Speculations which are Untrue ATW Scores - Research Paper Example

QUESTIONS ON RESEARCH METHODS IN PSYCHOLOGY RESEARCH METHODS IN PSYCHOLOGY 9th May, 2013 Part A: Research Scenarios (34 points) Scenario 1 The Dr. Anne Sullivan wondered whether the gender of an examiner influences responses by male subjects on the Attitudes Towards Women Scale (ATW). (This scale measures whether an individual has liberal or conservative attitudes toward women’s roles.) She asked Dr. Tatshuiko Toyota to administer half of the questionnaires so that she could compare the ATW scores for males tested by a male versus a female examiner. Dr. Sullivan found that her respondents had much more liberal ATW scores than Dr. Toyota’s. She concluded that men ‘act’ more liberal to gain approval from women, whereas they reveal their true ‘macho’ selves to other men. Ho (2mks) Ho the null hypothesis is untrue and therefore rejected, because the research indicate a significant difference in the ATW scores submitted by the male respondents. H1 (2mks) H1 the alternative hypothesis is true and accepted, because of the significant differences in the ATW scores given by the male respondents. This is because the female examiner (Dr. Sullivan) found that, her respondents had much more liberal ATW scores than Dr. Toyota’s. This gives a conclusion that men acted more liberal to gain approval from women, whereas they reveal their true ‘macho’ selves to other men. IV (2mks) These are the variables which are varied or manipulated by a reseracher. For our experiment they are ATW scores, because the Attitudes Towards Women roles Scale varies with the gender of an examiner DV (2mks) These are variables with a pressumed effect, and for instance in this particular experiment its the gender of an examiner Describe the possible confounds in the experimental design: (2mks) The confounds are the valuables which vary with the independent variables (IV) in a statistical experiment and may share attributes to the variable changes. For instance in our experiment they can vary the ATW scores given, and for an effective study the effects of confounds must be reduced. They include; - Questionnaire type, the type of questions in the half that was conducted by each examiner may have influenced the outcome of the ATW scores given by the male respondents. This is because there is a significant difference indicated by the two halves of the questionnaire as a dministered by the two examiners, will not exactly be attributed to gender of examiner or the type of questions stipulated in each half. - The level of liberality or conservativeness in ATW scores given by the male respondents. The liberal or conservative attitude levels usually will have significant changes with changes in examiner gender. It is not easy to determine if the ATW scores were influenced by the gender of the examiner for each half, or is it because of individual understanding of liberal or conservative female roles. Scenario 2 Tom Rogers wanted to test a new ‘singalong’ method to teach math to 4th graders. (e.g., multiplying to the tune of Waltzing Matilda.) He used the singalong method in his first period class. His sixth period students continued solving math problems with the old method. At the end of the term, Mr. Rogers found that the first period class scored significantly lower than the sixth period class on a mathematics achievement test. He concluded that his singalong method was a total failure. H1 (2mks) the alternative hypothesis is true and accepted, because the math scores for the first class was relatively low as compared to the sixth class which used the old method in solving the math problem. This was indicated by the performance in math of the first class, and was low as compared to that one of the sixth class Ho (2mks) The null hypothesis was void, because of the low math score for the class taught using the new math method. IV (2mks) Independent variables were math score and this was found to be low in the first class which was taught using a new formular and high in the sixth class which was taught using the old formular DV (2mks) The dependent valuables for the study was the method used in the solving of math problem, that is the new Singalong method for the first class and the old method for the sixth class. Briefly describe the two main confounds in the experiment (5mks) Confounds for this experiment includes; - The class or group of students. This