GOLDSMITHSUniversity of London

PSYCHOLOGY DEPARTMENT

MSc in RESEARCH METHODS IN PSYCHOLOGY 2003You are reminded that you may not present substantially the same material in any two pieces of work submitted for assessment, regardless of the form of assessment. For instance, you may not repeat substantially the same material in a formal written examination or in a dissertation if it has already formed part of an essay submitted for assessment. This does not prevent you from referring to the same texts, examples or case studies as appropriate, provided you do not merely duplicate the same material.

PS 71020A STATISTICAL METHODS3 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 is attached SPSS printouts for questions 1, 3, 5, 7 & 9

SECTION A

1. Most studies of schizotypal personality have identified 4 aspects of schizotypy: reality distortion (RD) schizotypy; cognitively disorganised (CD) schizotypy; anhedonic (Anh) schizotypy; and impulsive nonconformist (IN) schizotypy. Eysenck argued that measures of these facets of schizotypy were just reflections of his big three personality dimensions: Anh schizotypy being an inverse reflection of extraversion; CD schizotypy reflecting neuroticism; and with RD and IN schizotypy reflecting psychoticism. A researcher decided to test these ideas using factor analysis by taking measures of the 4 kinds of schizotypy and the 3 Eysenck dimensions from the same participants. The researcher carried out a principal components analysis (PCA) asking for 3 components to be retained in the solution.(i) How does a PCA differ from a factor analysis? (5 marks)

The researcher requested the following options during the analysis: a varimax rotation; a scree plot; exclusion of cases listwise; KMO measure of sampling adequacy and Bartlett’s test of sphericity; and an anti-image correlation matrix.(ii) Explain what each of these options is and what purpose they serve in

PCA. (15 marks)

The results of this analysis on 211 participants are shown in the printout.

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(iii) Go through the printout and comment on what is shown in each section of output with particular reference to the answers given in (ii) above (15 marks)

The researcher decided that two further trait anxiety-neuroticism scores (Anx1 and Anx2), which she had for each participant, should be added to the analysis. The results of the second analysis are also shown in the printout.

(iv)To what extent do the results from the second PCA support or reject Eysenck’s explanation of schizotypal personality structure? (10 marks)

(v) From the information given briefly consider whether the second analysis is likely to represent a “good” PCA and suggest further analyses that might be attempted. (5 marks)

2. An applied psychologist recruited a representative sample of 1000 primary and secondary schoolchildren and administered to each participant a road-crossing test using a virtual reality simulator. An important dependent variable was the “risky road use” index (1-100; 100 =maximum risk-taking). This was computer-scored during the session in the simulator. The researcher hypothesised that the personality traits of impulsiveness and anxiety, plus performance measures of psychomotor speed and distance estimation, would each make an independent contribution to the amount of risky road use observed. These hypotheses were expected to be true even after any contribution of the child’s sex, age, socio-economic status, and road use experience were accounted for. From the point where the researcher had gathered the DV and the 8 predictors from each child, explain step-by-step, in detail, how she would test her hypotheses using multiple regression. Describe the statistical outcomes which would confirm or reject her predictions.

3. Chickens have a high mortality rate between hatching and point of sale. A chicken farmer suspected that some chicken feeds and some breeds may affect chick survival quite dramatically. The farmer consulted his neighbour, a psychologist, and they carried out 2 studies of chicken survival. Both studies used two feeds: Food 1 and Food 2. In the first study 100 chicks from each of 3 breeds (Breeds 1 to 3) were quasi-randomly allocated at hatching to receive one of the two feeds (ensuring that each feed was given to 50 chicks of each breed). The farmer recorded how many chicks survived to the point of sale (18 weeks) in each cell of the design. The result is shown in the cross-tabulation printout. The variable survived has a value of 1 if the chick survived and 0 otherwise. The psychologist used multinomial logistic regression (and some chi-squared statistics) to analyse the data, producing the printout shown below. In the second study, 100 chicks from each of 3 further breeds (Breeds 4 to 6) were quasi-randomly allocated as before to receive one or other of the same two feeds. The results of study 2 are also shown below. Comment in detail about what the various parts of the printout show, what models the psychologist fitted to the data in each study, and end up with the recommendations that the psychologist would

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make to the farmer about the breeds, and about what foods he should use with which breeds.

