthogonal, and one set which are not. (10
marks)
SECTION C
7.
A researcher investigated learning performance in schizophrenic patients and
matched controls. She tested half the patients and half the controls under standard
learning conditions and the remaining halves of the two groups under special
learning conditions. These were designed to reduce the learning deficit (relative to
the controls) that is usually observed in the patients under standard learning
conditions. She was therefore predicting an interaction between learning condition
6
PTO
and subject group in a 2x2 independent groups ANOVA carried out on the learning
scores. Unfortunately, she could not use parametric methods because her dependent
variable (the learning measure) was bimodally distributed in all cells of her design.
(i)
Explain in detail how the researcher might investigate her key hypothesis using
nonparametric (i.e., rank-based) statistics. (15 marks)
(ii)
Explain in detail how the researcher might investigate her key hypothesis using
resampling methods (i.e., bootstrapping and/or randomisation). (15 marks)
The same researcher was interested in whether individual schizophrenic patients
were able to generate random sequences of digits, suspecting that some might not
be able to do so because of a tendency to perseverate, thereby generating some
long sequences of the same digit. Participants were asked to generate random
sequences of 1s and 2s, and sequences of 50 responses were recorded for each
participant. The researcher wanted to test whether the sequences generated by an
individual were random but knew of no standard way to test this. She decided that
she would calculate the longest run of the same digit produced by each participant
(let this be N digits long). She then needed to estimate the probability that a
genuinely random process could generate a run of N digits or longer within a set of
50 digits.
To do this she used a resampling method: she randomised the 50-item sequence
produced by an individual participant 10000 times, and calculated the longest run of
the same digit within each of the 10000 randomised sequences. She then used these
10000 “longest run values” to estimate the probability of generating a sequence N
digits or longer by chance.
(iii) Explain this method more fully, giving details of the logic behind it, alternative
methods of randomisation that might be employed, and how she would use the
10000 “longest run values” to estimate the probability she needed. (10 marks)
(iv) The researcher also knew that some schizophrenic patients might show the
opposite tendency: i.e., to generate non-random sequences by avoiding runs of
the same digit. How would she need to use her 10000 “longest run values”
above to also evaluate whether an individual patient was showing this
alternative tendency to a significant extent? (5 marks)
An alternative method would have been to take a computer and programme it to
generate 10000 sequences randomly (i.e., where each digit in each sequence could,
with equal probability, be a 1 or 2).
(v)
8.
Why would this alternative method be likely to generate different findings from
the resampling method which she employed? (5 marks)
A research team had collected data from 50 male participants and 50 female
participants. Each participant had provided a rating of their self-esteem and a recent
body-mass index (BMI); a high body-mass index indicates someone who tends
towards obesity. The research team wanted to investigate the hypothesis that the
negative relationship between BMI and self-esteem would be stronger in female
participants than in males. Two research assistants A and B were given the task of
7
PTO
carrying out the analysis to address the hypothesis of interest. Assistant A used a
Fisher-Z transformation to compare the correlation in males with that in females,
writing some SPSS syntax (attached) to carry this out.
Assistant B took a different approach. He knew that if one tests the correlation
coefficient between variables X and Y, or tests the regression coefficient for a simple
linear regression of X on Y, then both methods would generate an identical
probability value. Assistant B therefore reasoned that he could address the
hypothesis of interest by statistically comparing the regression coefficient for BMI as
a predictor of self-esteem in females with the same coefficient in males. Moreover, he
knew how to carry out this “homogeneity of regression” analysis in SPSS and thus
could avoid using syntax commands. Confronted with these different analyses, the
project leader consulted her statistics advisor to find out who had used the correct
method. Was it A or B, or perhaps they had both used valid alternatives?
(i)
If you were the statistics advisor what would you have said about which
methods were appropriate, and give detailed arguments to support your advice.
(20 marks)
The observed correlation in the males was -0.24 and it was -0.55 in females. Using
the syntax file create by Assistant A, these values gave Fisher-transformed
correlations of
-0.245 and -0.618. (These are the values rfisherm and rfisherf in
the syntax file.)
