Null and Alternative Hypotheses - Assignment Example

Part A: Research Scenarios (30 points) In each of the research scenarios presented below, you are required to provide answers to various questions that ask about the Null and Alternative hypotheses, independent and dependent variables (if any). You are also asked to comment on what might be wrong with the experimental design employed in the study and you may be asked to provide a solution or asked to re-design the study. In answering the questions, It is in your best interests to be as precise and succinct as possible; Please write your answers legibly, preferably in pen. Scenario 1 (10 points) Karsh (1983) designed an experiment on the relationship between early handling and friendliness in cats. She randomly assigned kittens to one of three groups which differed according to the age of first handling. The first group was handled daily from 3 to 14 weeks of age, the second from 7 to 14 weeks of age and the third received no handling from birth to 14 weeks. Different experimenters handled each kitten on different days. "Handling" was defined as an experimenter holding a kitten on his/her lap while petting it for 15 minutes. Friendliness was measured by how long each kitten stayed with the experimenter when not restrained, and by how long it took each kitten to reach the experimenter. Karsh found that the kittens handled from 3 to 14 weeks stayed longer with the experimenter and ran more quickly to the experimenter than kittens handled from 7 to 14 weeks. Also, kittens handled from 7 to 14 weeks were more "friendly" than kittens who received no handling at all. Karsh concluded that kittens should be handled as early as possible to ensure life-long friendliness towards humans. Ho: There are no differences in friendliness in cats based on the age of handling. (1 point) Ha: There are differences in friendliness in cats based on the age of handling (At least one mean is different). (1 point) IV: The independent variable is the age of first handling (three different age groups of handling) (1 point) DV: The dependent variable is friendliness (duration of stay with experimenter and to reach the experimenter) (1 point) What are the possible confounds in Karsh’s experimental design: The possible confounds in Karsh’s experimental design are the way experimenters handled each kitten on different days. Each experimenter will hold a kitten on his/her lap while petting it differently for 15 minutes. The petting a kitten will be different for each experimenter and each kitten will feel it differently. This have affect on both handling and friendliness in cats. (6 points) Scenario 2 (10 points) An investigator believed that sensory deprivation inhibits the intellectual development of animals. He ran an experiment to examine this thesis in the following way. He used two rats, each of which had just given birth to eight pups. One rat and her litter were placed in a large cage. This cage had ample space and lots of objects to explore. The baby rats in this cage breastfed off their mother. The second rat’s pups were separated from the mother, and each was placed in a separate cage. These cages were quite small, and the only objects they could see or hear were the four walls and the food and water dispensers. After five months, both groups were tested in a multiple-T maze using food as a reward. Following 20 trials, all of the nondeprived pups were running the maze without error, but the deprived pups were still making several errors. This latter group frequently froze and had to be prodded to move. The experimenter concluded that sensory deprivation inhibits intellectual development such that deprived rats did not have the intellectual ability to learn even a simple maze. Ho: There is no difference in the error made in running the maze between non-deprived pups and deprived pups. (1 point) Ha: The error made in running the maze is less for non-deprived pups as compared to deprived pups. (1 point) IV: The independent variable is two groups of baby rats (deprived and non-deprived). (1 point) DV: The dependent variable is errors made by pups in running the maze. (1 point) Briefly describe the two confounds in the experiment: The two confounds in the experiment are pups breastfed off their mother and the size of cage. There may be possibility that breastfeeding have an effect on intellectual development of rats. Also because of the less space in cage rats were not able to move much, therefore, they became lazy. (3 points) Devise a solution to control the confounds in this experiment: The possible solution to control the confounds in this experiment is have the same size of cage and breastfeeding pups by any means. (3 points) Scenario 3 (10 points) A psychology lecturer wanted to compare two methods of teaching introductory statistics. One method relied heavily on teaching the