EXAM 1 HDFS 503L SPRING 2008 1 EXAM 1 ...

EXAM 1

HDFS 503L SPRING 2008 EXAM 1 SOLUTIONS

QUESTION 1: (4 points) Using PsychInfo, list any two articles that Frederick O. Lorenz wrote on which he was the first author. Please write down the standard APA citation: author(s), year, title, journal, volume(issue), and page numbers. Solution: Many possibilities. Here are two examples: The effects of context on the correspondence between observational ratings and questionnaire reports of hostile behavior: A multitrait, multimethod approach. Lorenz, Frederick O.1; Melby, Janet N.2; Conger, Rand D.3; Xu, Xia4 Journal of Family Psychology. Vol 21(3), Sep 2007, pp. 498-509 The Short-Term and Decade-Long Effects of Divorce on Women's Midlife Health. Lorenz, Frederick O.1; Wickrama, K. A. S.1; Conger, Rand D.2; Elder, Glen H., Jr.3 Journal of Health and Social Behavior. Vol 47(2), Jun 2006, pp. 111-125

QUESTION 2: (6 point) Your professor tells you, “Go find me that article by Frederick O. Lorenz published sometime around November, 1995.” Which article is your professor talking about? Write out the APA citation of the article (author(s), year, title, journal, volume(issue), and page numbers). How many times was this article cited? Write down the name(s) of the author(s) of the most recent article that cited Dr. Lorenz’s November, 1995 article. Solution: The effects of unequal covariances and reliabilities on contemporaneous inference: The case of hostility and marital happiness. Lorenz, Frederick O.1; Conger, Rand D.; Simons, Ronald L.; Whitbeck, Les B. Journal of Marriage & the Family. Vol 57(4), Nov 1995, pp. 1049-1064 Times Cited: 12, 13 Most recently cited by: Work orientation and wives' employment careers: An evaluation of Hakim's preference theory. Kan, Man Yee1 Work and Occupations. Vol 34(4), Nov 2007, pp. 430-462

QUESTION 3: (3 points) Using the grades.sav data file that I provided you at the beginning of this class period, search for a student who got 113 for their TOTAL score. Write down this person’s name. Solution: To find the person with Total score of 113, highlight the TOTAL variable column in the Data View and go to Edit Æ Find and enter 113. This technique was covered on page 40 of your textbook. Three students got 113 Total: Sandra Park, Deanna Webster, Don Prado. 4. (3 points) In the grades.sav data file that I provided you at the beginning of this class period, what is the lowest FINAL exam score in the class? Write down the lowest score and write down the name(s) of the people who got this score. Solution: There are many ways to find the answer to this. One way is to simply SORT the cases in descending order from lowest to highest score. To sort cases, go to DATA Æ SORT CASES. Then select the FINAL variable to sort in ascending order. The lowest scores will be listed at the top. The lowest score is 40. One student got this score: Joe Huang. 1

EXAM 1 HDFS 503L SPRING 2008 5. (6 points) A group of 21 people participated in a fast food taste test. The participants were asked to rate the taste of their cheeseburgers on a scale from 1 to 5 (1 = “tastes extremely bad” to 5 = “tastes excellent”). They were assigned to one of three fast food groups: (1) Dairy Queen, (2) Burger King, or (3) Wendy’s. There are seven people in each experimental condition. You want to set up the data so that you can compare the mean scores among the groups. These scores were recorded for each participant: Dairy Queen: 5, 4, 2, 4, 5, 5, 3 Burger King: 4, 5, 3, 3, 2, 4, 2 Wendy’s: 1, 1, 2, 4, 4, 3, 2 Set up the appropriate variables and data structure and enter the data into SPSS. Give me a screenshot of your Data View AND your Variable View. Solution: This is the same idea as Question 7 from your Chapter 3 homework questions. There, you had two groups of people based on ground type. Here you have three groups of people, but the setup is the same. You must set it up like this to do comparisons of group means later on.