(ii)
Employing the formulae given in A’s syntax file, use a calculator to work out the
Z-value for the difference between these transformed correlations and comment
on whether the difference would be significant at the 5% level. (10 marks)
(iii)
How would one change the line of syntax used for computing zprob (the
probability value associated with the Z statistic) if one were conducting a twotailed test? (5 marks)
Another part of this research involved comparing, for just the female participants, the
correlation between BMI and self-esteem (denoted r1) with that between trait anxiety
and self-esteem (denoted r2). Based on work by Steiger, the research team found
formulae for comparing these correlations in a single sample. To use the formulae,
they also needed to compute the correlation between BMI and trait anxiety (denoted
r3).
After doing this successfully, a member of the research team then asked if it were
possible to use the same formulae to compare multiple correlations in a single
sample in a similar fashion. Specifically, they wanted to compare the multiple
correlation (denoted R1) from a regression predicting self esteem using two “bodyrelated” variables (BMI and percentage body fat), with the multiple correlation
(denoted R2) found when predicting self esteem using two personality measures
(trait anxiety and extraversion). The research team correctly wanted to put the values
for the multiple correlations R1 and R2 into the formulae in the same places that they
had put the values for the simple correlations r1 and r2. However, they could not
work out how to compute R3, the multiple correlation equivalent of r3.
8
PTO
(iv) Explain in detail how the researchers could calculate R3 and therefore make the
desired comparison between R1 and R2. (15 marks)
9.
(i)
(ii)
(iii)
(iv)
(v)
Define statistical power and describe its relationships with Type II errors and
non-central distributions. (10 marks)
In planning and executing your experiment what can you do to increase power?
(5 marks)
What are conventionally regarded as adequate levels of power? (5 marks)
What information do you need in order to estimate sample sizes before
embarking upon a study? (5 marks)
How might you obtain the information required for the sample size estimation
process in (iv)? (5 marks)
A study compared smokers and nonsmokers on their lung capacity (SPSS variable =
lungcap) and their scores on a smell identification test (SPSS variable = smellid). The
descriptive statistics for the two measures in each group are included in the printout.
The hypotheses were that smokers would have a smaller lung capacity than
nonsmokers, and would perform more poorly on the smell identification test. The
researcher ran separate one-way ANOVAs on each of the two dependent variables
and selected the “observed power” option. The resulting printout is attached.
(vi)
Cohen’s d is a measure of effect size, which can be used in this two-group
situation. Define d and, showing your working, use a calculator to calculate this
measure of effect size for lung capacity and smell identification performance.
You should check your calculations against the ANOVA printout, as the noncentrality parameter shown is equal to (0.5*d 2*n), where n is the number of
participants in each group. (10 marks)
(vii) Explain in words what information is conveyed by the observed power results.
How might this information bear on the researcher’s conclusions about the
hypotheses under test? (5 marks)
(viii) Given the current sample sizes, and the observed standard deviations of scores
on the smell test in each group, what approximate mean difference between the
groups in smell identification performance could have been detected with >80%
power in this study? Base your estimate on alpha = 0.05 and a two-tailed test
(i.e., ignore the directional nature of the hypotheses). Use a calculator and
show your working. (5 marks)
9
PTO
Printout for Question 1
Satisfaction Predicted by Dummy-Coded Teacher Variables
Coeffi cientsa
Model
1
Unstandardized
Coeffic ient s
B
St d. Error
45.226
2.068
8.107
2.912
1.472
2.925
(Const ant)
TDUMCOD1
TDUMCOD2
St andardi
zed
Coeffic ien
ts
Beta
.250
.045
t
21.866
2.784
.503
Sig.
.000
.006
.616
Collinearity Statistics
Tolerance
VIF
.748
.748
1.338
1.338
a. Dependent Variable: SATIS
Satisfaction Predicted by Effect-Coded Teacher Variables
Coeffi cientsa
Model
1
Unstandardized
Coeffic ients
B
St d. Error
48.419
1.190
4.914
1.678
-1. 721
1.686
(Const ant)
TEFFCOD1
TEFFCOD2
St andardi
zed
Coeffic ien
ts
Beta
.262
-.091
t
40.672
2.928
-1. 021
Sig.