is because the significant difference in math score between the two classes can not be identifiable to be due to the type of students in each particular class in terms of academic abilities or due to the math solving. - Type of questions subjected to the students in each group. This is because it is not easy to determine if the math score for each class were due to the method used to solve the questions or simplicity of the submitted math questions in the achievement test. Devise a solution to control for the confounds in this experiment (5mks) The examiner should have used one of the class teach them the new method and find performance in achievement test to allow similarity of students. This will give a constant in terms of level of understanding and performance as well as hardworking levels. Secondly, Rogers should use similar math questions in the test achievement and allow the students to work out them using both methods. The two classes should have attempted similar math questions to enable the examiner decide on the effeciency of the new method of solving math problems. Part B: Research Analyses (t-tests) (81 points) Part B, Section 1: Background One of the basic tenets of child psychology is that human development across the lifespan may be conceived as achieving a series of developmental tasks that individuals have to master. An important milestone in the cognitive development of infants is their ability to form attachment relationships to primary social caregivers. One way that this attachment relationship is demonstrated is by observing the number of eye contacts made by babies with their mothers. A child psychologist arranged for a group of babies born on the same day to be observed at age 6 months and later at 9 months. The purpose of the study was to examine whether the amount of eye contact with their mother changes between these ages. The data for the children in the study are provided below. Part B, Section 2: Assignment Task 1. Analyse the data in Table 1. (a) Conduct the appropriate t-test manually to compare 6-months and 9-months age groups. Make sure you attach a copy of your calculations. (5 points) Degree of Freedom is given by N-1, and this is equal to (8-1) is equals to 7 Case. At 6 month At 9 month d d2 1. 5 9 -4 16 2. 7 8 -1 1 3. 7 5 2 4 4. 6 10 -4 16 5. 5 7 -2 4 6. 9 11 -2 4 7. 10 9 1 1 8. 9 11 -2 4 N-8 Sum d is -12 Sum d2 is 50 (b) Conduct the appropriate t-test using SPSS and attach a copy of your output. (1 point) 2. Write up the analyses as you would in the Results and Discussion section of a journal article, including the report of the central tendency, variability measures, and the outcome of the t-test (20points). You should also create a graph of the data (15 points). Follow this up with a discussion (40points) in which you: Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference Lower Upper Pair 1 Eye contact at 6 month - Eye contact at 9 month -1.500 2.138 .756 -3.287 .287 -1.984 7 .088 Descriptive Statistics N Sum Mean Std. Deviation Variance Skewness Kurtosis Statistic Statistic Statistic Statistic Statistic Statistic Std. Error Statistic Std. Error Eye contact at 6 month 8 58 7.25 1.909 3.643 .185 .752 -1.550 1.481 Eye contact at 9 month 8 70 8.75 2.053 4.214 -.743 .752 .142 1.481 Valid N (listwise) 8 Statistics Eye contact at 6 month Eye contact at 9 month N Valid 8 8 Missing 0 0 Mean 7.25 8.75 Std. Deviation 1.909 2.053 Variance 3.643 4.214 Skewness .185 -.743 Std. Error of Skewness .752 .752 Kurtosis -1.550 .142 Std. Error of Kurtosis 1.481 1.481 Sum 58 70 Percentiles 10 5.00 5.00 90 10.00 11.00 (a) present some conclusions about whether or not the amount of eye contact varies between the ages of 6 and 9 months; There is a general increase in frequency of eye contacts from the age of 6 months to 9 months, due to a development of baby cognition to their primary providers. This indicates that the psychology of children develop gradually and this is indicated by the development of attachment relationship as the children grows from the age of 6 months to 9 months. (b) Discuss the shortcomings of the study’s design and the possible confounds; Lack of gender differentiation in terms of level of psychological development. The significant changes in the rate of eye contacts as age progressed would have also been attributed to gender of the baby other than the cognitive development based on growth. Also the role and level of primary