SECTION B

4. Two researchers, A and B, consulted a statistician about the use of analysis of covariance (ANCOVA) to analyse their findings. Researcher A was carrying out a word memory study using a between-groups design with participants randomised to one of the two groups involved. Each group received a different type of study instructions and the researcher wanted to know if word study instructions affected subsequent memory performance. He had also taken a quick measure of IQ from all participants. He found that the groups differed significantly in mean level of IQ and that IQ was significantly correlated with memory performance across all participants combined. Researcher B was looking at the performance on a particular test of attention in a group of chronic schizophrenic patients and age-matched controls. She had also taken a quick measure of IQ from all the participants and found a significantly lower mean IQ in the chronic schizophrenic patients. IQ was also significantly correlated with attention performance across all participants combined. Both researchers wanted to control for the IQ difference in analysing the possible between-groups difference in memory, or attention, performance.

(i) What advice should the statistician have given to researchers A and B about whether ANCOVA was a suitable technique for achieving their desired aim, and what reasons should the statistician have given? Give some other (hypothetical) research example(s) that the statistician might have used to clarify his explanations further. (25 marks)

(ii) In fact the statistician was also able to advise correctly about another use that ANCOVA might have been put to in their studies. What is the logic behind this application of ANCOVA? How did he advise researchers A and B on whether to use ANCOVA for this second purpose? (10 marks)

(iii) Apart from general checks that need to be made before carrying out an analysis of variance, what additional specific test would the statistician have advised the researchers to employ before carrying out any ANCOVA he might have recommended? What diagram did he draw to illustrate the need for this test? (10 marks)

(iv)Briefly describe one other use to which ANCOVA can be put. (5 marks)

5. A researcher tested 3 groups of participants: non-smokers (NS); ex-smokers who have successfully quit and remained abstinent for over a year (Quit-1yr); and smokers who have quit smoking during the last week (Quit-1wk). The participants were coded with the SPSS variable subjtype (1=NS; 2=Quit-1yr; and 3 =Quit-1wk). Within each group participants were randomly assigned to receive nicotine or placebo tablets in a double-blind design. The drug group of a participant was coded after the experiment was completed using the SPSS variable druggrp (1=placebo; 2=nicotine). The dependent variable was performance on a reaction time (RT) task used to

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measure concentration. The mean RTs in each cell of the design and the results of a two-way between-subjects ANOVA are shown in the printout below.

The researcher believed in a withdrawal model of the effects of nicotine on cognitive performance. This model predicts a particular pattern of mean RTs across the cells of the design. Specifically, RT task performance is predicted to be slower for the Quit-1wk participants receiving placebo relative to the other groups receiving placebo (and the other two groups should not differ from one another under placebo); the RT under nicotine should be faster than the RT under placebo for the Quit-1wk group but the same comparison for the other 2 groups should not be significant.

(i) Do the pattern of means and the overall ANOVA lend any support to this model? (5 marks)

The researcher decided to test the model using planned between-subjects t-tests each based on pairs of cell from the overall experiment (data from the other 4 cells were not used). The researcher correctly deemed that 9 such tests were needed and the results are shown in the table below.

Comparison Restricted to participants in the following groups/conditions

T-value df p-value

Nicotine (N) vs placebo (P)

Quit-1wk 3.50 20 0.002

N vs P Quit-1yr 0.28 23 0.78N vs P NS 0.29 21 0.78Quit-1yr vs NS

Nicotine 0.77 21 0.45

Quit-1yr vs NS

Placebo 0.70 23 0.49

Quit-1wk vs NS

Nicotine 0.22 21 0.83

Quit-1wk vs NS

Placebo 3.04 20 0.006

Quit-1yr vs Quit-1wk

Nicotine 0.62 20 0.54

Quit-1yr vs Quit-1wk

Placebo 2.74 23 0.01

(ii) Based on the table above, do the various comparisons between mean RTs show the pattern of significant and nonsignificant effects predicted by the withdrawal model. Explain your reasoning carefully. (10 marks)

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A statistician was consulted and she suggested that the use of t-tests was probably not the most powerful way to check if the pattern of mean RTs was as predicted. She advised the researcher to use specific planned contrasts within the overall ANOVA and prepared appropriate syntax commands using SPSS (the GLM procedure and the LMATRIX option). The results of these contrasts are shown in the table below.

(iii) What is name for the effects being studied in the first 3 rows of the table below? (5 marks)

(iv)Do the results of the contrast analyses reveal the pattern of significant and nonsignificant differences predicted by the withdrawal model? Explain your reasoning carefully. (10 marks)

(v) What are the features of the contrast analyses which render them more powerful than the t-test approach originally used by the researcher? Are there any particular assumptions upon which the use of these contrasts critically depends? (10 marks)

Contrast Restricted to participants in the following groups/conditions

T-value df p-value

Nicotine (N) vs. placebo (P)

Quit-1wk 3.51 64 <0.001

N vs. P Quit-1yr 0.26 64 0.80N vs. P NS 0.31 64 0.76Quit-1wk vs. rest

Nicotine 0.19 64 0.85

Quit-1wk vs. rest

Placebo 3.61 64 <0.001

Quit-1yr vs. NS Nicotine 0.74 64 0.46Quit-1yr vs. NS Placebo 0.71 64 0.48

The syntax commands which the statistician provided are shown in the printout below.

