Student Behavior: Babysitting, Delinquency & Academic Performance, Exams of School management&administration

Download Student Behavior: Babysitting, Delinquency & Academic Performance and more Exams School management&administration in PDF only on Docsity! Student ID______KEY_______________ CEP 933 Fall 2001 Midterm Exam Part I. Data analysis (25 points – 5 points per item). Each of the following scenarios describes some data that would be analyzed using procedures that you have learned about so far in CEP 933. For each scenario, select the most appropriate analysis from those listed just below. There is only one correct answer for each scenario. Write the letter for the analysis in the blank just before the number for each scenario, then write one sentence to briefly justify your answer. Choose from the following analyses: A. one-way ANOVA B. planned comparisons C. Spearman’s rho D. Pearson’s r E. point-biserial correlation F. phi coefficient __A__ 1. A school counselor is interested in the effects of three different treatments on student absenteeism. Each student will be randomly assigned to one of three treatment groups. The effectiveness of the treatments will be measured by counting the number of days each student is absent during a school year. This could be addressed using one-way ANOVA because the treatments constitute a categorical predictor and the outcome is quantitative. __E__ 2. A researcher is interested in the relationship between age and incidence of skin cancer. The study will examine several hundred subjects, who will be asked if they have ever had skin cancer and who will also indicate their age. This is represented by the point-biserial r because presence of skin cancer is a true dichotomy and age is a continuous variable. __B__ 3. A researcher is interested examining the relationship between academic major and a quantitative score on a liberalism scale. She believes that students who are English and journalism majors are more liberal than science and mathematics majors; however, she also believes that English majors are more liberal than journalism majors. This must be addressed using planned comparisons because the outcome is a quantitative variable (liberalism) and the researcher has very specific directional hypotheses that cannot be confirmed or denied using a simple ANOVA. 1 D or C 4. A researcher is interested in the alcohol consumption of students. The participants in the study will be asked how many drinks they typically consume during a 30-day period. The researcher thinks that the subjects’ academic performance may relate to their consumption so students will also report their cumulative grade point averages. Both of the variables here are quantitative so the preferred approach would be to examine Pearson’s r if you are willing to treat GPA as an interval-scale variable. However, if you believe that GPA is ordinal, or if you think it is interval but the relationship does not appear to be bivariate normal we might use Spearman’s rho. There is no reason to expect to use rho because of nonnormality, given the description in the scenario. __F__ 5. A state school administrator is interested in whether charter or traditional public schools have higher student achievement. The administrator will measure achievement in terms of whether more than half of the students from each school receive Michigan Merit Scholarships. So each school will report yes/no answers to these two questions: “A) Is this school a charter school?” and “B) Did more than half of the students in this school receive a Michigan Merit Scholarship?” Here we are examining two dichotomous variables, as they are measured. Technically one might argue that question B represents an artificial dichotomy, but even in that case the analysis of choice would be the phi coefficient. Part II. Problem Solving (75 points) On the next 2 pages is an SPSS analysis of the NELS student-level data. The researcher who ran this analysis was interested in scores on newhood1 – a measure of delinquent or problematic behavior – for students with differing levels of childcare responsibilities (either for the students siblings or their own children). Here are the relevant variables: babysit “Does student babysit own children or younger siblings?” The variable is coded 1= yes, 2=no, or 3=does not apply. The label “does not apply” means that the student has no children and no younger siblings. These students might babysit other children, but this is not clear from the babysit variable. However, the researcher is still interested in these students’ levels of delinquent behavior. newhood1 is a continuous outcome variable, where students who had higher scores reported experiencing a higher number of problematic