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6. (6 points) Suppose that you are analyzing data from self-report assessments to examine how well a parenting education course works. You want to compare the self-assessments on two different groups of people: (1) those who took the parenting education course, and (2) those who did not take the course (instead, they only got informational brochures about the parenting education program). There are 5 people in each group. To assess whether the parenting education course works, everyone in both groups took the self-assessments twice: (1) once before the course started, and (2) a second time, when the course was completed. Set up the appropriate variables and enter the data into SPSS. Give me a screenshot of your Data View AND your Variable View. Their data are as follows:

Sex Before After

Took Parenting Ed. Course Person Person Person Person 1 2 3 4 M M F M 171 180 201 174 192 185 215 196

Did NOT Take Parenting Ed. Course Person Person Person Person Person Person 5 6 7 8 9 10 F F M M F F 165 154 182 195 203 176 184 152 190 192 200 180

Solution: This is the same idea as Questions 7 & 8 from your Chapter 3 homework questions. It is actually a combination of the two questions. In Question 7, you had two groups of people based on ground type. In Question 8, you had a repeated measures design—the same people measured twice. Here you have two groups of people who are also measured twice. 3

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7. (5 points) Using the grades.sav data file that I provided you at the beginning of this class period, list the ID number, firstname and lastname of all students who got the same scores across all five quizzes: quiz1, quiz2, quiz3, quiz4, quiz5. (I’m not asking for students who got the same scores with each other; instead, I want to know the students who got the same score for all five quizzes). ‘Copy objects’ to get your Case 4

EXAM 1 HDFS 503L SPRING 2008 nd Summaries output (2 table only) into a Word document. Also paste your syntax for any SPSS commands you run into the Word document. Solution: This topic was covered on slide 48 of Lab 4 notes, under the topic of SELECT CASES. You need to go to DATA Æ SELECT CASES, then click on “IF” button under “IF CONDITION IS SATISFIED”. Then do the following:

Note that you cannot do: quiz1=quiz2=quiz3=quiz4=quiz5. Case Summaries 1 2 3 Total

Case Number 36 59 99 N

id 420327 616095 972678 3

lastname BADGER SPRINGER KAHRS

firstname SUZANNA ANNELIES JANN 3

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SYNTAX: USE ALL. COMPUTE filter_$=(quiz1=quiz2 AND quiz2=quiz3 AND quiz3=quiz4 AND quiz4=quiz5). VARIABLE LABEL filter_$ 'quiz1=quiz2 AND quiz2=quiz3 AND quiz3=quiz4 AND'+ 'quiz4=quiz5 (FILTER)'. VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'. FORMAT filter_$ (f1.0). FILTER BY filter_$. EXECUTE . SUMMARIZE /TABLES=id lastname firstname /FORMAT=LIST CASENUM TOTAL /TITLE='Case Summaries' /MISSING=VARIABLE /CELLS=COUNT .