(vi)Explain the choice and ordering of the numbers (coefficients) which appear in the LMATRIX lines of the syntax. (10 marks)

6. Two clinical psychologists (Janet and John) were recruited by a university to conduct a preliminary trial of anxiety-management counselling for high trait

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anxious students prior to their exams. The psychologists were first invited to carry out a brief intervention study to see if such counselling was likely to be beneficial. Two weeks before the first exam the anxious students were randomly allocated either to receive a series of general talks on revision skills (control group) or the 1-week anxiety-management training sessions. The following week, each student in the trial completed state anxiety checklists while thinking about their forthcoming exams. Two different well-known and reliable checklists (A and B) were used and each student completed both checklists on 3 occasions: 7 days before the exam, 4 days before the exam, and the day before the exam. For each participant, the psychologist also calculated two mean state anxiety scores (by averaging the 3 scores obtained with A and B checklists respectively).

Several different research questions were potentially of interest to the psychologists. The first research question was whether the treatment had a beneficial effect on state anxiety in the 1-week period prior to an exam. This question was analysed using the 2 averaged state anxiety measures from each subject.

(i) The psychologist called John was not interested in the differences in the results from the two anxiety measures (A and B). Obviously John could average the two state anxiety measures into a single measure and carry out an appropriate analysis of variance. Other than doing this, there are two other simple ways John could use analysis of variance to address the first research question. Explain in detail the pros and cons of these two analytic approaches, including any differences in their underlying assumptions? What advantages and disadvantages might each have over the simple averaging method? (20 marks)

(ii) The psychologist called Janet was more interested in whether the effects of the anxiety management training were different for the physical aspects of anxiety (more strongly measured by checklist A than by B) or for the cognitive aspects of anxiety (more strongly measured by checklist B than by A). In this case, explain what type of analysis of variance approach she should use and say which of the statistical results would be most critical in answering her question? (10 marks)

The second research question concerned the effects of time until the exam, and was addressed using the A and B anxiety measures taken at each time point.

(iii) If John were interested in whether the effect of the anxiety management training on state anxiety varied as a function of the time before the exam, how might he extend each of the analytic methods, described in answer to (i), to investigate this secondary question? (10 marks)

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(iv) Janet was also interested in the effects of the anxiety management training as a function of when the anxiety measures were taken, but she had a more specific prediction than John. She felt that the effect of anxiety management would be greatest 7 days before the exam (and shortly after completing training), somewhat less 4 days before the exam, and least the day before the exam. What is the best way for Janet to extend the analysis, described in answer to (iii), to address her secondary research question? (10 marks)

SECTION C

7. A researcher measured a particular aspect of cognitive performance in two age groups (45-60 years old; 70-85 years old). She wanted to compare the mean scores of the two groups on the cognitive test concerned, and also explore the influence of gender on any age effect. She also wanted to correlate cognitive test performance in the older group with a number of other variables which she had recorded. After conducting the study, she found that her data strongly violated the assumptions and requirements for employing the appropriate parametric statistical techniques.

(i) Discuss in detail the various resampling methods that she might use as an alternative analytic approach in pursuit or her research questions. Explain the principles behind these methods. (20 marks)

(ii) Discuss some other nonparametric alternative analytic approaches that might be tried with your data. (10 marks)

In another related study, the researcher wanted to investigate executive (or “frontal lobe” function) in a group of 100 healthy elderly participants. In younger participants there is evidence for a number of separate executive function factors. She collected scores on 9 well-known tests of executive function. These tests sample the range of processes usually considered to be “executive” functions. She wanted to conduct a principal components analysis or factor analysis to see how many executive function factors were present in her data. She tried maximum likelihood factor analysis, which tests statistically how many factors are needed, but warning messages indicated the analysis was likely to be untrustworthy. She used a scree plot with PCA but found this to be unhelpful (see printout). So, she decided to use a resampling approach to the problem. She randomised the data 10000 times and generated empirical sampling distributions for the eigenvalues of the 9 principal components (these were stored as variables eigval1-eigval9).