caregiver may differ, and it is not definite whether the significant attributable change in cognitive behaviour development was due to the role of mothers in the baby. (c) Provide some suggestions for improvement of the study’s design. Therefore the designer of this psychological study would have looked for a uniform gender of babies for the study, or separate male from female babies before doing the test. Also the designer would have subjected the babies to a similar treatment by the mothers under similar conditions. Results Mean - 7.25, 8.75 Standard deviation-1.909, 2.053 P-value 0.088 Discussion The P-value is statistically significant and this shows that as the children grew from 6 months to 9 months of age there was an increase in frequency of a minute-eye contact with their mothers. Since the mothers were the primary social providers to the developing babies, there was a high level of cognitive development for the baby. As the babies matured there is a high reliability for them to form attachment relationship with their mothers indicating a respective in improved cognitive development. This is usually achievable as the childs finds that the ultimate provider of social care is the mother or the person who takes a lot of the baby basic needs stuff. Part C: Research Analyses (between-groups ANOVA) (106 points) Part C, Section 1: Background A police researcher, influenced by Paul Ekman’s work on the psychology of lying, wanted to investigate the ability of members of various law enforcement groups to detect when someone is lying. In Ekman’s original experiments, different university-aged women were filmed while they answered questions about a video that they had watched. Ekman created 10 video presentations. In half the videos, the woman was asked to lie about what they had seen. In the other half, they were asked to tell the truth about what they had seen. The police researcher replicated parts of Ekman’s original studies by involving some of his fellow police robbery detectives, agents from the Secret Service, police officers who specialised in conducting lie detector tests (polygraphers), magistrates, lawyers, and psychiatrists. The researcher asked the participants to watch each of the videos and then simply decide whether or not the woman in the video was telling the truth or lying. The data below in Table 1 represent the total number of correct identifications (in either detecting a lie or detecting truth telling) that each participant made. Based on his data, the police researcher made the startling conclusion that all law enforcement groups – that is, the participants of his experiment who represented law enforcement groups – were superior compared to members of the general public when it came to detecting who was telling a lie compared to who was telling the truth. Statistics Number of correct Identification 1 Number of correct Identification 2 Number of correct Identification 3 Number of correct Identification 4 Number of correct Identification 5 Number of correct Identification 6 Number of correct Identification 7 Number of correct Identification 8 Number of correct Identification 9 Number of correct Identification 10 N Valid 6 6 6 6 6 6 6 6 6 6 Missing 0 0 0 0 0 0 0 0 0 0 Mean 5.17 5.83 5.83 5.17 5.33 5.33 6.00 5.33 5.83 3.67 Std. Deviation 1.169 .753 .408 1.472 .816 1.033 1.549 1.211 .983 2.944 Skewness .668 .313 -2.449 .711 -.857 .666 .000 .075 .456 -.711 Std. Error of Skewness .845 .845 .845 .845 .845 .845 .845 .845 .845 .845 Kurtosis -.446 -.104 6.000 -2.052 -.300 .586 -1.875 -1.550 -2.390 -2.052 Std. Error of Kurtosis 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 Sum 31 35 35 31 32 32 36 32 35 22 Statistics Number of correct Identification 1 Number of correct Identification 3 N Valid 6 6 Missing 0 0 Mean 5.17 5.83 Std. Deviation 1.169 .408 Skewness .668 -2.449 Std. Error of Skewness .845 .845 Kurtosis -.446 6.000 Std. Error of Kurtosis 1.741 1.741 Sum 31 35 Reliability Statistics Cronbach's Alpha Cronbach's Alpha Based on Standardized Items N of Items .749 .810 10 Statistics Number of correct Identification 1 Number of correct Identification 2 Number of correct Identification 3 Number of correct Identification 4 Number of correct Identification 5 Number of correct Identification 6 Number of correct Identification 7 Number of correct Identification 8 Number of correct Identification 9 Number of correct Identification 10 N Valid 6 6 6 6 6 6 