A Visual Approach to SPSS for Windows

Answers to Odd-Numbered Exercises

for

A Visual Approach to SPSS for Windows

Leonard Stern Eastern Washington University

TABLE OF CONTENTS

INTRODUCTION ......................................................................................................................................... I CHAPTER 2 ................................................................................................................................................. 2

Exercise 2.1........................................................................................................................................... 2 Exercise 2.3........................................................................................................................................... 2 Exercise 2.5........................................................................................................................................... 3 Exercise 2.7........................................................................................................................................... 5

CHAPTER 3 ............................................................................................................................................... 10 Exercise 3.1......................................................................................................................................... 10

CHAPTER 4 ............................................................................................................................................... 12 Exercise 4.1......................................................................................................................................... 12 Exercise 4.3......................................................................................................................................... 13


CHAPTER 6 ............................................................................................................................................... 20 Exercise 6.1......................................................................................................................................... 20 Exercise 6.3a....................................................................................................................................... 22 Exercise 6.3b....................................................................................................................................... 25 Exercise 6.5......................................................................................................................................... 29




CHAPTER 10 ............................................................................................................................................. 51 Exercise 10.1....................................................................................................................................... 51 Exercise 10.3....................................................................................................................................... 55 Exercise 10.5....................................................................................................................................... 59



CHAPTER 13 ............................................................................................................................................. 72

Exercise 13.1....................................................................................................................................... 72 Exercise 13.3....................................................................................................................................... 74 Exercise 13.5....................................................................................................................................... 76

CHAPTER 14 ............................................................................................................................................. 79 Exercise 14.1....................................................................................................................................... 79 Exercise 14.3....................................................................................................................................... 85



CHAPTER 17 ........................................................................................................................................... 104 Exercise 17.1..................................................................................................................................... 104 Exercise 17.3..................................................................................................................................... 105 Exercise 17.5..................................................................................................................................... 106 Exercise 17.7..................................................................................................................................... 108

CHAPTER 18 ........................................................................................................................................... 110 Exercise 18.1..................................................................................................................................... 110 Exercise 18.3..................................................................................................................................... 117

CHAPTER 19 ........................................................................................................................................... 125 Exercise 19.1..................................................................................................................................... 125 Exercise 19.3..................................................................................................................................... 130

i

Introduction This supplement provides answers to the odd-numbered exercises presented at the end of each chapter. Because some exercises require that tasks be completed (rather than answers given), the material provided here sometimes demonstrates how the tasks are to be accomplished and may provide all or some of the resulting output produced. The answers shown here often include a screen shot of the Data View of the SPSS Statistics Data Editor and may be preceded by a display of the screens required to produce the data. Exercises that call for exploration such as the exercises at the end of Chapters 1 and 2 are not answered.

2

Chapter 2 Exercise 2.1 The data are type in using the keyboard to look like this:

Exercise 2.3 The data file is opened starting from the File menu:

1. From the File menu select Open and then Data.

The drop-down menu is used to select Excel as the file type.

The desired file can be opened by double-clicking it.

3

The file is read into SPSS by clicking OK in the window that opens:

The file is then read in:

Exercise 2.5 A blank file is opened from the File menu by selecting New and then Data:

4

The data are typed in using the keyboard. The finished file is shown below:

The file is saved using the name EX25.sav.

The file name is entered here.

5

Exercise 2.7 The file is opened from the File menu by selecting Open and then Data:

A .dat file is selected as the file type:

6

The Text Import Wizard is used to read in the file:

The file name is entered here.

10

Chapter 3 Exercise 3.1 1. The data are typed in using the keyboard to look like this:

2. The new variable is created by highlighting column 2 and choosing Insert Variable from the Edit menu.

3. The variable is deleted by highlighting the column and choosing Cut from the Edit menu.

11

4. Values of the new variable are type in from the keyboard.

5. The file is saved using Save As from the File menu.

12

Chapter 4 Exercise 4.1 1. The file looks like this:

The file’s Variable View looks like this:

13

Exercise 4.3 The file contains 20 cases. Its Data View looks like this:

The file’s Variable View looks like this:

14

Chapter 5 Exercise 5.1 1. The file is opened using the File menu and selecting Open and then Data. 2. Select Descriptive Statistics then Frequencies from the Analyze menu and run the

Frequencies procedure.

Select the variable tvhrs and transfer it to the Variables pane

Click OK.