(iii) Explain at least one way that she could have done the randomisation. (10 marks)

(iv) By looking at the results of the resampling analysis explain what conclusion the researcher should come to regarding the number of executive function factors in her data. (10 marks)

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8. (i) Using clearly defined equations, or precise verbal descriptions, outline the basic features of classical test theory. (15 marks)

(ii) Using the ideas and assumptions of the theory, define the reliability of a measure and explain why a test-retest correlation estimates the reliability. (10 marks)

(iii) A researcher has two measures, A and B, which he feels are measuring the same construct. Using the ideas and assumptions of the theory, explain why the average (or sum) of two measures is a more reliable measure of the construct than either A or B alone. (10 marks)

In a study, the researcher measured mean reaction times (RTs) for each participant to various stimuli in 3 separate conditions. In condition 1 the participant pressed a key marked A whenever the letter A appeared on the screen. The researcher argued that a participant’s mean RT in condition 1 (RT1) is largely determined by basic psychomotor speed. In condition two the participant pressed one of two keys (marked A and B) whenever the corresponding letter appeared on the screen. The researcher argued that RT2 measures basic psychomotor speed plus the time taken for a simple response choice. In condition 3 the participant had to press key B whenever letter A appeared, and vice versa. The researcher argued that RT3 measures basic psychomotor speed plus the time taken for a simple response choice plus the added time needed to inhibit a “natural” but incorrect response. The researcher wanted to investigate the hypothesis that individual differences in the RT costs of response inhibition will be inversely related to individual differences in the time taken to make a simple response choice. He proposed to do this by correlating two RT difference measures with one another: i.e., (RT3-RT2) with (RT2 - RT1).

(iv) Assume the researcher’s characterisation of the task processes in each condition is correct. However, explain in detail using classical test theory why the proposed correlation is unlikely to answer the researcher’s research question definitively. (10 marks)

(v) Under some conditions (or assumptions) the proposed correlation would be more likely to be informative. Briefly note what these conditions or assumptions are. (5 marks)

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9. (i) Explain, using a suitably annotated table or graph, the relationship between Type II error rate and power. Also explain the influence of Type I error rate on power. (10 marks)

A researcher wanted to estimate the sample size in each group, m, that would be required to compare job satisfaction in two occupational groups. A statistician had given him a set of simple syntax commands for SPSS which would estimate (approximately) the required sample size for a given power (see Syntax 1 in the printout). The researcher was told that, to use the syntax commands, he would need to create an SPSS dataset in which the required type I error rate should be entered as a variable alpha, along with the desired power (as a variable power). The researcher was told that he would also need to estimate an effect size of interest and enter it into his dataset as a variable called d. The researcher knew that he was interested in a mean difference between his two groups of 2.5 units or more on his job satisfaction scale. He also knew that the standard deviation of job satisfaction scores in each group would be around 7.5 units.

(ii) How would the researcher calculate the effect size to be entered as variable d? (5 marks)

(iii) What conventional values of power could the researcher use? (5 marks)(iv) The syntax commands assumed that the researcher was going to use

a two-tailed between-groups t-test to analyse his data. What change would need to be made to the syntax commands if a one-tailed t-test was anticipated? (5 marks)

(v) If the researcher anticipated obtaining twice as many participants from one group as the other would this increase or decrease (or leave unaffected) the total number of subjects required? (5 marks)

Another researcher had already completed her study of job stress levels in two occupational groups. She had analysed her findings using a two-tailed between-groups t-test. She wanted to measure the actual power of her study. She was given another set of syntax commands (see Syntax 2) by the same statistician. She was told that, in order to use the syntax, she needed to create an SPSS dataset in which the number of participants tested in each group was entered under a variable called mtested. She was also told that she would need to enter the type I error rate that she had used as a variable alpha, along with the degrees of freedom for her t-test (as a variable called tdf). Finally, she needed to enter an effect size in the variable called d. the syntax would give her the level of power in a variable called power.

(vi) The first line of the syntax calculates a parameter called delta. What is this parameter known as? (5 marks)

(vii) The final line of the syntax uses the NCDF.T function from SPSS. This is the cumulative distribution function for a non-central t-distribution. Explain, with the aid of a suitably labelled diagram, how the non-central t-distribution is related to the standard t-distribution,

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indicating which feature of the diagram corresponds to delta. (10 marks)

(viii) The middle line of the syntax computes a critical t-value (as a variable called tcrit). Mark on the diagram approximately where tcrit would appear for a standard value of alpha (=0.05). Shade and label areas under the distributions corresponding to alpha and power. (5 marks)

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Printout for Question 1

KMO and Bartlett's Test

.673

521.25521

.000

Kaiser-Meyer-Olkin Measure of SamplingAdequacy.

Approx. Chi-SquaredfSig.