.000
.004
.309
Collinearity Statistic s
Tolerance
VIF
.752
.752
1.329
1.329
a. Dependent Variable: SATIS
Descriptive Statistics and Bivariate Correlations
Descriptive Statistics
N
SATIS
DIFFICUL
HOWLONG
SIZECLAS
Valid N (lis twise)
160
160
160
160
160
Minimum
15.00
30.00
1.00
5.00
Maximum
90.00
90.00
132.00
20.00
10
Mean
48.4500
55.0313
69.7000
12.8563
Std.
Deviation
15.3778
12.4690
28.3262
2.9691
PTO
Correl ations
SATIS
DIFFICUL
HOWLONG
SIZECLAS
SATIS
DIFFICUL
1.000
-.455**
.
.000
160
160
-.455**
1.000
.000
.
160
160
.373**
-.840**
.000
.000
160
160
-.015
.046
.854
.565
160
160
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
Pearson Correlation
Sig. (2-tailed)
N
HOWLON
G
SIZECLAS
.373**
-.015
.000
.854
160
160
-.840**
.046
.000
.565
160
160
1.000
-.080
.
.316
160
160
-.080
1.000
.316
.
160
160
**. Correlation is s ignificant at the 0.01 level (2-tailed).
Main Regression Output
Model Summaryb
Model
1
R
.531a
R Square
.281
Adjusted
R Square
.258
Std. Error
of the
Es timate
13.2449
a. Predictors: (Constant), TDUMCOD2, SIZECLAS,
DIFFICUL, TDUMCOD1, HOWLONG
b. Dependent Variable: SATIS
ANOVAb
Model
1
Regres sion
Residual
Total
Sum of
Squares
10583.770
27015.830
37599.600
df
5
154
159
Mean
Square
2116.754
175.427
F
12.066
Sig.
.000a
a. Predictors: (Constant), TDUMCOD2, SIZECLAS, DIFFICUL, TDUMCOD1,
HOWLONG
b. Dependent Variable: SATIS
11
PTO
Coefficientsa
Model
1
(Constant)
DIFFICUL
HOWLONG
SIZECLAS
TDUMCOD1
TDUMCOD2
Unstandardized
Coefficients
B
Std. Error
83.128
13.896
-.648
.156
-2.97E-02
.069
-2.73E-02
.356
9.258
2.574
.761
2.580
Standardi
zed
Coefficien
ts
Beta
-.525
-.055
-.005
.286
.023
t
5.982
-4.156
-.432
-.077
3.597
.295
Sig.
.000
.000
.666
.939
.000
.769
Collinearity Statistics
Tolerance
VIF
.292
.292
.989
.740
.743
3.422
3.424
1.011
1.351
1.345
a. Dependent Variable: SATIS
12
PTO
Printout for Question 2
Crosstabs
Crosstab
Count
CANSEE3
2.000
25
15
11
12
63
1.000
EXTRA4
1
2
3
4
1
9
11
10
31
Total
3.000
Total
7
8
3
8
26
33
32
25
30
120
Crosstab
Count
GENDER
male
female
Total
1.000
22
9
31
CANSEE3
2.000
20
43
63
3.000
18
8
26
Total
60
60
120
Logistic Regression Output (DV= cansee)
Case Processing Summary
Males are coded 0
Females are coded 1
N
CANSEE
CANSEE is coded 1=always sees;
2=sometimes sees; 3= never sees
GENDER
Valid
Mis sing
Total
always sees
sometimes sees
never s ees
male
female
31
63
26
60
60
120
0
120
Model Fitting Information
Model
Intercept Only
Final
-2 Log
Likelihood
82.301
54.200
Chi-Squa
re
28.101
df
Sig.
4
13
.000
PTO
Goodness-of-Fit
Pearson
Deviance
Chi-Squa
re
15.199
14.848
Likelihood Ra tio Tests
df
10
10
Sig.
.125
.138
Effect
Int ercept
EXTRA4
GENDER
-2 Log
Lik elihood
of
Reduc ed
Model
54.200
64.136
71.519
Chi-Squa
re
.000
9.936
17.319
df
Sig.
0
2
2
.
.007
.000
The chi-square stat istic is t he difference in -2 log-likelihoods
between the final model and a reduc ed model. The reduc ed
model is formed by omitting an effec t from the final model. The
null hy pothesis is t hat all parameters of that effect are 0.