15

3. Replace the word Frequencies in the title with the new title information.

4. In the outline pane, click and drag the icon for the table to the icon for Title.

16

5. Click on the table to be copied then select Copy from the Edit menu.

Paste the table into the desired word-processing document.

17

Exercise 5.3 The file can be opened using the File menu in SPSS and selecting Open and then Output. Then the Export procedure is selected from the File menu.

18

Select All

Verify this option is selected.

Insert the proper file name here.

19

Note that the wide table has been shrunk to fit the page.

20

Chapter 6 Exercise 6.1

2. Using the Chart Builder select Line as the chart type.

3. Drag the icon of the multiple line plot from here to here.

1. From the Graphs menu choose the Chart Builder.

21

4. Drag the nominal scale condition to the x-axis drop zone.

6. Drag the scale variable dv to the y-axis drop zone.

5. Drag the nominal scale gender to the grouping drop

7. Click OK.

22

Exercise 6.3a


23

2. From the Gallery tab select Bar.

3. Drag the icon of the bar chart from here to the chart canvas.

4. Drag the nominal scale highest to the x-axis drop zone.

24

5. Press the Titles/Footnotes tab and select Title 1.

6. Type the title in the Content field of the Element Properties window.

7. Click Apply then OK.

25

Exercise 6.3b


26

2. From the Gallery tab select Bar.

3. Drag the icon of the clustered bar chart from here to the chart canvas.

4. Drag the nominal scale highest to the x-axis drop zone.

5. Drag the nominal scale gender to the grouping drop zone.

27

6. From the Statistic drop-down menu of the Element Properties window select Percentage ().

7. Click Set Parameters.

8. Select Total for Each Legend Variable Category (same fill color).

10. Click Apply. 9. Click Continue.

11. Press the Titles/Footnotes tab and select Title 1.

28


13. Click Apply then OK.

29

Exercise 6.5


2. From the Gallery tab select Scatter/Dot.

3. Drag the icon of the simple scatterplot from here to the chart canvas.

30

4. Drag homework1 to the x-axis drop zone.

5. Drag quiz1 to the y-axis drop zone.

6. Click the Titles/Footnotes tab and select Title 1.

31


13. Click Apply then OK on the Chart Builder window shown on the previous page.

32

Chapter 7 Exercise 7.1 1. The Recode into Different Variables procedure provides the requested

transformations:

33

2. The Compute Variable expression shown below creates the variable newsfreq1.

The variables created in problems 1 and 2 are shown below:

34

Exercise 7.3 1. The variable seqnum is set to the value of $Casenum that is obtained from the

miscellaneous function group.

35

2. The variable nocontact can be created in two steps. First, all cases can be declared to have the value of nocontact be 0:

Then, using the IF button, nocontact can be set to 1 when two other conditions are met

36

3. The desired selections are made using the Select Cases procedure with an IF statement:

Press this button to form the If statement.

Select this option.

37

This statement appears after the window below has been filled out.

38

These 2 cases are selected (as well as case 347).

39

Chapter 8 Exercise 8.1 1. The Frequencies procedure provides the requested output:

40

The output is shown next:

Statistics Body temperature (degrees Fahrenheit)

Valid 130N

Missing 0Mean 98.249Std. Deviation .7332Skewness -.004Std. Error of Skewness .212Kurtosis .780Std. Error of Kurtosis .422

25 97.80050 98.300

Percentiles

75 98.700

41

Body temperature (degrees Fahrenheit)

