Click on the Problem Number to find the answer: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 Problem set 2 Answer 1: Match the research problems listed on the left with the most appropriate statistical procedure listed on the right.     b.   comparison of the average IQ's of 10 brothers and their sisters. a. independent-samples t test.    e.   comparison of the variability of IQ distributions of Hispanics and Blacks. b. dependent-samples t test.    a.   comparison of the average number of sick leave days taken by managers and production workers. c. Mann-Whitney U.    c.   comparison of ranked levels of aggression of male and female basketball players. d. Wilcoxon T.    d.   comparison of high school class rankings of male graduates and their girlfriends. e. chi-square test for two independent samples. Explanation: Brothers and sisters are related by virtue of family. The data (IQ scores) are usually considered to be interval. The distributions are being compared. Actually what is being examined is the independence of the two distributions. Hence the chi-square test for independence is called for. The number of sick days taken by managers would normally be considered independent of the number of sick days taken by production workers. Ranks are ordinal data. Aggression in males is unrelated to aggression in females. The girlfriends are related to their boyfriends. The data are ranks. Back Problem set 2 Answer 2: Ten males and ten females were interviewed for a mid-level management position. Given below, for each candidate, is the duration of the interview in minutes. Is there a significant difference in interview duration for male and female applicants? How strong is the association between sex and interview duration? Males Females 27 35 41 42 42 45 45 32 33 40 30 30 35 36 37 42 40 32 38 38 Answer. To answer the first part of this item we need to run an independent sample t test. First enter the data into SPSS, making sure to include a group identifier variable. Then run SPSS®Analyze®Compare Means®Independent Samples t-test. This yields the following table of descriptive statistics: Next, we note that the Levene's test for equality of variances is NOT significant, meaning that we can conclude that the variances across groups are not unequal (i.e., they are homogeneous). This allows us to use the usual t test. The remainder of the table showing the results of the independent samples t test is shown below: Assuming equal variances, t, with 18 degrees of freedom, is -.421. The table shows that the probability of a t this large, assuming the null hypothesis of no differences between group means is true, is .679. In other words, if the null hypothesis is true (i.e., that in the population there is no difference between males and females in interview duration) than we could expect a difference as large as that observed (a difference of 1.0) to occur by chance, due to sampling error, nearly 7 out of ten times. We thus conclude that there is NO difference between the males and females in terms of length of interview. To answer the second part of the question we need to compute omega-squared (page 217), where omega-squared = (t2 - 1)/ (t2 + df + 1)                         = .177 / 19.177                         = .009. Omega-squared is interpreted as a correlation coefficient where 0.0 indicates no relationship and 1.0 indicates a perfect relationship. Hence, we conclude that there is virtually no relationship between sex and length of interview duration. Back Problem set2 Answer 3: An elementary school teacher rank-ordered pupils based on their levels of playground aggression (the more aggressive the higher the rank). She then asked students whether or not they watch professional wrestling on TV. Given below are the rank-orders of pupils who do watch wrestling and those who do not. Do children who watch professional wrestling on TV show higher levels of aggression than those who do not? Aggression Rank-orders Watch Professional Wrestling on TV Do NOT watch Professional Wrestling on TV 9 3 5 12 6 7 11 2 14 4 8 10 15 1 13 Answer. The data are ranks. The groups are independent. Hence the Mann-Whitney U statistic (non-parametric test analogous to the independent t test) is indicated. This is easily computed by hand following the the details given in the following resource:  Mann-Whitney U Test. U = N1N2 + [N1(N1 + 1) / 2] - ER     = 8 x 7 + [8(9)/2] - 81     = 56 - 45     = 11 You get the same result with SPSS after entering the data (and a group identifier, or course) and running Analyze®Nonparametric Test®2 Independent Samples®Mann-Whitney U. From a table of critical values for U (with N1 = 8, and N2 = 7) we learn that the value obtained for U is NOT statistically significant at the at the .05 level (recall that to be statistically significant, U