SAMPLE QUESTIONS. Research Methods II - HCS 6313

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1 SAMPLE QUESTIONS Research Methods II - HCS 6313 This is a (small) set of sample questions. Please, note that the exam comprises more questions that this sample. Social Security Number: NAME: IMPORTANT NOTICE. Your answers must be placed in the spaces provided. Use the amount of space provided as an indication of the maximum length of answer that is expected. You may want to do your your calculations on scratch paper before writing them on the exam, but make sure you show all your calculations on the exam paper. Answers placed elsewhere than in the spaces provided will not be considered. Do not add any loose sheets. 1. (00 points) For each of the following multiple choice questions, select one and only one response (the response that is the most correct). a. A frequency distribution in which most scores are concentrated around the center and which is symmetrical about its midpoint is a: a. probability distribution b. normal distribution c. Monte Carlo distribution d. all of the above b. Gamay, who is a developmental psychologist, measures how responsive a mother is to her infant by rating the mother s behavior on a 10-point scale. He also obtains a measure of the strength of the infant attachment to the mother. He finds that as maternal responsiveness increases so does the strength of attachment. This an example of a(n) a. negative correlation b. inverse experimental mortality c. prediction d. positive correlation 2. (00 points) The staff at the student coffee house are trying to find out whether to continue serving their current blend of morning coffee. The customers are asked to indicate whether they like the particular roast of coffee beans or not by checking off either the YES or NO box on a slip of paper they are given when ordering their coffee. A total of 100 people completed the survey. The results shows that 60 people indicated their preference for the coffee that is currently being served. Can the staff of the coffee house consider this results reliable? This asks for a Binomial Test. The problem is to find the probability of finding 60 or more out of 100. If you are really patient you can compute this probability using the binomial distribution. A shorter way of doing is to use the normal approximation. This means that we need to transform the score of 60 out of 100 into a Z-score and then use the Z to N(Z) Table in order to find the probability associated with this Z-score. The Z-score is obtained as Z = (Y.5) M Y σ Y (Y.5) N P = = (60.5) = N P (1 P) = = 1.9. If we use the Z to N(Z) Table, we find that p(z) =.0287, (for a one-tail test) and therefore we can reject the null hypothesis at α =.05, but not at α =.01. So at α =.05 we accept that the customers prefer the new coffee. (food for thought: Would we have reached the same conclusions with a two-tailed test? What would have happened without the correction of continuity?). But at α =.01, we are forced to suspend the judgement and declare that we don t know. 1

2 2 3. (00 points ) Marie & Antoinette (1789) report that the correlation between wages and number of cakes eaten is equal to.6. Their data came from 15 (revolutionary) subjects. What statistical conclusion can they draw? Justify your answer. The question here is to find if this correlation is statistically significant. So we need to compute an F-ratio and find if this F ratio is large enough (i.e., rare enough ) to allow for the rejection of the null hypothesis. The F is computed with the usual formula (signal to noise... ): r df = 13 = 13 = 1 r = 9 13 = From the F table, we find that the critical values are (for ν 1 = 1 and ν 2 = 13): 4.67 (for α =.05) and 9.07 (for α =.01). Therefore,we can reject the null hypothesis at α =.05 but not at α =.01. So at α =.05, we can conclude that wealthier you are, the more cakes you eat (maybe because you can afford them!). 4. (00 points) There is (at best) one outstanding musician (a.k.a. a genius) per million human beings. According to some estimation (given by the Official Bureau of Improbable Statistics) 95% of musicians composed their first work before the age of 10 (their first work was in general good but not outstanding). However, composing at an earlier age is, by itself, quite remarkable, because less than one child in a thousand composes music of any interest before the age of 10. Your child, who is 9 year old, has just composed a nice song. This piece, actually is good enough to compare with the first work of a genius. Should you start thinking that you have a genius at home (justify your answer!)? This is a Bayes theorem question. What we want to find is the probability of begin a genius knowing that we have composed before the age of 10. Call A the even A = {Being a genius} and Call B the even B = {Composing before the age of 10}. This means that we want to find Pr{A B} from the probabilities that are given in text. From the text, we find that Pr{A} =.000,001, Pr{B} =.001, and Pr{B A} =.95. We need to plug all these values in the equation and we get: Pr{A B} = Pr{B A} Pr{A} Pr{B} = , = = = So we have less than a chance in one thousand that your child is a musical genius (Good Bye Mozart!).

3 3 5. (00 points ) Paire & Ternel (2000) are working on aggressive behaviors in children, and are particularly interested in the possibility of a relationship between relational aggression and overt aggression. They ask a group of 3rd and 5th graders to complete a peer assessment questionnaire in which the children are asked to describe a what the other children are doing during recess. From these reports, the investigators computed the number of overt and relational aggressive acts observed at recess during a two-week period. We have extracted the number of aggressive acts reported for six children. The results are given in the following table: Child A Child B Child C Child D Child E Child F W=Relational Y=Overt a. Draw a scatterplot of these data. 20 Overt and Relational Agressive Acts 18 A D Y (Overt Agression) C B 6 E F 4 2 Et voilà: W (Relational Agression) b. What is the value of the coefficient of correlation between the number of relational aggressive acts and the number of overt aggressive acts? An easy way of computing the coefficient of correlation is to create a table with all the information that we need. Here it is (NB: M W = 66 6 = 11, M Y = 72 6 = 12): Obs Name W w w 2 Y y y 2 y w 1 A B C D E F W SSW Y SSY SCP WY From this table, we can now compute the value of the coefficient of correlation ( rectangle divided by the square-root of the squares ): (W MW )(Y M Y ) r WY = (W MW ) 2 (Y M Y ) = SCP WY 90 = = 90 2 SSW SS Y = 3 4 =.75 c. How much variance do the variables (relational and overt aggression) have in common? The proportion of common variance between 2 variables is equal to the square of their coefficient of correlation. So the proportion of common variance is equal to r 2 =.75 2 = 32 4 = =.5625.