Frequency Percent Valid Percent

Cumulative Percent

96.3 1 .8 .8 .8

96.4 1 .8 .8 1.5

96.7 2 1.5 1.5 3.1

96.8 1 .8 .8 3.8

96.9 1 .8 .8 4.6

97.0 1 .8 .8 5.4

97.1 3 2.3 2.3 7.7

97.2 3 2.3 2.3 10.0

97.3 1 .8 .8 10.8

97.4 5 3.8 3.8 14.6

97.5 2 1.5 1.5 16.2

97.6 4 3.1 3.1 19.2

97.7 3 2.3 2.3 21.5

97.8 7 5.4 5.4 26.9

97.9 5 3.8 3.8 30.8

98.0 11 8.5 8.5 39.2

98.1 3 2.3 2.3 41.5

98.2 10 7.7 7.7 49.2

98.3 5 3.8 3.8 53.1

98.4 9 6.9 6.9 60.0

98.5 3 2.3 2.3 62.3

98.6 10 7.7 7.7 70.0

98.7 8 6.2 6.2 76.2

98.8 10 7.7 7.7 83.8

98.9 2 1.5 1.5 85.4

99.0 5 3.8 3.8 89.2

99.1 3 2.3 2.3 91.5

99.2 3 2.3 2.3 93.8

99.3 2 1.5 1.5 95.4

99.4 2 1.5 1.5 96.9

99.5 1 .8 .8 97.7

99.9 1 .8 .8 98.5

100.0 1 .8 .8 99.2

100.8 1 .8 .8 100.0

Valid

Total 130 100.0 100.0

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2. The percentile rank of the temperature 98.6 is 70. 3. Examination of the histogram and comparison of the index of skewness and kurtosis

with each measure’s standard error indicates that, although there are too many scores in the center of the distribution, the distribution is approximately normal.

Exercise 8.3 1. The Frequencies:Statistics window is used to request the table:

The Frequencies: Format window is completed as shown below:

Selecting this option puts the variables in the same table.

43

Statistics

MeanTime2003 MeanTime2002 MeanTime2001

Valid 65 65 65 N

Missing 35 35 35 Mean 23.328 23.551 23.54 Std. Deviation 4.0805 3.9571 3.899

25 20.650 21.100 21.05 50 22.600 23.300 22.80

Percentiles

75 25.550 25.300 25.65

2. Mean commute times do not seem to have appreciably changed over the 3-year

period.

44

Chapter 9 Exercise 9.1 1. The requested output can be obtained using the Explore procedure:

The output is shown below:

Case Processing Summary

Cases

Valid Missing Total

N Percent N Percent N Percent

Annual Rate of Change 16 100.0% 0 .0% 16 100.0%

Descriptives Statistic Std. Error

Mean -.0089 .01904

Lower Bound -.0495 95% Confidence Interval for Mean Upper Bound .0317

5% Trimmed Mean -.0121

Median -.0136

Variance .006

Std. Deviation .07618

Minimum -.14

Maximum .18

Range .32

Interquartile Range .09

Skewness .606 .564

Annual Rate of Change

Kurtosis 1.453 1.091

46

2. The shape of the distribution, although not perfectly normal, does not deviate significantly from normal according to measures of skewness and kurtosis (the z value for each measure, obtained by dividing its value by its standard error, is close to 1). The mean change is -.01 and the standard deviation is .08.

3. The year with the highest rate of change in water usage was 1994. The data point

represents an outlier, one that falls from 1.5 to 3 units of interquartile range beyond the upper edge of the box.

4. Based on the output for the variable usageperc that is shown below, the distribution

can be described as being approximately normal, having a mean of 114,841.30, and a standard deviation of 8,685.66.

Descriptives

Statistic Std. Error Mean 114841.3394 2171.41593

Lower Bound 110213.0759 95% Confidence Interval for Mean Upper Bound 119469.6029 5% Trimmed Mean 114805.6318 Median 113709.9997 Variance 7.544E7 Std. Deviation 8685.66373 Minimum 99311.40 Maximum 131014.01 Range 31702.61 Interquartile Range 12810.62 Skewness .265 .564

Water Usage per Capita

Kurtosis -.347 1.091

48

Exercise 9.3 1. Z-scores are requested using the Descriptives procedure:

2. Histograms can be requested from the Frequencies procedure.

50

The distributions are identical in shape.

51

Chapter 10 Exercise 10.1 1. The scatterplot of car weight as a function of wheelbase is shown below:

2. The output of the regression analysis is shown below:

Descriptive Statistics

Mean Std. Deviation N

Weight 3581.2146 759.98462 424WheelBase 108.1722 8.34544 424

Correlations

Weight WheelBase

Weight 1.000 .761 Pearson Correlation

WheelBase .761 1.000 Weight . .000 Sig. (1-tailed) WheelBase .000 . Weight 424 424 N WheelBase 424 424

52

Variables Entered/Removedb

Model Variables Entered

Variables Removed Method

1 WheelBasea . Enter a. All requested variables entered. b. Dependent Variable: Weight

\

Model Summaryb

Model R R Square Adjusted R

Square Std. Error of the

Estimate

1 .761a .579 .578 493.80747 a. Predictors: (Constant), WheelBase b. Dependent Variable: Weight

ANOVAb

Model Sum of Squares df Mean Square F Sig.