must be equal to or smaller than the critical value listed in the table). Since 11 is greater than 10 we RETAIN the null hypothesis. On the basis of a 2-tailed test, we have no basis for concluding that students who watch TV wrestling are more likely to engage in playground aggression. On the other hand, there may be merit in an argument for a 1-tailed test. In this case, the critical value for U, given in the table, is 13 (at the .05 level). Since 11 is less than 13, the result is statistically significant. If you have already computed the Mann-Whitney U, you can determine its level of significance HERE. Back Problem set 2 Answer 4: Samples of freshmen, sophomore, junior, and senior college woman were surveyed as to the color of their hair. The results were as shown below. Is there evidence to support the statement, "The color of college women's hair is dependent upon their level of academic classification"? How strong is the relationship between level of academic classification and color of hair?     Black Brown Red Blond Freshmen 6 2 2 12 Sophomores 6 2 3 12 Juniors 5 9 3 9 Seniors 2 11 2 3 Answer. The data are categorical (i.e., nominal); hence the chi-square test of independence (chi-square test for two independent samples) is called for. Details can be found in Chi-square Test of Equality of of Distributions (Independence) or in A  Chi Square Calculator. Applying the Chi-square formula yelds  chi-square = 20.1 The degrees of freedom are (number of rows -1) x (number of columns -1), or (4-1) x (4-1) = 9. A chi-square table would provide critical values for chi-square statistic. When using such a table we find that the critical value for a chi-square, with 9 degrees of freedom and alpha = .05, is 16.919. Since our obtained value of chi-square exceeds the critical value we REJECT the null hypothesis and conclude that hair color is NOT independent of academic classification. In other words there is a relationship between hair color and academic classification. It appears from the data that blond and black hair is more prevalent among lower classmen. The strength of the relationship between academic classification and hair color is given by Cramer's V statistic (see When to use Cramer's V  for a discussion on when to use it, and Phi vs Cramer's V  for a comparison of Phi and V ). The solution for V  yields V = .274, indicating a mild relationship. You can get te same result using SPSS. In the SPSS Data Editor set up three columns: AcadClass (for Academic Classification), Color (for Hair Color), and Frequencies (for the number of cases in each AcadClass by Color cell). Next, enter 16 rows of data. The first row would contain the values (1, 1, 6) for the 6 cases in AcadClass 1 and Color 1. The next row would contain the values (1, 2, 2) for the 2 cases in AcadClass1 and Color 2. The LAST row would contain the values (4, 4, 3) for the 3 cases in AcadClass 4, Color 4. Next, click Analyze®Descriptioves®Crosstabs.... Once the data are entered, click on Data®Weight Cases..., select Weight Cases by and move Frequency into the Frequency Variable window. Click OK. Move AdadClass (or Academic Classification) into the Rows window and Color (or Hair Color) into the Columns window. Click on Statistics and select Chi Square and Phi and Cramer's V. Click Continue. This should yield three tables that look something like the following. From the second table, obtain the value of Chi-Square (Pearson Chi-Square) = 20.1 with 9 degrees of freedom. The probability of obtaining a Chi-Square statistic this large by chance, when the null hypothesis is true, is given as 0.017, which is less than 0.05. We reject the null hypothesis. Again, in the third table, Cramer's V is given as 0.274. (See the earlier discussion for an interpretation.) Back Problem set 2 Answer 5: Below are given the GPA of 25 members of the Class of 2001. Assign rank-orders to each individual, giving a rank of 1 to the lowest GPA. GPAs for Members of the Class of '01 GPA Rank GPA Rank GPA Rank GPA Rank GPA Rank 2.4 6 3.2 16 3.0 13.5 3.6 20 2.7 12 3.5 18 3.8 23 2.1 3 3.9 24 3.7 21.5 2.5 8.5 3.7 21.5 4.0 25 3.5 18 2.3 5 2.5 8.5 2.5 8.5 3.1 15 2.0 2 3.0 13.5 3.5 18 2.2 4 2.6 11 1.8 1 2.5 8.5 Answers given in bold. Back Problem set 2 Answer 6: Match the research problems listed on the left with the most appropriate statistical procedure listed on the right.   b.     Comparison of 10 children's average levels of depression 6 mos., 12 mos., and 24 mos. following the divorce of their parents. a. Completely randomized one-way ANOVA. b. Randomized blocks ANOVA   a.    Comparison of the average IQ's of 5-year-olds from low-, middle-, and upper-socioeconomic classes. c. Kruskal-Wallis one-way ANOVA by ranks.    a.    Comparison of average anxiety levels of males and females who received either cognitive therapy or relaxation training.    