Bartlett's Test ofSphericity

Total Variance Explained

2.696 38.510 38.510 2.471 35.305 35.3051.781 25.436 63.946 1.663 23.760 59.0651.069 15.269 79.215 1.411 20.150 79.215.457 6.530 85.745.420 6.003 91.748.331 4.728 96.476.247 3.524 100.000

Component1234567

Total% of

VarianceCumulativ

e % Total% of

VarianceCumulativ

e %

Initial Eigenvalues Rotation Sums of Squared Loadings

Extraction Method: Principal Component Analysis.

Scree Plot

Component Number

7654321

Eige

nval

ue

3.0

2.5

2.0

1.5

1.0

.5

0.0

Rotated Component Matrixa

.892 .119 2.147E-02-7.73E-03 -.881 .176

.851 .259 -.1994.340E-03 1.532E-02 .936

.655 -.132 .529

.714 -.144 .410

.111 .876 .124

COGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypy

1 2 3Component

Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Rotation converged in 6 iterations.a.

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Anti-image Matrices

.389 5.053E-02 -.223 -8.06E-03 -7.348E-02 -.144 2.913E-025.053E-02 .572 7.523E-02 1.035E-02 -.112 -.113 .329

-.223 7.523E-02 .415 .162 -9.283E-02 -3.35E-02 -6.139E-02-8.059E-03 1.035E-02 .162 .729 -.217 -.111 -6.673E-02-7.348E-02 -.112 -9.28E-02 -.217 .516 -.134 -2.214E-02

-.144 -.113 -3.35E-02 -.111 -.134 .528 -3.682E-02

2.913E-02 .329 -6.14E-02 -6.67E-02 -2.214E-02 -3.68E-02 .636.712a .107 -.557 -1.51E-02 -.164 -.319 5.862E-02.107 .530a .154 1.602E-02 -.206 -.206 .546

-.557 .154 .668a .295 -.201 -7.17E-02 -.120-1.515E-02 1.602E-02 .295 .529a -.353 -.178 -9.805E-02

-.164 -.206 -.201 -.353 .755a -.258 -3.867E-02

-.319 -.206 -7.17E-02 -.178 -.258 .787a

-6.357E-02

5.862E-02 .546 -.120 -9.80E-02 -3.867E-02 -6.36E-02 .566a

COGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypyCOGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypy

Anti-image Covariance

Anti-image Correlation

COGDIS1 CD

schizotypy

EPQE extraversi

on

EPQN neuroticis

m

EPQP psychotici

smIN1 IN

Schizotypy

UNUEXPER RD

schizotypyIA1 Anh

schizotypy

Measures of Sampling Adequacy(MSA)a.

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Second AnalysisKMO and Bartlett's Test

.762

748.64936

.000

Kaiser-Meyer-Olkin Measure of SamplingAdequacy.

Approx. Chi-SquaredfSig.

Bartlett's Test ofSphericity

Total Variance Explained

3.343 37.147 37.147 2.876 31.952 31.9522.229 24.768 61.915 2.330 25.885 57.8371.220 13.554 75.469 1.587 17.632 75.469.554 6.159 81.628.448 4.981 86.609.358 3.978 90.587.331 3.680 94.267.291 3.230 97.497.225 2.503 100.000

Component123456789

Total% of

VarianceCumulativ

e % Total% of

VarianceCumulativ

e %

Initial Eigenvalues Rotation Sums of Squared Loadings

Extraction Method: Principal Component Analysis.

Scree Plot

Component Number

987654321

Eige

nval

ue

4.0

3.5

3.0

2.5

2.0

1.5

1.0

.5

0.0

Rotated Component Matrixa

.692 .555 4.855E-02-.258 .254 -.816.803 .340 .179

-.450 .710 .106.180 .827 -9.23E-02

.230 .799 -.119

7.746E-02 9.411E-02 .888.853 -2.52E-03 5.976E-02.815 -6.93E-02 .247

COGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypyBIS anx1HATOTAL anx2

1 2 3Component

Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Rotation converged in 5 iterations.a.

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Anti-image Matrices

.351 4.651E-03 -.126 -6.44E-02 -5.435E-02 -.138 2.034E-02 -9.23E-02 -5.891E-024.651E-03 .523 5.713E-03 9.527E-03 -6.618E-02 -9.53E-02 .312 -5.29E-03 .146

-.126 5.713E-03 .337 7.921E-02 -8.729E-02 -4.31E-02 -7.649E-02 -6.03E-02 -.124-6.445E-02 9.527E-03 7.921E-02 .644 -.192 -7.41E-02 -5.315E-02 .161 1.768E-02-5.435E-02 -6.62E-02 -8.73E-02 -.192 .500 -.140 -3.713E-03 -5.73E-02 6.999E-02