4 4 d. Is this result reliable? This means do a test! So we compute our usual F-ratio ( signal to noise... ): r2 9/16 df = 1 r2 7/16 4 = = If you are scared of fractions, you can get the same result with a bit more effort (and a pocket calculator): r2.752 df = 1 r = = When we compare this value with the critical value found in the table (critical F for ν 1 = 1 and ν 2 = 4; with α =.05, critical 7.71, with α =.01, critical 21.20). So we cannot reject the null hypothesis for any α level) e. Write the conclusion apa style. This is a possible way of writing:... We found that children who were reported (by their peers) to engage in overt aggressive behavior were also reported to engage in relational aggressive behavior r =.75. But this trend was not significant: r =.75, F(1,4) = 5.14, ns. Another possible way of writing is... We did not find a significant relations (correlation) between overt aggressive behaviors and relational aggressive behavior (as reported by the children s peers): r =.75, F(1,4) = 5.14, ns. (Incidentally, most experimenters would comment on the fact that their design had very low power because of the very small number of children participating in the experiment). 6. (00 points) To see if there is an age related change in altruism in children, Grunberg, Meycock, and Authery (1985) measured the number of pennies children donated to UNICEF when left alone. Six groups of respectively 3, 5, 7, 9, 11, and 13 year old children were given an opportunity to donate any or all of 10 pennies given to them by the experimenter. The number of pennies given by each child was recorded and analyzed. The results of a fictitious replication of this experiment are presented below. For simplicity, the number of children per group in this fictitious replication was equal to two and only four age groups were used: 3, 5, 7, 9. 3 yrs 5 yrs 7 yrs 9 yrs Child one Child two a. Knowing that the Pearson coefficient of correlation between age (X) and number of pennies given (Y ) is equal to.72 and that σ Y = 2.0 and σ X = 2.4, find the equation that predicts the number of pennies given from the age of the children. Here we need to use the alternative formula for b, namely: b = σ Y r XY = =.60 σ X 2.4 In order to compute a, we need to compute first the means of X and Y : M X = and then we need to plug these values in the formula for a: = 48 8 = 6 M Y = = = 3, a = M y b M x = 3 (0.60 6) = = So, the equation that gives the predicted number of pennies from the age of the child is Y = X or Number of Pennies = (0.60 Age ). (The older the children, the more they give; and they give.60 penny for each year of age).

5 5 b. Use the regression line equation to predict the number of pennies given by a 6 year old child We just need to use the previous equation with X = 6: Y = X = = 3. (i.e., we predict that a 6 year old will give 3 pennies). c. Evaluate if this equation gives a better prediction than chance alone at α =.01. (Take into account the fact that rŷ.y = r Y.X ) This means once again compute an F-ratio and perform a statistical test. The F ratio is computed with the standard ( signal to noise... ): r2.722 df = 1 r = = = When we compare this value with the critical value found in the table (critical F for ν 1 = 1 and ν 2 = 6; with α =.05, critical 5.99, with α =.01, critical 13.74). So, we can reject the null hypothesis for α =.05, but not for α =.01. d. Indicate the proportion of variance in number of pennies given explained by the age of the children This means computes r 2. Here it is: r 2 =.72 2 =.5184 So we will say that 52% of the variance of the number of pennies given can be explained by the age of the child.

6 6 7. (00 points ) de Ci and de La (2004) analyzed their data with SAS PROC REG. Below is the listing given by the program. **** SAS Listing **** A simple regression example The REG Procedure Model: MODEL1 Dependent Variable: Y Percent remembered Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model <.0001 Error Corrected Total Root MSE R-Square Dependent Mean Adj R-Sq Coeff Var Parameter Estimates Parameter Standard Variable Label DF Estimate Error t Value Intercept Intercept X Number of years passed Parameter Estimates Squared Semi-partial Variable Label DF Pr > t Type II SS Corr Type II Intercept Intercept 1 < X Number of years passed 1 < a. What is the equation predicting the dependent variable from the independent variable(s) (use their names): Percent Remembered = (Number of years passed ). b. With the alpha level of.01, is the prediction of the dependent variable better than chance? (write your answer apa style). One way of writing could be The number of years passed was found to predict the percent of material remembered by the participants: r 2 = 91, F(1,12) = , p <.01. An alternative way of writing: The number of years passed was found to predict the percent of material remembered by the participants: r 2 =.91, t(12) = 11.15, p <.01. c. What is the proportion of variance of the dependent variable explained by the independent variable(s)? The independent variable (Number of years passed) explains 91% of the variance of the dependent variable (percent remembered).