c.    Comparison of high school class rankings of White, Black, and Hispanic students.    b.   Comparison of levels of social alienation among junior vs. senior high school gang members vs. nonmembers. Back Problem set 2 Answer 7: In an experimental evaluation of heart rate biofeedback, 30 individuals were instructed to lower their heart rates. Ten were provided with no feedback, ten were given false feedback, and ten received accurate feedback. Listed below are their heart rates in beats-per-minute during a 5-minute target period one hour after the beginning of the training. Do the groups differ significantly? Which group means differ significantly? How strong is the association between the independent and dependent variables? No-Feedback False-Feedback Accurate Feedback 57 65 52 66 67 63 60 58 58 65 63 62 65 70 62 67 63 60 58 62 55 60 58 51 67 65 50 60 60 57 Answer. The first question here calls for a one-way analysis of variance. Enter the data into SPSS including a group-id variable (1, 2, 3). Then run Analyze®Compare Means®One-way ANOVA. Enter the group id variable into the "factor" and the heartbeat data as the dependent variable. Then click on Post-hoc ®Tukey (Note, we could have used Sheffe)®Continue ®OK. This results in the ANOVA table given below. We see immediately that the F ratio is statistically significant (Sig. = .005, which is definitely smaller than .05). We could also have compared the obtained value of F with the critical value of F obtained from a table of the F distribution (or from a calculator for computing critical values of F. In such a table we find that the critical value for F , at alpha = .05 with 2 degrees of freedom for the numerator and 27 degrees of freedom for the denominator, is 3.35. Since our F of 6.393 exceeds the critical value of F we REJECT the null hypothesis which stated that the means of all three groups are equal. Rejecting the null hypothesis does not tell us where (i.e., between which pairs of means) the significant differences occur. To determine this we need the post-hoc test (Tukey). SPSS gives the following table. The table tells us that the no-feedback and false-feedback groups can be considered as two samples drawn from the same population. Only the accurate feedback group can be considered unique. Thus we conclude that the accurate feedback group differs significantly from both no-feedback group and the false-feedback group. The strength of the relationship between independent and dependent variables is given by eta-squared, which, in our text (p. 337 of Srinthall, 8th Ed.) is computed as, η2 = SSbetween / SStota = 226.067 / 703.467 = .32 a strong association. Another way to compute the strength of the relationship is to compute omega-squared. This is computed as follows: omega-squared = [SSbetween - (k -1)MSwithin] / (SStotal + MSwithin)                         = (226.067 - 2 x 17.681) / (703.467 + 17.681)                         = 209.882 / 721.148                         = .291, a strong association. Back Problem set 2 Answer 8: The Braxton Corporation interviewed 30 employees (10 managers, 10 line supervisors, and 10 production workers) to assess job satisfaction levels. Based on these interviews, employees were rank-ordered from least satisfied to most satisfied. These ranks are listed below. Do the groups differ significantly in job satisfaction? Which groups differ significantly? Managers Line Supervisors Production Workers 24 26 1 17 20 12 30 4 7 22 28 10 29 8 2 16 21 18 25 5 19 23 14 3 27 5 13 11 6 9 Answer. Here we are examining the differences in perceived job satisfaction among three independent groups of individuals. The data to be analyzed are ranks. The Kruskal-Wallis One-Way Analysis of Variance by Ranks is the appropriate test procedure. We can use SPSS to compute the statistic (H). Enter the data, along with a group id variable. One column for the group id, one column for the data. Then run Analyze®Nonparametric Tests®K Independent Samples. Enter the ranks as the test variable and the group id as the grouping variable. Define the range of the grouping variable (e.g., min = 1, max = 3). Select Kruskal-Wallis H ® OK. You should get a table like the following. In the table the Chi-Square IS the Kruskal-Wallis H. We see immediately that the test is significant with p = .005, which obviously is less than .05. (Note, we could also have determined whether the Kruskal-Wallis test was significant by consulting the Chi-square table with two degrees of freedom.) Since the test is statistically significant we REJECT the null hypothesis and conclude that the three groups are not all equal in their ratings of job satisfaction. To determine which groups differ from which other groups we need to compute a post hoc test. Our book recommends the Tukey