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EXAM 1 HDFS 503L SPRING 2008 QUESTION 8. (6 points) Using the grades.sav data file that I provided you at the beginning of this class period, answer each of the following questions (you do not need to give me your syntax, just write down your answers): a. How many male Black students are there in each class (i.e. freshmen, sophomores, juniors, and seniors)? List each class and the number of Black male students in each class separately. (Note: If there is no output for a certain class then it means there are no cases in that class). b. What is the overall average (mean) Final Exam score of these male Black students? c. What is the average Quiz 2 score for just the male Black students in the junior class? d. Which group has the higher overall GPA: female Hispanic students or male Hispanic students? What are their GPAs? e. Rank order from highest to lowest, the average GPA for the five ethnic groups, and write down the ethnic group names in that order. You can either use OLAP Cubes, SELECT CASES, or CASE SUMMARIES to answer these questions. The fastest way is by using OLAP CUBES. Answer: A) Male Black freshman=0, sophomore=3, junior=6, senior=1 B) Average Final Exam score for Black males = 59.40 C) Average Quiz 2 score for Black male juniors =8.00 D) Female Hispanic GPA = 2.8257; Male Hispanic GPA = 2.9975 E) Native = 2.9660; Hispanic=2.8882; White = 2.8729; Black = 2.6983; Asian = 2.5570 9. (5 points) Suppose that the grades.sav data file that I provided you at the beginning of this class period was entered by hand, so it may contain some errors. As a researcher, you want to check to make sure that none of your assistants made a mistake entering the data. Specifically, check to see whether the TOTAL and PERCENT variables were calculated correctly. If you find any errors, write down the names of the students for whom their TOTAL and PERCENT variables are incorrectly calculated. Also write down what the correct scores (i.e., TOTAL & PERCENT) should be. (See pg. 56 in your book if you don’t remember how to calculate TOTAL and PERCENT – make sure to use your new total variable when you calculate your new percent variable). Don’t just check by eye; use SPSS to identify the incorrect cases. Give me your syntax for all SPSS commands you use. Solution: First you need to COMPUTE new variables called TOTAL2 and PERCENT2. Then SELECT CASES where this new total does not equal the old total (total ~= TOTAL2). Then run Case Summaries on the selected cases. SYNTAX: COMPUTE TOTAL2 = sum(quiz1,quiz2,quiz3,quiz4,quiz5,final) . EXECUTE . COMPUTE PERCENT2 = RND(100*TOTAL2/125). EXECUTE . USE ALL. COMPUTE filter_$=(total ~= TOTAL2). VARIABLE LABEL filter_$ 'total ~= TOTAL2 (FILTER)'. VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'. FORMAT filter_$ (f1.0). FILTER BY filter_$. EXECUTE .

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SUMMARIZE /TABLES=id lastname firstname total TOTAL2 /FORMAT=LIST CASENUM TOTAL /TITLE='Case Summaries' /MISSING=VARIABLE /CELLS=COUNT . Case Summaries 1 2 Total

Case Number 62 89 N

id 380157 972678 2

lastname LUTZ KAHRS 2

firstname WILLIAM JANN 2

total 118 93 2

TOTAL2 108.00 103.00 2

There were two students in the dataset whose TOTAL score was computed incorrectly: William Lutz and Jann Kahrs. Their correct TOTAL scores are 108 and 103 respectively. SYNTAX: FILTER OFF. USE ALL. EXECUTE . USE ALL. COMPUTE filter_$=(percent ~= PERCENT2). VARIABLE LABEL filter_$ 'percent ~= PERCENT2 (FILTER)'. VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'. FORMAT filter_$ (f1.0). FILTER BY filter_$. EXECUTE . SUMMARIZE /TABLES=id lastname firstname percent PERCENT2 /FORMAT=LIST CASENUM TOTAL /TITLE='Case Summaries' /MISSING=VARIABLE /CELLS=COUNT .

There is only one student, Jann Kahr, whose PERCENT was incorrectly calculated. The correct PERCENT score should be 82, not 74. Case Summaries 1 Total

Case Number 89 N

id 972678 1

lastname KAHRS 1

firstname JANN 1

percent 74 1

PERCENT2 82.00 1

10. (6 points) What is the SPSS default method (most commonly used) method for replacing missing values for continuous variables? When might this procedure not be a good idea to use? Give me an example. What other missing data replacement method might you use? Explain why and how you would do it. Answer: Page 50 of your textbook addresses this issue. Replace with SERIES MEAN is the default method and most commonly used. This may not be a good idea if there is a lot of missing values in your dataset or if it doesn’t makes sense to replace with the mean value of other subjects. For example, it 7

EXAM 1 HDFS 503L SPRING 2008 doesn’t make sense to replace a student’s missing QUIZ5 score with the average QUIZ5 score of the rest of the class. Instead, it is better to replace with the average score of this student’s other quizzes. The better method of data replacement is to create a regression equation with the variables of interest. Then replace the missing values with the predicted scores from the regression equation.

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