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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