6 6 6 6 Missing 0 0 0 0 0 0 0 0 0 0 Mean 5.17 5.83 5.83 5.17 5.33 5.33 6.00 5.33 5.83 3.67 Std. Deviation 1.169 .753 .408 1.472 .816 1.033 1.549 1.211 .983 2.944 Variance 1.367 .567 .167 2.167 .667 1.067 2.400 1.467 .967 8.667 Skewness .668 .313 -2.449 .711 -.857 .666 .000 .075 .456 -.711 Std. Error of Skewness .845 .845 .845 .845 .845 .845 .845 .845 .845 .845 Kurtosis -.446 -.104 6.000 -2.052 -.300 .586 -1.875 -1.550 -2.390 -2.052 Std. Error of Kurtosis 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 1.741 Sum 31 35 35 31 32 32 36 32 35 22 Percentiles 10 4.00 5.00 5.00 4.00 4.00 4.00 4.00 4.00 5.00 .00 90 7.00 7.00 6.00 7.00 6.00 7.00 8.00 7.00 7.00 6.00 One-Sample Test Test Value = 0 95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Lower Upper Number of correct Identification 1 10.826 5 .000 5.167 3.94 6.39 Number of correct Identification 2 18.981 5 .000 5.833 5.04 6.62 Number of correct Identification 3 35.000 5 .000 5.833 5.40 6.26 Number of correct Identification 4 8.598 5 .000 5.167 3.62 6.71 Number of correct Identification 5 16.000 5 .000 5.333 4.48 6.19 Number of correct Identification 6 12.649 5 .000 5.333 4.25 6.42 Number of correct Identification 7 9.487 5 .000 6.000 4.37 7.63 Number of correct Identification 8 10.787 5 .000 5.333 4.06 6.60 Number of correct Identification 9 14.533 5 .000 5.833 4.80 6.87 Number of correct Identification 10 3.051 5 .028 3.667 .58 6.76 Part C, Section 2: Assignment Task Perform appropriate analyses to determine whether or not the data supports the claim that different law enforcement officers differ when it comes to detecting detects truth telling and lying. As part of your analyses, you should perform two planned orthogonal comparisons: One that compares the Secret Service agents with all of the other groups and another that compares the law enforcement officers who are in the ‘front line’ (police robbery detectives, Secret Service agents) versus the others. You should also perform post hoc comparisons to check whether or not particular pairs of groups are significantly different. You must attach your SPSS output (1 point). ANOVA Sum of Squares df Mean Square F Sig. Number of correct Identification 2 Between Groups 2.333 3 .778 3.111 .253 Within Groups .500 2 .250 Total 2.833 5 Number of correct Identification 3 Between Groups .333 3 .111 .444 .747 Within Groups .500 2 .250 Total .833 5 Number of correct Identification 4 Between Groups 8.833 3 2.944 2.944 .264 Within Groups 2.000 2 1.000 Total 10.833 5 Number of correct Identification 5 Between Groups .833 3 .278 .222 .875 Within Groups 2.500 2 1.250 Total 3.333 5 Number of correct Identification 6 Between Groups 4.333 3 1.444 2.889 .268 Within Groups 1.000 2 .500 Total 5.333 5 Number of correct Identification 7 Between Groups 9.500 3 3.167 2.533 .296 Within Groups 2.500 2 1.250 Total 12.000 5 Number of correct Identification 8 Between Groups 6.833 3 2.278 9.111 .101 Within Groups .500 2 .250 Total 7.333 5 Number of correct Identification 9 Between Groups 2.833 3 .944 .944 .551 Within Groups 2.000 2 1.000 Total 4.833 5 Number of correct Identification 10 Between Groups 23.333 3 7.778 .778 .605 Within Groups 20.000 2 10.000 Total 43.333 5 One-Sample Statistics N Mean Std. Deviation Std. Error Mean Number of correct Identification 1 6 5.17 1.169 .477 Number of correct Identification 2 6 5.83 .753 .307 Number of correct Identification 3 6 5.83 .408 .167 Number of correct Identification 4 6 5.17 1.472 .601 Number of correct Identification 5 6 5.33 .816 .333 Number of correct Identification 6 6 5.33 1.033 .422 Number of correct Identification 7 6 6.00 1.549 .632 Number of correct Identification 8 6 5.33 1.211 .494 Number of correct Identification 9 6 5.83 .983 .401 Number of correct Identification 10 6 3.67 2.944 1.202 One-Sample Test Test Value = 0 95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Lower Upper Number of correct Identification 1 10.826 5 .000 5.167 3.94 6.39 Number of correct Identification 2 18.981 5 .000 5.833 5.04 6.62 Number of correct Identification 3 35.000 5 .000 5.833 5.40 6.26 Number of correct Identification 4 8.598 5 .000 5.167 3.62 6.71 Number of correct Identification 5 16.000 5 .000 5.333 4.48 6.19 Number of correct Identification 6 12.649 5 .000 5.333 4.25 6.42 Number of correct Identification 7 9.487 5 .000 6.000 4.37 7.63 Number of correct Identification 8 10.787 5 .000 5.333 4.06 6.60 Number of correct Identification 9 14.533 5 .000 