HSD (Tukey in SPSS) test. To compute this test compute a one-way analysis of variance on the data like was done in Problem 7 above. Have SPSS compute the Tukey procedure. You should get a table like this. The table tells us that production workers and line workers form a homogeneous (i.e., similar) group, but that managers are unique. Hence we conclude that managers' satisfaction ratings are different (higher) that those of both production workers and line workers, but that these two latter groups do not differ from each other. Back Problem set 2 Answer 9: A therapist is interested in comparing the effectiveness of three types of biofeedback therapy (muscle relaxation, hand warming, and alpha brainwave training) in the treatment of two types of headaches (migraine and tension). Shown in the table below are the data from this study. Factor A (rows) is type of headache. Factor B (columns) is type of biofeedback therapy. Six patients have been assigned randomly to each treatment combination (cell). The dependent variable is the number of prescription pain pills consumed by each patient during the 30-day treatment period.   Type of Therapy (B) b1:  Muscle Relaxation b2: Hand Warming b3: Alpha Brain Wave Type of headache (A) a1: Migraine 15, 10, 12, 18, 17, 18 8, 6, 10, 9, 7, 8 12, 15, 9, 12, 14, 10 a2: Tension 4, 8, 0, 4, 2, 6 7, 5, 9, 7, 3, 11 6, 2, 10, 6, 3, 9 Determine whether this is a completely randomized or a randomized blocks ANOVA design. Compute an analysis of variance to evaluate the main effects and interaction effects. Plot a line graph of the cell means. Use the eta-square statistic to assess the strength of the Factor A, Factor B, and interaction effect. Give a brief interpretation of the results of the overall analysis. Answers. This is a completely randomized factorial ANOVA design. Each treatment combination (cell) has a different group of individuals in it. Enter the data into the SPSS spreadsheet. You will need one variable (column) for coding Factor A; one variable for coding Factor B, and one variable for the data. Assuming you have entered the data correctly you then run Analyze®General Linear Model®Univeriate. When the the univariate dialog window opens, move the factor coding variables (headache and therapy) into the Fixed Factors window and the data variable into the Dependent Variable window. Click Options®Descriptive statistics®Display®Continue®Plots. Decide which variable you want to use on the horizontal axis and which you want to plot as separate lines. Click Add®Continue. When you are returned to the univariate dialog window click OK.  You should first obtain a table of descriptive statistics similar to the one below. Next you should see a table showing "Tests of Between-Subjects Effects." Double click in this table and edit it so that you are left with a sufficient ANOVA summary table like the following: c, d, and e. The ANOVA table shows that the Factor A (type of headache) main effect is significant (p < .05), but that the Factor B (type of therapy) is NOT significant (p > .05). It appears that the number of prescription pain pills consumed was greater for sufferers of migraine headaches than for sufferers of tension headaches, but that type of therapy made no difference. The presence of a significant interaction (p < .05), however, complicates this interpretation. Since the interaction is significant, we examine the plot of cell means. I have chosen to plot type of therapy (separate lines) across type of headache. This plot is shown below: From the From the plot we conclude the following: (1) hand warming has not appreciable effect for either type of headache; (2) alpha brain wave training is more effective for tension headaches than for migraine headaches (fewer prescription drugs were consumed among sufferers of tension headaches than among migraine sufferers when they receive brain wave training); (3) muscle relaxation is dramatically more effective for tension headache sufferers than for migraine sufferer.   The size of the eta-squared statistics suggest that the significant main effect for headache and the headache by type of therapy are very strong.   