theory behind statistics (the theory method). The other method consisted of teaching the students various statistical tests and explaining when to use each test (the cookbook method). The lecturer found that the engineering faculty was using the theory method in all its introductory statistics classes and that the education department was using the cookbook method in all its classes. At the end of the semester, the lecturer administered a standardised statistics test (the SST) to the engineering and education classes. The results indicated that the classes that received the theory method performed far better than did the classes that received the cookbook method. The lecturer concluded that the theory method was the superior method and that it should be adopted by anyone who wanted to teach statistics. Ho: There is no difference in the SST score between the engineering and education classes’ students. (1 point) Ha: There is a difference in the SST score between the engineering and education classes’ students . (1 point) IV: The independent variable is method of teaching (the theory method, and the cookbook method). (1 point) DV: The dependent variable is the students score in a standardised statistics test (the SST) . (1 point) Briefly describe the major confound in the experiment: The major confound in the experiment is the type of students (engineering and education classes’ students). The ability to understand any subject is different for different type of students (based on their stream). (3 points) Devise a solution to control the confounds in this experiment: A possible solution to control the confound in this experiment is to have two group of students from same classes (either engineering or education classes’ students) and than using the theory method of teaching statistics to one group and the cookbook method of teaching statistics to another group of students. (3 points) End of Part A Part B: Research Analyses (t-tests) (91 points) There is an old adage: ‘Absence leads the heart to wander’. In psychological terms, this adage is represented as the physical reinforcement theory. Briefly, the theory proposes that attraction to another person depends on physical reinforcement achieved through interaction with them. A researcher has conducted an experiment to test this theory. In her experiment, participants are asked to make attractiveness ratings of two different people – one with whom they have interacted (Person I, for Interaction) and one with whom they have not (Person N, for No interaction). The physical reinforcement theory predicts that attraction ratings will be greater for Person I than for Person N. The dependent variable is the rating of the attractiveness of these two people on a nine-point scale from 1 to 9 (where 1 = ‘do not like at all’ and 9 = ‘like a great deal’). The attractiveness ratings data for the study are provided below. Table 1 Rating of attractiveness for Person I or Person N Subject Person I Person N 1 7 5 2 6 6 3 5 2 4 6 3 5 5 4 6 7 5 7 3 6 1. Analyse the data in Table 1. (a) Conduct the appropriate t-test manually to compare the ratings for Person I and Person N. Make sure you attach a copy of your calculations. (5 points) Table 2 Mean and standard deviation of difference of rating of attractiveness for Person I and Person N Subject Person I Person N D = Person I-Person N 1 7 5 2 0.734 2 6 6 0 1.306 3 5 2 3 3.448 4 6 3 3 3.448 5 5 4 1 0.020 6 7 5 2 0.734 7 3 6 -3 17.164 = 8 = 26.857 = 1.143 = 2.116 The null and alternate hypotheses are The appropriate test will be right-tailed paired sample t-test. The selected significance level, α is .05. The degrees of freedom are, ν = n – 1 = 7 – 1 = 6 At level of significance of .05 with 6 degrees of freedom, the right-tailed critical value of t is 1.943. Therefore, the decision rule will be Reject H0 if t > 1.943, otherwise do not reject H0. The test statistic is Since, the test statistic 1.429 is less than the critical value 1.943. Therefore, do not reject H0. (b) Conduct the appropriate t-test using SPSS and attach a copy of your output. (1 point) Table 3 SPSS output Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 Person I 5.57 7 1.397 .528 Person N 4.43 7 1.512 .571 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 Person I - Person N 1.143 2.116 .800 -.814 3.100 1.429 6 .203 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 (30 points). You should also create a graph of the data (5 points). Follow this up with a discussion (50 points) in which you: (a) present some conclusions about whether or not the attractiveness ratings vary for Person I versus Person N and whether or not the physical reinforcement theory is supported by the data; (b) discuss the shortcomings of the study’s design and the possible confounds; (c) provide some suggestions for improvement of the study’s design. Results The average rating of