5.833 4.80 6.87 Number of correct Identification 10 3.051 5 .028 3.667 .58 6.76 Write up the analyses as you would in the Results section of a journal article, including the report of the central tendency and variability measures and the outcome of the analyses (40 points). Include a graph of the data in your Results section (15 points). (This means that the graph should be formatted as part of the Results section.) Follow this up with a Discussion section (50 points) in which you present some conclusions about the relationship between reinforcement and learning. Discussion Mean scores for specialised officers are generally higher than the rest of the respondents. From the study results indicates that P-value is less than 0.05 and this indicates a statistical signifince, showing that there is a difference between the scores of the general public and the scores of specialised officers. The results shows that on average the identification of lies and truth telling is averagely higher for the specialised officers who were involved in the test, as compared to the general public. This cane be attributed to the regular exposures they have encountered as well as the trainings they frequently attend. The members of the public have low abilities to determine whether somebody is telling a truth or a lie, because they have little exposure and usually expects most people being honest to what they do and say. Also there is lack of concentration by subjects involved, from the general public and thisrenders them to develop low skills of truth identification in this experiment and in their natural environment. Part D: Research Analyses (within-groups ANOVA) (106 points) Part D, Section 1: Background Researchers have long suspected that exercise has a positive effect on one’s psychological wellbeing. Dr Richard Simmons decided to investigate the effects of exercise in a longitudinal study. He recruited five male volunteers to test a four-week exercise regime in which the participants did aerobic exercise for four hours per day, five days per week under the supervision of a fitness instructor. At the end of each week, the participants filled in a measure of psychological well-being known as the Psychological Well-Being Scale (PWBS). Scores on the PWBS could range between 1 to 70 with higher values indicating greater self-reported psychological well-being. Dr Simmonds carefully recruited his participants. All his participants were born in October, 1962. At the beginning of the study, all participants weighed 77 kilos and all were the same height (177 cm tall). None of the participants had exercised regularly in the past 12 months. The well-being scores for the four assessments are listed in Table 1. At the end of the four-week research study, Dr Simmons wrote up his results with the conclusion that a four-week exercise programme was necessary to significantly improve one’s psychological well-being for people of all ages. However, when he submitted his paper to a journal, the peer reviewers pointed out a number of flaws with his study and his conclusions and asked him to rewrite the Results and Discussion section, taking into account these flaws. Part D, Section 2: Assignment Task 1. Conduct a repeated-measures ANOVA by hand on the data and complete the following summary table (1 point). (You must include all of your hand calculations (10 points) as an Appendix to this assignment.) 2. Conduct Tukey’s HSD analyses to make pairwise comparisons between all of the cell means. What is the q-comparison value required for the Tukey’s HSD test (assume ∝ = 0.05)? (3 points) Show your manual calculations here. ANOVA Sum of Squares df Mean Square F Sig. Psychological wellbeing score for week 1 Between Groups 250.000 4 62.500 . . Within Groups .000 0 . Total 250.000 4 Psychological wellbeing score for week 2 Between Groups 265.200 4 66.300 . . Within Groups .000 0 . Total 265.200 4 Psychological wellbeing score for week 3 Between Groups 222.800 4 55.700 . . Within Groups .000 0 . Total 222.800 4 Descriptives N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum Lower Bound Upper Bound Psychological wellbeing score for week 1 50 1 44.00 . . . . 44 44 61 1 52.00 . . . . 52 52 67 1 63.00 . . . . 63 63 68 1 60.00 . . . . 60 60 69 1 61.00 . . . . 61 61 Total 5 56.00 7.906 3.536 46.18 65.82 44 63 Psychological wellbeing score for week 2 50 1 46.00 . . . . 46 46 61 1 53.00 . . . . 53 53 67 1 62.00 . . . . 