Back The first step is to read in the existing data file which should be found HERE. Click on the Variable View tab at the bottom (if you are not already on the Variable View page. Check the labels given to the variables (Ability Group {factor_a| and Textbook Conceptual Level {factor_b}) and the value given in the value column. By clicking on the ellipses in the value column for factor_a you will see that the Low Ability group is coded "1" and the High Ability group, "2." Similarly, you can determine the codes given to the levels of textbook conceptual levels (factor_b). Since this is a 2 x 4, factorial ANOVA design (two levels of ability by four levels of conceptual complexity) the appropriate analytical procedure is a univariate general linear model analysis. Click Analysis®General Linear Model®Univariate to open the Univariate dialog window. The dependent variable is tst_scr (Algebra Test Score) so move tst_scr to the Dependent Variable window. factor_a and factor_b are the independent variables (in the SPSS General Linear Model procedure, these are called "Factors"), and both, in this example, are Fixed. Move both of these to the Fixed Factor(s) window. Under Options select Descriptive Statistics, since we always want a table of descriptive statistics. There are only two ability groups. Hence, if the test for factor_a is significant we will conclude that the means of the two population from which these two group are drawn are different. We won't need a post-hoc test. For factor_b, however, there are four conditions. If we get a significant main effect for this factor we won't know which group(s) are different from witch other groups(s). For this factor, then, we will need a post-hoc test. Some textbooks recommend the Sheffe test. Click on Post Hoc..., move factor_b into the  Post Hoc Tests for: window, and then click on Sheffe®Continue. This brings you back to the Univariate dialog window. Since, in a factorial ANOVA, one of the effects tested is an interaction effect (factor_a by factor_b) it will be useful to have a plot of the interaction, should it prove statistically significant. Click on Plots.... This opens the Univariate: Profile Plots window. We have two choices: we can either plot ability groups (factor_a) as separate lines or type of textbook (factor_b) as separate lines. Let's try both. First move factor_a to the Separate Lines: window, then move factor_b to the Horizontal Axis: window and click Add. This moves factor_b*factor_a to the Plots: window. Next reverse the procedure to move factor_a*factor_b to the Plots: window. Click Continue and then OK. After a few seconds the output window will open. The first table you get is the following, which shows the design of the study. This table should be self-explanatory. The next table you get is a table of descriptive statistics. Edit this table in SPSS so that its heading reads: "Means and Standard Deviations of Algebra Test Scores." You can accomplish this as follows. Double click somewhere in the table. This results in a hatched border around the table. (You may also open a Pivoting Tray window and/or an editing window. If either of these open, close it or both). Then double click on "Descriptive Statistics" and replace it with "Means and Standard Deviations of Algebra Test Scores." Next, while the table is still has a hatched border around it, hold the control and alt keys down an click on the heading, "Mean." This should highlight the "Mean" column. We want to change the number of decimal places for reporting means to two decimal places. Right-click somewhere down in the highlighted area and then left-click on Cell Properties®Number and change the Decimals: to "2," then click OK. Use a similar procedure to change the number of decimal places for the standard deviations to "3." When finished, click outside the table. The table should now look like this: The next table you see is the Tests of Between Subjects Effects table. This is our ANOVA table. However, we do not want everything shown in this table; we have to edit it. Like before, double click in the table. Change the title of the table to 2 x 4 ANOVA of Algebra Test Scores. Don't leave the table. We want to hide some of the rows in the table. Specifically we want to hide the rows headed by "Corrected Model," "Intercept," and "Total." After doing so, we want to change the "Corrected Total" heading to "Total." Follow these steps: While simultaneously holding the Ctrl and Alt keys down, click on the heading, "Corrected Model." The row will be highlighted. Point to View on the menu bar above and click Hide. This hides the Corrected Total row. Follow the same procedures for the rows headed "Intercept," and "Total." Next, double click on "Corrected Total" and change it to "Total." We need to make one final edit before leaving the hatched area. Double click the column heading, "Type III Sums of Squares" and change it to "SS." Click outside the hatched border. The table should now look something like the following. The analysis yields a statistically significant main effect for Ability Level (Factor_A), F(1,56) = 121.437, p < .01, with the High Ability group (M = 30.47) out-scoring the Low Ability group (M = 20.59); a statistically significant main effect due to conceptual level of textbook, F(3,56) = 4.827, p < .01; AND a statistically significant interaction, F(3,56) = 48.518, p< .01. Since the main effect due to textbook was significant, we need to explore further where