-.138 -9.53E-02 -4.31E-02 -7.41E-02 -.140 .507 -3.307E-02 4.355E-02 1.823E-02

2.034E-02 .312 -7.65E-02 -5.32E-02 -3.713E-03 -3.31E-02 .652 -1.47E-02 7.269E-02-9.228E-02 -5.29E-03 -6.03E-02 .161 -5.731E-02 4.355E-02 -1.473E-02 .472 -.132-5.891E-02 .146 -.124 1.768E-02 6.999E-02 1.823E-02 7.269E-02 -.132 .419

.821a 1.086E-02 -.366 -.136 -.130 -.327 4.255E-02 -.227 -.1541.086E-02 .625a 1.361E-02 1.642E-02 -.130 -.185 .535 -1.07E-02 .312

-.366 1.361E-02 .818a .170 -.213 -.104 -.163 -.151 -.329-.136 1.642E-02 .170 .640a -.338 -.130 -8.207E-02 .291 3.404E-02-.130 -.130 -.213 -.338 .761a -.279 -6.508E-03 -.118 .153

-.327 -.185 -.104 -.130 -.279 .792a

-5.751E-02 8.895E-02 3.954E-02

4.255E-02 .535 -.163 -8.21E-02 -6.508E-03 -5.75E-02 .566a -2.66E-02 .139-.227 -1.07E-02 -.151 .291 -.118 8.895E-02 -2.656E-02 .824a -.296-.154 .312 -.329 3.404E-02 .153 3.954E-02 .139 -.296 .784a

COGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypyBIS anx1HATOTAL anx2COGDIS1 CD schizotypyEPQE extraversionEPQN neuroticismEPQP psychoticismIN1 IN SchizotypyUNUEXPER RDschizotypyIA1 Anh schizotypyBIS anx1HATOTAL anx2

Anti-image Covariance

Anti-image Correlation

COGDIS1 CD

schizotypy

EPQE extraversi

on

EPQN neuroticis

m

EPQP psychotici

smIN1 IN

Schizotypy

UNUEXPER RD

schizotypyIA1 Anh

schizotypy BIS anx1HATOTAL

anx2

Measures of Sampling Adequacy(MSA)a.

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Printout for Question 3

Study 1: Logistic Regression

BREED * SURVIVED * FOOD Crosstabulation

Count

23 27 5030 20 5017 33 5070 80 15031 19 5013 37 501 49 50

45 105 150

1.002.003.00

BREED

Total1.002.003.00

BREED

Total

FOOD1.00

2.00

.00 1.00SURVIVED

Total

Model Fitting Information

88.86723.378 65.488 5 .000

ModelIntercept OnlyFinal

-2 LogLikelihood

Chi-Square df Sig.

Goodness-of-Fit

.000 0 .

.000 0 .PearsonDeviance

Chi-Square df Sig.

Likelihood Ratio Tests

23.378 .000 0 .23.378 .000 0 .23.378 .000 0 .48.562 25.184 2 .000

EffectInterceptBREEDFOODBREED * FOOD

-2 LogLikelihood

ofReduced

ModelChi-Squa

re df Sig.

The chi-square statistic is the difference in -2 log-likelihoods betweenthe final model and a reduced model. The reduced model is formed byomitting an effect from the final model. The null hypothesis is that allparameters of that effect are 0.

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Study 1: Food -by- Survived Crosstabs and Chi-Square Analyses for Each Breed Separately

FOOD * SURVIVED Crosstabulation for Breed 1

Count

23 27 5031 19 5054 46 100

1.002.00

FOOD

Total

.00 1.00SURVIVED

Total

FOOD * SURVIVED Crosstabulation for Breed 2

Count

30 20 5013 37 5043 57 100

1.002.00

FOOD

Total

.00 1.00SURVIVED

Total

FOOD * SURVIVED Crosstabulation for Breed 3

Count

17 33 501 49 50

18 82 100

1.002.00

FOOD

Total

.00 1.00SURVIVED

Total

Chi-Square Tests for Breed 1

2.576b 1 .1081.973 1 .1602.588 1 .108

100

Pearson Chi-SquareContinuity Correctiona

Likelihood RatioN of Valid Cases

Value df

Asymp.Sig.

(2-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than5. The minimum expected count is 23.00.

b.

Chi-Square Tests for Breed 2

11.791b 1 .00110.445 1 .00112.056 1 .001

100

Pearson Chi-SquareContinuity Correctiona

Likelihood RatioN of Valid Cases

Value df

Asymp.Sig.

(2-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5.The minimum expected count is 21.50.

b.