Parameter Estimates
CANSEE
always sees
sometimes sees
Intercept
EXTRA4
[GENDER=.000]
[GENDER=1.000]
Intercept
EXTRA4
[GENDER=.000]
[GENDER=1.000]
B
-1.045
.422
.107
0a
2.270
-.253
-1.582
0a
Std. Error
.842
.247
.589
0
.663
.222
.507
0
Wald
1.541
2.901
.033
.
11.728
1.299
9.727
.
df
1
1
1
0
1
1
1
0
Sig.
.215
.089
.856
.
.001
.254
.002
.
Exp(B)
95% Confidence
Interval for Exp(B)
Lower
Upper
Bound
Bound
1.524
1.113
.
.938
.351
.
2.476
3.530
.
.776
.206
.
.502
7.609E-02
.
1.200
.556
.
a. This parameter is set to zero because it is redundant.
14
PTO
Logistic Regression Parameter Estimates (DV= cansee2)
Parameter Estimates
CANSEE2
always sees
never s ees
Intercept
EXTRA4
[GENDER=.000]
[GENDER=1.000]
Intercept
EXTRA4
[GENDER=.000]
[GENDER=1.000]
B
-3.315
.675
1.689
0a
-2.270
.253
1.582
0a
Std. Error
.743
.224
.504
0
.663
.222
.507
0
Wald
19.910
9.045
11.237
.
11.728
1.299
9.727
.
df
1
1
1
0
1
1
1
0
Sig.
.000
.003
.001
.
.001
.254
.002
.
Exp(B)
95% Confidence
Interval for Exp(B)
Lower
Upper
Bound
Bound
1.963
5.413
.
1.265
2.017
.
3.048
14.530
.
1.288
4.864
.
.833
1.800
.
1.991
13.143
.
a. This parameter is set to zero because it is redundant.
15
PTO
Printout for Question 3
Correlation Matrix
Correlation
ANGRY
ANXIOUS
FRIENDLY
HAPPY
HOSTILE
IMPULSIV
NERVOUS
STRESSED
WARM
ANGRY
1.000
-.009
-.018
-.113
.509
.524
.072
.120
-.251
ANXIOUS
-.009
1.000
.101
-.078
.038
.066
.379
.479
.035
FRIENDLY
-.018
.101
1.000
.448
-.048
-.019
-.134
.016
.497
HAPPY
-.113
-.078
.448
1.000
.086
-.031
-.061
-.132
.481
HOSTILE
.509
.038
-.048
.086
1.000
.550
.083
.079
-.247
IMPULSIV
.524
.066
-.019
-.031
.550
1.000
.209
.057
-.133
NERVOUS
.072
.379
-.134
-.061
.083
.209
1.000
.496
-.199
STRESS
ED
.120
.479
.016
-.132
.079
.057
.496
1.000
-.174
WARM
-.251
.035
.497
.481
-.247
-.133
-.199
-.174
1.000
KMO a nd Bartlett's Te st
Kaiser-Mey er-Olkin Measure of Sampling
Adequacy.
Bartlet t's Test of
Sphericity
Approx . Chi-Square
df
Sig.
.622
229.153
36
.000
16
PTO
Comm una litie s
ANGRY
ANXIOUS
FRIENDLY
HA PPY
HOSTILE
IMPULSIV
NE RVOUS
STRES SED
W ARM
Initial
.390
.303
.351
.378
.442
.428
.357
.387
.440
Ex trac tion Method: Unweighted Leas t Squares.
Total Vari ance Ex pla ined
Factor
1
2
3
4
5
6
7
8
9
Initial Eigenvalues
Rotation Sums of Squared Loadings
% of
Cumulativ
% of
Cumulativ
Total
Variance
e%
Total
Variance
e%
2.498
27.760
27.760
1.645
18.280
18.280
1.768
19.644
47.404
1.485
16.496
34.775
1.741
19.349
66.753
1.408
15.649
50.424
.733
8.147
74.899
.606
6.728
81.628
.560
6.219
87.846
.414
4.601
92.448
.382
4.247
96.695
.297
3.305
100.000
Ex trac tion Met hod: Unweighted Least Squares.