the differences in means lie. This is why we specified a post-hoc analysis. The table we want is the following. This table of homogeneous subsets tells us that the Very high and Moderately high textbook groups are statistically distinguishable (the form a homogeneous subgroup). The table tells us, also, that the Moderately high, Moderately easy, and Very easy textbook groups are not statistically distinguishable. From this table, we can conclude that the Very high conceptual level textbook group, as a whole, had a lower mean Algebra Test score than both the Very easy and Moderately easy groups. But wait! The interpretation is more complicated than this because of the significant interaction. We examine the interaction by looking at the profile plots. The plots reveal the story. The High Ability group scored higher on the Algebra Test for all the textbooks EXCEPT the Very easy one. With this textbook they scored below the Low ability group. What happened?  Recall that the Low Conceptual Level textbook was structured around drill and practice. Apparently, High ability students are "turned off" by this type of instruction. They may not have paid much attention to it. On the other hand, the drill and practice approach appeared to favor low-ability students. BACK   In a study of reading comprehension, a researcher examines a group of 50 second-graders for whom she has scores on the Test of Phonemic Awareness (PA) that were originally collected when the children were in Kindergarten. She has just administered two standardized tests to the children: a reading comprehension (RC) test, and a word analysis (WA) test. The data are in the SPSS data file Reading Example Dataset, which you can download. The researcher has hypothesized a relationship between early phonemic awareness and later reading achievement.  Help the researcher analyze her data and draw a conclusion. Answer: This problem call for a correlational analysis. First download the data file to the desktop and then start SPSS. To perform a preliminary inspection of the data consider producing scatter graphs of the data. Click Graphics®Scatter... to open the Scater/Dot dialog window. Select Simple and then click on Define.  Move one of the three variables (Phonemic Awareness, say) into the X Axis: window, and another variable (Word Analysis, say) into the Y Axis: widow. Click OK. This generates the scatter graph shown below. The graph shows a general, if not well-defined, trend: As phonemic awareness scores increase so do word analysis scores. You can construct similar graphs for phonemic awareness and reading comprehension; and for word analysis and reading comprehension. In each case you would find a similar trend, although it is most clearly defined in the scatter graph of word analysis with reading comprehension. The next step is to compute a correlation matrix. Click Analyze®Correlate®Bivariate and move all three variables into the Variables: window. Make sure Pearson, Two-tailed, and Flag significant correlations are all checked. Click Options®Means and standard deviations®Continue®OK.  You will get the following correlation table: The table indicates statistically significant (p < .01) correlations between phonemic awareness and reading comprehension (r = .367), between phonemic awareness and word analysis (r = .421), and between reading comprehension and word analysis (r = .586). It can be concluded that there is a moderately strong relationship between kindergarten phonemic awareness and both word analysis and reading comprehension in second grade.  BACK   Given below are GPA distributions for the population of all college sophomores and for a sample of 60 sophomores. Population Sample GPA Percent Frequency 3.0-4.0 18.5 17 2.0-2.9 52.3 23 1.0-1.9 22.7 15 0-.9 6.5 5 Perform a statistical test to determine if the sample of sophomores differs significantly from the population of sophomores. Interpret your results. ANSWER: This problem calls for a chi-square goodness of fit test, Chi-square = sum{ (freq. obs. - freq. exp.)2 / freq. exp}. fo fe = (pop. prop) x (total N) fo - fe (fo - fe)2 (fo - fe)2 / fe 17 .185 x 60 = 11.10 5.90 34.810 3.136 23 .523 x 60 = 31.38 -8.38 70.224 2.288 15 .227 x 60 = 13.62 1.38 1.904 .140 5 .065 x 60 = 1.10 1.10 1.260 .310 SUM (X2)= 5.824 Chi-square = 5.825. Using the Chi-square table (Table 3 on page 404 of our text) we find the critical value for a chi-square with (k - 1) = (4 - 1) = 3 degrees of freedom and level of significance = .05 to be 7.815. Since our obtained value for the chi-square is less than the critical value we RETAIN the null hypothesis (i.e., that the sample was drawn from the hypothesized, or target, population) and conclude that the the sample of sophomores does not differ significantly from the population distribution of sophomores. BACK