Chi-Square Tests for Breed 3

17.344b 1 .00015.244 1 .00020.371 1 .000

100

Pearson Chi-SquareContinuity Correctiona

Likelihood RatioN of Valid Cases

Value df

Asymp.Sig.

(2-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5.The minimum expected count is 9.00.

b.

16P.T.O.

Study 2: Logistic Regression -First Step

Model Fitting Information

46.34125.652 20.689 5 .001

ModelIntercept OnlyFinal

-2 LogLikelihood

Chi-Square df Sig.

Goodness-of-Fit

.000 0 .

.000 0 .PearsonDeviance

Chi-Square df Sig.

Likelihood Ratio Tests

25.652 .000 0 .25.652 .000 0 .25.652 .000 0 .25.909 .257 2 .880

EffectInterceptBREEDFOODBREED * FOOD

-2 LogLikelihood

ofReduced

ModelChi-Squa

re df Sig.

The chi-square statistic is the difference in -2 log-likelihoods betweenthe final model and a reduced model. The reduced model is formed byomitting an effect from the final model. The null hypothesis is that allparameters of that effect are 0.

17P.T.O.

Study 2: Logistic Regression - Second Step

Model Fitting Information

46.34125.909 20.432 3 .000

ModelIntercept OnlyFinal

-2 LogLikelihood

Chi-Square df Sig.

Goodness-of-Fit

.256 2 .880

.257 2 .880PearsonDeviance

Chi-Square df Sig.

Likelihood Ratio Tests

25.909 .000 0 .34.776 8.867 2 .01237.825 11.916 1 .001

EffectInterceptBREEDFOOD

-2 LogLikelihood

ofReduced

ModelChi-Squa

re df Sig.

The chi-square statistic is the difference in -2 log-likelihoodsbetween the final model and a reduced model. The reducedmodel is formed by omitting an effect from the final model. Thenull hypothesis is that all parameters of that effect are 0.

18P.T.O.

Parameter Estimates

-.247 .236 1.094 1 .296-.678 .294 5.313 1 .021 .508 .285 .904-.812 .297 7.494 1 .006 .444 .248 .794

0a 0 . 0 . . . ..829 .243 11.625 1 .001 2.290 1.422 3.687

0a 0 . 0 . . . .

Intercept[BREED=4.00][BREED=5.00][BREED=6.00][FOOD=1.00][FOOD=2.00]

SURVIVED.00

B Std. Error Wald df Sig. Exp(B)LowerBound

UpperBound

95% ConfidenceInterval for Exp(B)

This parameter is set to zero because it is redundant.a.

Study 2: Crosstabs

FOOD * SURVIVED Crosstabulation

Count

78 72 15049 101 150

127 173 300

1.002.00

FOOD

Total

.00 1.00SURVIVED

Total

BREED * SURVIVED Crosstabulation

Count

38 62 10035 65 10054 46 100

127 173 300

4.005.006.00

BREED

Total

.00 1.00SURVIVED

Total

19P.T.O.

Printout for Question 5 RT1

493.05212 11 60.58625485.91013 12 58.59337489.32587 23 58.29576509.08330 14 54.35774503.27363 11 49.24684506.52705 25 51.18282574.14582 11 64.38248490.64980 11 45.82952532.39781 22 69.28142524.06510 36 67.16312493.06116 34 50.70588509.00604 70 61.33005

DRUGGRP1.0002.000Total1.0002.000Total1.0002.000Total1.0002.000Total

SUBJTYPE1.000

2.000

3.000

Total

Mean NStd.

Deviation

Tests of Between-Subjects Effects

Dependent Variable: RT1

59943.961a 5 11988.792 3.844 .0041.8E+07 1 1.8E+07 5777.169 .000

17944.027 1 17944.027 5.754 .01920986.230 2 10493.115 3.365 .04122332.004 2 11166.002 3.580 .034199590.9 64 3118.6081.8E+07 70

259534.8 69

SourceCorrected ModelInterceptDRUGGRPSUBJTYPEDRUGGRP * SUBJTYPEErrorTotalCorrected Total

Type IIISum ofSquares df

MeanSquare F Sig.

R Squared = .231 (Adjusted R Squared = .171)a.