62 62 68 1 59.00 . . . . 59 59 69 1 67.00 . . . . 67 67 Total 5 57.40 8.142 3.641 47.29 67.51 46 67 Psychological wellbeing score for week 3 50 1 50.00 . . . . 50 50 61 1 58.00 . . . . 58 58 67 1 66.00 . . . . 66 66 68 1 63.00 . . . . 63 63 69 1 69.00 . . . . 69 69 Total 5 61.20 7.463 3.338 51.93 70.47 50 69 Fill in the following table with the appropriate numbers that reflect the pairwise comparisons for the Tukey’s HSD tests. (1 point) Put a * next to the comparisons that are statistically significant. (1 point) 2. Write up the analyses as you would in the Results section of a journal article, including the report of the central tendency and variability measures and the outcome of the analyses (30 points). Include a graph of your data (15 points) and make sure you report the percentage of variability in PWBS scores that is accounted for by the ‘time’ variable (5 points). Follow this up with a Discussion section (40 points) in which you Statistics Psychological wellbeing score for week 1 Psychological wellbeing score for week 2 Psychological wellbeing score for week 3 Psychological wellbeing score for week 4 N Valid 5 5 5 5 Missing 0 0 0 0 Mean 56.00 57.40 61.20 63.00 Std. Deviation 7.906 8.142 7.463 7.906 Skewness -1.063 -.448 -.849 -1.518 Std. Error of Skewness .913 .913 .913 .913 Kurtosis -.232 -.546 .091 1.841 Std. Error of Kurtosis 2.000 2.000 2.000 2.000 Sum 280 287 306 315 3. (a) present some conclusions about the relationship between exercise and psychological well-being (b) discuss the shortcomings of Dr Simmond’s design and the possible confounds What is the q-comparison value required for the Tukey’s HSD test (assume ∝ = 0.05)? (3 points) Show your manual calculations here. Summary of statistics Week 1. Week 2. Week 3. Week 4. Sum total Sum of X. 280. 287. 306. 315. 1188 Sum of X2. 78400. 82369. 93636. 99225. 353630 Mean. 56.0. 57.4. 61.2. 63.0 Summation of components; SSt _ 353630 - (1188 x 1188) / (5 x 4) _ 283062.8 SSs _ 1/4 (62500+50176+70756+36100+66564) x (1188 x 1188)/(5x4) _ (286096/4) - 70567.2 _ 71524 - 70567.2 _ 956.8 SSb _ (78400+82369+93636+99225) - 70567.2 _ (353630/5) - 70567.2 _ 70726 - 70567.2 _ 158.8 Finally SSo or F is given by _ SSt - SSs - SSe _ 283062.8 - (956.8 + 158.8) _ 283062.8 - 1115.6 _ 281947.2 Part D, Section 2: Assignment Task 1. Conduct a repeated-measures ANOVA by hand on the data and complete the following summary table (1 point). (You must include all of your hand calculations (10 points) as an Appendix to this assignment.) Number of conditions is k and number of cases is n. dfs is given by the number of cases minus one - 5 minus 1 equals 4 dfb is number of times minus 1 --- 4-1 equals to 3 dfe equals to dfb x dfs dfT equals to (nK-1) 2. Conduct Tukey’s HSD analyses to make pairwise comparisons between all of the cell means. Source. df. SS. MS. F Time. 3. 956.8. 318.93. 634.7 Subjects. 4. 158.8. 39.45. Error. 12. 6.03. 0.5025 Total. 19. 1121.63 The computed F is 634.7 Fill in the following table with the appropriate numbers that reflect the pairwise comparisons for the Tukey’s HSD tests. (1 point) Put a * next to the comparisons that are statistically significant. (1 point) The computed F value is very critical in the understanding of this result, and this is done by comparing it with the critical value of F from the F Table. You look up the critical value by using the degrees of freedom , and for our case, the dfb is 3 and the dferror is 12. Because our obtained F exceeds this value, we reject the null hypothesis and conclude that there is a significant difference between the conditions. Therefore we accept the alternative hypothesis, which states that ' exercises or physical activity exposure by an individual usually is associated with an improvement in the psychological health of that individual. 