Regression 1.414E8 1 1.414E8 579.924 .000a

Residual 1.029E8 422 243845.818 1

Total 2.443E8 423 a. Predictors: (Constant), WheelBase b. Dependent Variable: Weight

Coefficientsa

Unstandardized Coefficients Standardized Coefficients

Model B Std. Error Beta t Sig.

(Constant) -3913.224 312.133 -12.537 .0001

WheelBase 69.282 2.877 .761 24.082 .000a. Dependent Variable: Weight

53

Casewise Diagnosticsa

Case Number Std. Residual Weight Predicted Value Residual

300 3.263 7190.00 5578.4787 1611.52125 305 3.628 6400.00 4608.5238 1791.47624 307 3.316 5969.00 4331.3938 1637.60624 323 3.531 5590.00 3846.4163 1743.58373 326 3.193 5423.00 3846.4163 1576.58373 338 3.126 5390.00 3846.4163 1543.58373 348 3.161 4576.00 3015.0263 1560.97373 418 -3.069 4548.00 6063.4562 -1515.45624 a. Dependent Variable: Weight

Residuals Statisticsa

Minimum Maximum Mean Std. Deviation N

Predicted Value 2252.9187 6063.4561 3581.2146 578.19311 424Residual -1515.45630 1791.47620 .00000 493.22343 424Std. Predicted Value -2.297 4.293 .000 1.000 424Std. Residual -3.069 3.628 .000 .999 424a. Dependent Variable: Weight

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Here are the answers to the questions: a. Based on the two scatterplots, the relation between car weight and length of

wheelbase appears to be linear. b. The r value is .76. c. From the ANOVA table, the relationship is statistically significant, F (1,422) =

579.92, p < .001. d. The equation is

22.3913 Wheelbase69.2 Weight −×= e. Based on the histogram showing the distribution of standardized residuals and the

normal P-P plot, the assumption of normality appears to hole. The plot showing the standardized predicted values on the x-axis and the standardized residuals on the y-axis does not reveal serious threats to the assumption of homogeneity of residuals.

Exercise 10.3 1. The Model Summary and ANOVA tables from the linear regression analysis provide

an answer:

Model Summary



Estimate

1 .081a .007 -.014 75.34596 a. Predictors: (Constant), Average pupil/teacher ratio

56

ANOVAb


Regression 1811.030 1 1811.030 .319 .575a

Residual 272496.650 48 5677.014 1

Total 274307.680 49 a. Predictors: (Constant), Average pupil/teacher ratio b. Dependent Variable: Mean total SAT in 1994-1995

The relation between mean SAT score and mean pupil/teacher ratio was not significant, r = .08, F < 1.

2. A new variable LOGCOST was created with the Compute Variable procedure.

From the Model Summary, ANOVA, and coefficients tables of a linear regression analysis, the necessary data are available to draw a conclusion about the relation.

Model Summaryb



Estimate

1 .392a .154 .136 69.55055 a. Predictors: (Constant), LOGCOST b. Dependent Variable: Mean total SAT in 1994-1995

57

ANOVAb


Regression 42118.295 1 42118.295 8.707 .005a

Residual 232189.385 48 4837.279 1

Total 274307.680 49 a. Predictors: (Constant), LOGCOST b. Dependent Variable: Mean total SAT in 1994-1995

Coefficientsa



(Constant) 1202.433 80.754 14.890 .0001

LOGCOST -310.851 105.346 -.392 -2.951 .005a. Dependent Variable: Mean total SAT in 1994-1995

There was in inverse relation between log cost and mean SAT, r = -.39, F (1, 48) = 8.71, p < .01. Thus, the more money states spent, the lower the mean SAT score was. The scatterplot below illustrates the relation:

The negative sign shows the relation is an inverse one.

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Histograms of the distributions of residuals before and after the log transformation of cost show a slight improvement in the degree to which the residuals are normally distributed around the regression line.

3. The Model Summary, ANOVA, and coefficients tables of a linear regression analysis

allow a conclusion to be drawn about the relation between the variables.

Model Summaryb



Estimate

1 .887a .787 .783 34.89065 a. Predictors: (Constant), Percent of all eligible students taking the SAT in 1994-1995 b. Dependent Variable: Mean total SAT in 1994-1995

ANOVAb


Regression 215874.533 1 215874.533 177.330 .000a

Residual 58433.147 48 1217.357 1

Total 274307.680 49 a. Predictors: (Constant), Percent of all eligible students taking the SAT in 1994-1995 b. Dependent Variable: Mean total SAT in 1994-1995

Coefficientsa



(Constant) 1053.320 8.211 128.278 .0001

Percent of all eligible students taking the SAT in 1994-1995

-2.480 .186 -.887 -13.317 .000

a. Dependent Variable: Mean total SAT in 1994-1995

Before log transform. After log transform.