events or behaviors (e.g., cutting class, being late for school, getting suspended from school or arrested). Scores on newhood1 ranged from 0 to 3.36, with a standard deviation of 0.73. The total score was an average across 7 items and the score scale for each item in this variable used the following labels: 0 = this event never happened to me 1 = this event happened 1 or 2 times 2 = this event happened 3 to 6 times 3 = this event happened 7 to 9 times 4 = this event happened 10 or more times 2 6. (8 points) Write the null hypothesis tested in this analysis in words and symbols. The null hypothesis here is that, in the population, the three groups of students with different babysitting activities have the same means on newhood1, the measure of delinquent behavior. Using the codes above for 1, 2, and 3 we may write Ho : 1 = 2 = 3 or we may write the hypothesis in terms of the treatment effects Ho : 1 = 2 = 3 = 0 or even Ho : j 2 = 0 5 7. (6 points) What do you conclude about this hypothesis? State exactly which part(s) of the output you are using to draw your conclusions. I will use a significance level of .05 for my significance test, shown in the ANOVA display on p. 3. There we find F(2, 257) = 3.029 which is just exactly significant because p = .05!! The tabled value Fc = 3.04 in our book is approximate because it is for dfe = 200. So we can conclude that at least one population group mean on delinquent behavior differs from the rest. Alternately we could decide to reject the mull model. This example shows the problem with hypothesis testing: our decisions are based on an arbitrary level of “unusualness.” Also if we compute E2 it is very low: E2 = 3.18/138.08 = .023. Only 2% of the variation in newhood1 scores is accounted for by the babysitting activities. If we decide to reject Ho we still have not done much to explain why delinquency levels differ. 8. (10 points) Is there anything in the output that would lead you to believe that the results are not valid? Please describe which part(s) of the output raise questions or concerns, and why. We can use the output to check two assumptions: normality of residuals and equal variances in the population. We can be quite comfortable that variances are equal across groups because Levene’s test is not significant (p = .86) and the scatterplot of predicted values versus residuals (in plot B) shows equal spread within groups. However, plot C shows that the residuals are somewhat skewed which raises concerns about the normality assumption. 9. (8 points) The researcher had two planned contrasts (results are shown in the output). Describe which means are being compared by each of these 2 contrasts (hint: you may want to give the null hypotheses that are tested). In contrast 1 the researcher is comparing the mean for group 1 (babysitters) to the mean for groups 2 and 3 combined (non-babysitters). The null hypothesis is Ho : 1 = (2 + 3) /2 or 2 1 = (2 + 3). In contrast 2 the two subgroups of non-babysitters are compared. The Ho is Ho : 2 = 3 . That is, regardless of whether the student said they had sibs or kids but did not babysit or that the question did not apply, the newhood1 population means are the same. 5 10. (8 points) Are the two contrasts that you described in your answer to item 9 orthogonal? Prove your answer. Contrast 1 weights 2 -1 -1 Contrast 2 weights 0 1 -1 Products of weights 0 -1 1 Sum of products = 0 + 1 (-1) + 1 = 0 Because the sum of the products equals zero, the contrasts are orthogonal. 11. (10 points) What are the results of the tests of the contrasts? What conclusions can we draw about babysitting and levels of delinquent behavior? Please state exactly which line(s) of the contrast-results table you are using to answer this item. Because according to Levene’s test the variances can be treated as equal, we examine the top 2 lines of the ‘contrast tests’ output. There we see that contrast 1 is not quite significant (t(257) = 1.94) and the second contrast is definitely not significant (t (257) = - 1.05, p = .30). This tells us that none of the groups differ at the .05 level!! The largest different (post hoc) seems to be between the mean for group 2 (which looks low) and those for groups 1 and 3 which are relatively higher. Had we compared (1 + 3)/ 2 to 2 we might have found a difference. A new graduate student did a pilot study of the learning of juggling where he developed two treatments to help people learn to juggle. His performance outcome was the amount of time it took for his subjects to be able to successfully juggle 3 balls for 2 minutes without dropping any of the balls. Note that for this outcome a LOWER mean represents BETTER performance. Output from his analysis of variance is shown here: Oneway 6