3. Write up the analyses as you would in the Results section of a journal article, including the report of the central tendency and variability measures and the outcome of the analyses (30 points). The results shows that there is a general increase in the mean values of the psychological wellbeing of the five subjects after that four week excercise experiment. This is indicated by the raise in the psychological wellbeing scores (PWBS) mean from an initial score of 56 to a mean score of 63 at the end of the four week exercise experiment. This shows that physical activity enhances the PWBS of an individuals and this gives these individuals positive attitutes towards life. Include a graph of your data (15 points) and make sure you report the percentage of variability in PWBS scores that is accounted for by the ‘time’ variable (5 points). See the graph above. The mean scores for the subsequent weeks are relatively high as compared to the initial ones before intervention, and this may be attributed to the experience and training they have been subjected to. From the data results we get that physical activity and exercises improves personal scores for PWBS, up to the optimum point where further exposure to exercise lacks a significant improvement. Follow this up with a Discussion section (40 points) in which you: (a) Present some conclusions about the relationship between exercise and psychological well-being Exercise or physical activity improves the psychological wellbeing of an individual as well as the physiological fitness especially by enhancing the cardiovascular function. This is because physical activity allows even body distribution and deposition of the adipose tissue, improving on the body mass index. There is burning of extra energy through regulated exercises and this prevents excessive accumulation of ernergy reservours in the body in terms of glycogen deposition. The frequent physical activity enhances proper body cellular metabolism, and this reduces the levels of plasma glucose. Low circulating glucose in the blood stream results in the utilization of energy reserves, and this further enhances cardiovascular function through a decreased load. Frequent exercises also reduces the kevels of stress hormones in plasma and this enhances an individual's psychological wellbeing. (b) discuss the shortcomings of Dr Simmond’s design and the possible confounds For this research study the possible confounds include level of physical activity involved by the five subjects. We can not be able to determine whether this increased PWBS was due to the experimental exercise or was it due to healthy occupational work conditions. Also we can't be able to determine whether the PWBS was due to the changes in financial status and stress levels attributed to the individual's environment. Individuals exposed to financial constraints and high stress levels may score loiwly on the PWBS as compared to the general population. (c) discuss what additional data you would have liked to have collected to clarify the meaning of the present results, especially in light of Dr Simmond’s conclusions. The designer of this experiment should have checked on; - the financial status of the individual suspect - living standards of the subjects such as expenditures per day as well as the income - occupational duties and working hours per day References Anderson, M.L. and Taylor, H.F. (2009). Sociology: The Essentials. Belmont, CA: Thomson Wadsworth. Anthony M. Glaziano and Michael L. Raulin (2013), Research methods 8th Edition T. Hastie, R. Tibshirani and J. Friedman (2001), The Elements of Statistical Learning, Springer Appendix Part D Manual Calculation of Repeated Measure Anova F is given bu MS time/ MS error or MS conditions/ MS error Where MS is sum of squares for the difference between groups Subject. Week 1. Week 2. Week 3. Week 4. Total 1. 60. 59. 63. 68. 250 2. 52. 53. 58. 61. 224 3. 61. 67. 69. 69. 266 4. 44. 46. 50. 50. 190 5. 63. 62. 66. 67. 258 Total X. 280. 287. 306. 315 Summary of statistics Week 1. Week 2. Week 3. Week 4. Sum total Sum of X. 280. 287. 306. 315. 1188 Sum of X2. 78400. 82369. 93636. 99225. 353630 Mean. 56.0. 57.4. 61.2. 63.0 Summation of components; SSt _ 353630 - (1188 x 1188) / (5 x 4) _ 283062.8 SSs _ 1/4 (62500+50176+70756+36100+66564) x (1188 x 1188)/(5x4) _ (286096/4) - 70567.2 _ 71524 - 70567.2 _ 956.8 SSb _ (78400+82369+93636+99225) - 70567.2 _ (353630/5) - 70567.2 _ 70726 - 70567.2 _ 158.8 Finally SSo is given by _ SSt - SSs - SSe _ 283062.8 - (956.8 + 158.8) _ 283062.8 - 1115.6 _ 281947.2 Read More

You must attach your SPSS output (1 point). ANOVA Sum of Squares df Mean Square F Sig. Number of correct Identification 2 Between Groups 2.333 3 .778 3.111 .253 Within Groups .500 2 .250 Total 2.833 5 Number of correct Identification 3 Between Groups .333 3 .111 .444 .747 Within Groups .500 2 .250 Total .833 5 Number of correct Identification 4 Between Groups 8.833 3 2.944 2.944 .264 Within Groups 2.000 2 1.000 Total 10.833 5 Number of correct Identification 5 Between Groups .833 3 .278 .222 .