17
PTO
Rotated Factor Matrixa
Scree Plot
3.0
ANGRY
ANXIOUS
FRIENDLY
HAPPY
HOSTILE
IMPULSIV
NERVOUS
STRESSED
WARM
2.5
2.0
1.5
Eigenvalue
1.0
1
.686
-6.09E-03
-7.44E-04
3.086E-02
.754
.730
.117
5.555E-02
-.237
Factor
2
-.110
7.684E-02
.689
.638
-2.39E-02
-8.78E-03
-.136
-8.61E-02
.747
Factor Score Coefficient Matrix
3
3.933E-02
.637
4.677E-02
-9.51E-02
3.254E-02
9.477E-02
.609
.774
-9.39E-02
Extraction Method: Unweighted Leas t Squares.
Rotation Method: Varimax with Kaiser Normalization.
a. Rotation converged in 4 iterations.
.5
0.0
1
2
3
4
5
6
7
8
ANGRY
ANXIOUS
FRIENDLY
HAPPY
HOSTILE
IMPULSIV
NERVOUS
STRESSED
WARM
1
.290
-.032
.047
.046
.395
.355
.010
-.026
-.066
Factor
2
-.010
.044
.337
.257
.073
.033
-.012
.003
.472
3
-.037
.289
.043
-.011
-.035
.035
.248
.517
.006
Extraction Method: Unweighted Leas t Squares.
Rotation Method: Varimax with Kaiser Normalization.
9
Factor Number
18
PTO
Printout for Question 4
Between-Subjects Factors
TEAM
1
2
3
Value
Label
ars enal
man utd
chelsea
W ithin -Su bjects F acto rs
N
40
40
32
Measure
HA PPY
EXCITE D
HOPE FUL
CONFIDNT
PHASE
1
2
3
1
2
3
1
2
3
1
2
3
19
Dependent
Variable
HA PPY 1
HA PPY 2
HA PPY 3
EXCITE D1
EXCITE D2
EXCITE D3
HOPE FUL1
HOPE FUL2
HOPE FUL3
CONFID1
CONFID2
CONFID3
PTO
Multivariate Testsc
Effect
Between
Subjects
Intercept
TEAM
Within Subjects
PHASE
PHASE * TEAM
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
Pillai's Trace
Wilks' Lambda
Hotelling's Trace
Roy's Largest Root
Value
.992
.008
118.216
118.216
.126
.877
.138
.118
.376
.624
.603
.603
.359
.660
.487
.421
F
3132.722a
3132.722a
3132.722a
3132.722a
1.791
1.804a
1.816
3.158b
7.688a
7.688a
7.688a
7.688a
2.813
2.946a
3.076
5.414b
Hypothesi
s df
4.000
4.000
4.000
4.000
8.000
8.000
8.000
4.000
8.000
8.000
8.000
8.000
16.000
16.000
16.000
8.000
Error df
106.000
106.000
106.000
106.000
214.000
212.000
210.000
107.000
102.000
102.000
102.000
102.000
206.000
204.000
202.000
103.000
Sig.
.000
.000
.000
.000
.080
.078
.076
.017
.000
.000
.000
.000
.000
.000
.000
.000
a. Exact s tatis tic
b. The statistic is an upper bound on F that yields a lower bound on the s ignificance level.
c.
Design: Intercept+TEAM
Within Subjects Des ign: PHASE
20
PTO
17.5
24.0
17.0
16.5
23.5
Mean Mood Rating
Mean Mood Rating
24.5
23.0
TEAM
22.5
arsenal
22.0
16.0
15.5
TEAM
15.0
arsenal
14.5
man utd
21.5
man utd
14.0
chelsea
21.0
1
2
1
PHA SE
2
3
PHA SE
EXCITED
HAPPY
24.0
22.0
23.5
20.0
23.0
22.5
TEAM
22.0
arsenal
21.5
man utd
Mean Mood Rating
Mean Mood Rating
chelsea
13.5
3
18.0
TEAM
16.0
arsenal
14.0
chelsea
21.0
1
2
man utd
chelsea
12.0
3
1
PHA SE
2
3
PHA SE
HOPEFUL
CONFIDENT
21
PTO
Printout for Question 6
STUDY 1
a
Le vene's Test of Equa lity of Error Va riances
Dependent Variable: RTMEAN
F
2.441
df1
df2
3
Sig.