Syntax CommandsGLM rt1 BY subjtype druggrp /METHOD = SSTYPE(3) /LMATRIX = "drug simple main effect for Quit-1wk subjtype" druggrp 1 -1 subjtype*druggrp 0 0 0 0 1 -1 /LMATRIX = "drug simple main effect for Quit-1yr subjtype" druggrp 1 -1 subjtype*druggrp 0 0 1 -1 0 0 /LMATRIX = "drug simple main effect for NS subjtype" druggrp 1 -1 subjtype*druggrp 1 -1 0 0 0 0 /DESIGN = subjtype druggrp subjtype*druggrp.GLM rt1 BY subjtype druggrp /METHOD = SSTYPE(3) /LMATRIX = "Quit-1wk vs others for NICO" subjtype 0.5 0.5 -1 subjtype*druggrp 0 0.5 0 0.5 0 -1 /LMATRIX = "Quit-1wk vs others for PLAC" subjtype 0.5 0.5 -1 subjtype*druggrp 0.5 0 0.5 0 -1 0 /LMATRIX = "Quit-1yr vs NS for NICO" subjtype 1 -1 0 subjtype*druggrp 0 1 0 -1 0 0 /LMATRIX = "Quit-1yr vs NS for PLAC" subjtype 1 -1 0 subjtype*druggrp 1 0 -1 0 0 0 /DESIGN = subjtype druggrp subjtype*druggrp.

P.T.O.

18

Printout for Question 7 Principal Components Analysis on Actual Data

Scree Plot

Component Number

987654321

Eige

nval

ue

3.0

2.5

2.0

1.5

1.0

.5

0.0

Total Variance Explained

2.620 29.106 29.106 2.620 29.106 29.1062.022 22.469 51.575 2.022 22.469 51.5751.334 14.827 66.402 1.334 14.827 66.4021.023 11.372 77.774 1.023 11.372 77.774.636 7.071 84.845 .636 7.071 84.845.493 5.476 90.321 .493 5.476 90.321.368 4.089 94.410 .368 4.089 94.410.314 3.485 97.895 .314 3.485 97.895.189 2.105 100.000 .189 2.105 100.000

Component123456789

Total% of

VarianceCumulativ

e % Total% of

VarianceCumulativ

e %

Initial EigenvaluesExtraction Sums of Squared

Loadings

Extraction Method: Principal Component Analysis.

P.T.O.

19

Resampled Eigenvectors: Descriptive Statistics Descriptive Statistics

10000 1.26 1.90 1.4873 8.694E-0210000 1.12 1.57 1.3155 6.021E-0210000 1.02 1.40 1.1872 4.961E-0210000 .92 1.25 1.0785 4.386E-0210000 .82 1.12 .9793 4.184E-0210000 .72 1.03 .8843 4.271E-0210000 .62 .94 .7901 4.446E-0210000 .48 .85 .6936 4.725E-0210000 .34 .76 .5841 5.471E-0210000

EIGVAL1EIGVAL2EIGVAL3EIGVAL4EIGVAL5EIGVAL6EIGVAL7EIGVAL8EIGVAL9Valid N (listwise)

N Minimum Maximum MeanStd.

Deviation

Percentiles

1.3565 1.3809 1.4260 1.4798 1.5401 1.6039 1.64231.2227 1.2410 1.2737 1.3120 1.3545 1.3955 1.42031.1083 1.1246 1.1526 1.1859 1.2189 1.2514 1.27151.0081 1.0228 1.0478 1.0779 1.1080 1.1362 1.1527.9104 .9251 .9517 .9795 1.0076 1.0326 1.0477.8124 .8288 .8563 .8847 .9132 .9390 .9530.7162 .7332 .7598 .7907 .8211 .8475 .8623.6142 .6314 .6626 .6948 .7259 .7539 .7704.4912 .5131 .5478 .5859 .6223 .6525 .6709

1.4261 1.4798 1.54011.2737 1.3120 1.35451.1526 1.1859 1.21891.0478 1.0779 1.1080.9517 .9795 1.0076.8563 .8847 .9132.7598 .7907 .8211.6627 .6948 .7258.5478 .5859 .6223

EIGVAL1EIGVAL2EIGVAL3EIGVAL4EIGVAL5EIGVAL6EIGVAL7EIGVAL8EIGVAL9EIGVAL1EIGVAL2EIGVAL3EIGVAL4EIGVAL5EIGVAL6EIGVAL7EIGVAL8EIGVAL9

WeightedAverage(Definition 1)

Tukey's Hinges

5 10 25 50 75 90 95Percentiles

P.T.O.

20

Printout for Question 9 Syntax 1COMPUTE zalpha=IDF.NORMAL(1-alpha/2,0,1) .EXECUTE .

COMPUTE zpower=IDF.NORMAL(power,0,1) .EXECUTE .

COMPUTE m=(2*(zalpha + zpower)**2)/(d**2) .EXECUTE.

Syntax 2COMPUTE delta= d *sqrt(mtested/2).EXECUTE .

COMPUTE tcrit=IDF.T(1-alpha/2,tdf) .EXECUTE .

COMPUTE power = 1-NCDF.T(tcrit,tdf,delta) .EXECUTE .

21