attractiveness on a nine-point scale from 1 to 9 (where 1 = ‘do not like at all’ and 9 = ‘like a great deal’) for Person I and Person N is about 5.57 (SD = 1.40), and 4.43 (SD = 1.51), respectively. As shown in figure 1, there appears that mean rating of attractiveness for Person I is greater than Person N. A paired-sample t-test (right-tailed) suggested that there is statistically no difference in mean rating of attractiveness between Person I and person N, t(6) = 1.43, p = .203 (>.05). The 95% confidence interval for the difference in mean rating of attractiveness between Person I and person N is -.81 to 3.1 points. Discussion The result suggested statistically non-significant difference in mean rating of attractiveness between Person I and person N. Therefore, the attractiveness ratings do not vary for Person I (interaction) versus Person N (no interaction) and thus, the physical reinforcement theory is not supported by the data. However, Person I reported higher rating of attractiveness as compared to Person N. The sample size taken for study is very small. The possible confounds of the study are type of interaction between two persons and knowing about the person with whom one is not interacted. A possible solution to control the confound in this study is to have same type of interaction between two persons. In addition, the sample size taken should be large so as to reach a definite conclusion about the physical reinforcement theory. Figure 1: Mean rating of attractiveness for Person I and Person N End of Part B Part C: Research Analyses (t-tests) (91 points) An educational psychologist believed that daily instruction in phonics (i.e., how to sound out letters phonetically) would help Year 4 primary school children improve some aspects of their reading ability. She arranged for a Year 4 class of 12 students to receive a daily 40-minute practice session in phonics by a registered phonics instructor for eight weeks. A control classroom of 10 Year 4 students followed the same curriculum but without the phonics sessions. At the end of the eight-week period, all students were given a Degree of Reading Ability (DRA) test designed to measure their reading fluency. The higher the DRA score, the higher the reading fluency. The data for the two groups are provided below. Table 1 DRA scores for Year 4 students 1. Analyse the data in Table 1. (a) Conduct the appropriate t-test manually to compare Phonics and Control groups. Make sure you attach a copy of your calculations. (5 points) Table 2 Mean and standard deviation for DRA scores for Year 4 students Phonics Group Control Group 12 2.008 6 12.25 18 21.004 9 0.25 14 0.340 4 30.25 10 11.676 13 12.25 19 31.170 14 20.25 8 29.344 9 0.25 15 2.506 8 2.25 11 5.842 12 6.25 10 11.676 11 2.25 13 0.174 9 0.25 15 2.506 16 6.672 = 161 = 124.917 = 95 = 86.5 = 12 = 10 = 13.417 = 3.370 = 9.500 = 3.100 The null and alternate hypotheses are The appropriate test will be right-tailed independent-samples t-test. The selected significance level, α is .05. The degrees of freedom are, ν = n1 + n2 – 1 = 12 + 10 – 1 = 20 At level of significance of .05 with 20 degrees of freedom, the right-tailed critical value of t is 1.725. Therefore, the decision rule will be Reject H0 if t > 1.725, otherwise do not reject H0. , and , and Assuming equal variances (or standard deviations), the pooled variance is The test statistic is Since, the test statistic 2.813 is greater than the critical value 1.725. Therefore, reject H0. (b) Conduct the appropriate t-test using SPSS and attach a copy of your output. (1 point) Table 3 SPSS output Group Statistics Group N Mean Std. Deviation Std. Error Mean DRA Score Phonics 12 13.42 3.370 .973 Control 10 9.50 3.100 .980 Independent Samples Test Levenes Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper DRA Score Equal variances assumed .212 .650 2.813 20 .011 3.917 1.392 1.013 6.821 Equal variances not assumed 2.836 19.768 .010 3.917 1.381 1.034 6.800 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 (30 points). You should also create a graph of the data (5 points). Follow this up with a discussion (50 points) in which you: (a) present some conclusions about the relationship between phonics and reading ability (b) discuss the shortcomings of the study’s design and the possible confounds (c) provide some suggestions for improvement of the study’s design. Results The average DRA scores for Year 4 students for phonics and control group is about 13.42 (SD = 3.37), and 9.50 (SD = 3.10), respectively. As shown in figure 1, there appears that mean DRA scores for Year 4 students for phonics group is greater than control group. A independent-sample t-test assuming equal variances (right-tailed) suggested that there is a statistically significant difference in mean DRA scores for Year 4 students between phonics and control groups, t(20) = 2.81, p = .005 ( Read More