875 Within Groups 2.500 2 1.250 Total 3.333 5 Number of correct Identification 6 Between Groups 4.333 3 1.444 2.889 .268 Within Groups 1.000 2 .500 Total 5.333 5 Number of correct Identification 7 Between Groups 9.500 3 3.167 2.533 .296 Within Groups 2.500 2 1.250 Total 12.000 5 Number of correct Identification 8 Between Groups 6.833 3 2.278 9.111 .101 Within Groups .500 2 .250 Total 7.333 5 Number of correct Identification 9 Between Groups 2.833 3 .944 .944 .551 Within Groups 2.000 2 1.

000 Total 4.833 5 Number of correct Identification 10 Between Groups 23.333 3 7.778 .778 .605 Within Groups 20.000 2 10.000 Total 43.333 5 One-Sample Statistics N Mean Std. Deviation Std. Error Mean Number of correct Identification 1 6 5.17 1.169 .477 Number of correct Identification 2 6 5.83 .753 .307 Number of correct Identification 3 6 5.83 .408 .167 Number of correct Identification 4 6 5.17 1.472 .601 Number of correct Identification 5 6 5.33 .816 .333 Number of correct Identification 6 6 5.33 1.033 .

422 Number of correct Identification 7 6 6.00 1.549 .632 Number of correct Identification 8 6 5.33 1.211 .494 Number of correct Identification 9 6 5.83 .983 .401 Number of correct Identification 10 6 3.67 2.944 1.202 One-Sample Test Test Value = 0 95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Lower Upper Number of correct Identification 1 10.826 5 .000 5.167 3.94 6.39 Number of correct Identification 2 18.981 5 .000 5.833 5.04 6.

62 Number of correct Identification 3 35.000 5 .000 5.833 5.40 6.26 Number of correct Identification 4 8.598 5 .000 5.167 3.62 6.71 Number of correct Identification 5 16.000 5 .000 5.333 4.48 6.19 Number of correct Identification 6 12.649 5 .000 5.333 4.25 6.42 Number of correct Identification 7 9.487 5 .000 6.000 4.37 7.63 Number of correct Identification 8 10.787 5 .000 5.333 4.06 6.60 Number of correct Identification 9 14.533 5 .000 5.833 4.80 6.87 Number of correct Identification 10 3.051 5 .028 3.667 .58 6.76 Write up the analyses as you would in the Results section of a journal article, including the report of the central tendency and variability measures and the outcome of the analyses (40 points).

Include a graph of the data in your Results section (15 points). (This means that the graph should be formatted as part of the Results section.) Follow this up with a Discussion section (50 points) in which you present some conclusions about the relationship between reinforcement and learning. Discussion Mean scores for specialised officers are generally higher than the rest of the respondents. From the study results indicates that P-value is less than 0.05 and this indicates a statistical signifince, showing that there is a difference between the scores of the general public and the scores of specialised officers.

The results shows that on average the identification of lies and truth telling is averagely higher for the specialised officers who were involved in the test, as compared to the general public. This cane be attributed to the regular exposures they have encountered as well as the trainings they frequently attend. The members of the public have low abilities to determine whether somebody is telling a truth or a lie, because they have little exposure and usually expects most people being honest to what they do and say.

Also there is lack of concentration by subjects involved, from the general public and thisrenders them to develop low skills of truth identification in this experiment and in their natural environment.

Read More