.088
25
Tests the null hypothes is that t he error variance of the
dependent variable is equal across groups.
a. Design: Int ercept+NOISE+EXTGP+ NOISE * EXTGP
Tests of Between-Subjects Effects
Dependent Variable: RTMEAN
Source
Corrected Model
Intercept
NOISE
EXTGP
NOISE * EXTGP
Error
Total
Corrected Total
Type III
Sum of
Squares
22855.907a
9494377
7717.001
786.979
12004.571
71964.987
9736880
94820.895
df
3
1
1
1
1
25
29
28
Mean
Square
7618.636
9494377
7717.001
786.979
12004.571
2878.599
F
2.647
3298.263
2.681
.273
4.170
Sig.
.071
.000
.114
.606
.052
a. R Squared = .241 (Adjusted R Squared = .150)
22
PTO
noise leve l * e xtra version group
Dependent Variable: RTMEAN
noise level
low
high
ex traversion group
ex travert
int rovert
ex travert
int rovert
Mean
618.375
566.905
544.542
575.028
St d. Error
18.969
20.279
18.969
21.904
95% Confidenc e
Int erval
Lower
Upper
Bound
Bound
579.308
657.442
525.140
608.670
505.474
583.609
529.917
620.139
23
PTO
Printout for Question 8
Syntax Commands with Notes
First transform the correlations for males (corrmale)
and females (corrfema) as described by Fisher
COMPUTE
EXECUTE
COMPUTE
EXECUTE
rfisherm = 0.5*LN(ABS((1 + corrmale)/(1 - corrmale))) .
.
rfisherf = 0.5*LN(ABS((1 + corrfema)/(1 - corrfema))) .
.
Compute the difference between the Fisher-transformed correlations
COMPUTE fisherz1 = rfisherm - rfisherf .
EXECUTE .
Calculate the standard error of the difference between Fisher-transformed correlations
as described by Fisher
COMPUTE fisherz2 = SQRT((1/(nmale-3)) + (1/(nfemale-3))) .
EXECUTE .
Fisher’s Z statistic is the difference in transformed correlation values
divided by the standard error
COMPUTE fisherz = fisherz1/fisherz2 .
EXECUTE .
Work out the one-tailed probability associated with the Z value
COMPUTE zprob = 1-CDF.NORMAL(fisherz,0,1) .
EXECUTE .
24
PTO
Printout for Question 9
Descriptive Statistics
GROUP = nonsmokers
De scri ptive Statisticsa
N
LUNGCAP
SMELLID
Valid N (lis twis e)
40
40
40
Minimum
12.365
2.632
Maximum
20.500
23.618
Mean
17.73728
12.09649
St d.
Deviation
2.57049
4.48954
Mean
15.98335
10.93400
St d.
Deviation
2.70604
3.88570
a. GROUP = nonsmokers
GROUP = smokers
De scri ptive Statisticsa
N
LUNGCAP
SMELLID
Valid N (lis twis e)
40
40
40
Minimum
9.878
2.396
Maximum
20.000
21.525
a. GROUP = smokers
25
PTO
ANOVA Results
Tests of Between-Subjects Effects
Dependent Variable: LUNGCAP
Source
Corrected Model
Intercept
GROUP
Error
Total
Corrected Total
Type III
Sum of
Squares
61.525 b
22741.623
61.525
543.271
23346.419
604.797
df
1
1
1
78
80
79
Mean
Square
61.525
22741.623
61.525
6.965
F
8.833
3265.122
8.833
Sig.
.004
.000
.004
Noncent.
Parameter
8.833
3265.122
8.833
Observed
a
Power
.835
1.000
.835
Sig.
.219
.000
.219
Noncent.
Parameter
1.533
601.798
1.533
Observed
a
Power
.231
1.000
.231
a. Computed using alpha = .05
b. R Squared = .102 (Adjus ted R Squared = .090)
Tests of Between-Subjects Effects
Dependent Variable: SMELLID
Source
Corrected Model
Intercept
GROUP
Error
Total
Corrected Total
Type III
Sum of
Squares
27.027 b
10608.070
27.027
1374.929
12010.026
1401.956
df
1
1
1
78
80
79
Mean
Square
27.027
10608.070
27.027
17.627
F
1.533
601.798
1.533
a. Computed using alpha = .05
b. R Squared = .019 (Adjus ted R Squared = .007)
26
PTO
27
PTO