59

There was an inverse relation between the percent of students taking the SAT and the mean SAT score, r = -.89, F (1, 48) = 177.33, p < .001.

If the percent of students taking the SAT is related to the other predictors, it could

considerably change their relation with the mean SAT.

Exercise 10.5 Using the Bivariate Correlations procedure, the following table of r values was obtained:

Correlations

Full-Scale IQ

Head Circumference

(cm) Total Surface Area (cm2)

Total Brain Volume (cm3)

Body Weight (kg)

Pearson Correlation 1 .138 -.291 -.063 -.003

Sig. (2-tailed) .562 .213 .791 .991

Full-Scale IQ

N 20 20 20 20 20Pearson Correlation .138 1 .337 .508* .240Sig. (2-tailed) .562 .147 .022 .308

Head Circumference (cm)

N 20 20 20 20 20Pearson Correlation -.291 .337 1 .601** .064Sig. (2-tailed) .213 .147 .005 .788

Total Surface Area (cm2)

N 20 20 20 20 20Pearson Correlation -.063 .508* .601** 1 .208Sig. (2-tailed) .791 .022 .005 .379

Total Brain Volume (cm3)

N 20 20 20 20 20Pearson Correlation -.003 .240 .064 .208 1Sig. (2-tailed) .991 .308 .788 .379

Body Weight (kg)

N 20 20 20 20 20*. Correlation is significant at the 0.05 level (2-tailed). **. Correlation is significant at the 0.01 level (2-tailed). From column 1 or row 1 it can be seen that none of the variables is significantly related to IQ.

No p < .05

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Chapter 11 Exercise 11.1

($CASENUM NE 69) AND ($CASENUM NE 70) AND ($CASENUM NE 94)

This statement in the Select Cases procedure omits cases 69, 70, and 94.

Use the Linear Regression procedure for the analysis.

61

The output requested in the Plots window are shown below:

The plots allow the first question to be answered:

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1. Yes, the distributional assumptions appear to have been met. The histogram showing the standardized residuals of HiwayMPG appears to be approximately normal, and the points in the normal P-P plot fall close to the diagonal. The scatterplot of predicted and observed standardized values of HiwayMPG show no serious violations of linearity or homogeneity of variance. However, there appear to be one or two outliers.

Questions 2 and 3 can be answered from the Correlations table:

Correlations

HiwayMPG Engine Cylinders HP Weight WheelBase Length

HiwayMPG 1.000 -.730 -.643 -.662 -.822 -.470 -.375

Engine -.730 1.000 .899 .768 .810 .628 .623

Cylinders -.643 .899 1.000 .747 .705 .525 .531

HP -.662 .768 .747 1.000 .623 .391 .370

Weight -.822 .810 .705 .623 1.000 .749 .648

WheelBase -.470 .628 .525 .391 .749 1.000 .865

Pearson Correlation

Length -.375 .623 .531 .370 .648 .865 1.000HiwayMPG . .000 .000 .000 .000 .000 .000Engine .000 . .000 .000 .000 .000 .000Cylinders .000 .000 . .000 .000 .000 .000HP .000 .000 .000 . .000 .000 .000Weight .000 .000 .000 .000 . .000 .000WheelBase .000 .000 .000 .000 .000 . .000

Sig. (1-tailed)

Length .000 .000 .000 .000 .000 .000 .HiwayMPG 386 386 386 386 386 386 386Engine 386 386 386 386 386 386 386Cylinders 386 386 386 386 386 386 386HP 386 386 386 386 386 386 386Weight 386 386 386 386 386 386 386WheelBase 386 386 386 386 386 386 386

N

Length 386 386 386 386 386 386 386

2.

r values for HiwayMPG and each predictor variable Engine -.730 Cylinders -.643 HP -.662 Weight -.822 WheelBase -.470 Length -.375

63

3. The two predictors that are most strongly related are Cylinders and Engine (r =

.899). Output needed to answer question 4 is shown below:

Model Summaryb



Estimate

1 .872a .760 .756 2.47007 a. Predictors: (Constant), Length, HP, Weight, Cylinders, WheelBase, Engine b. Dependent Variable: HiwayMPG

ANOVAb


Regression 7315.374 6 1219.229 199.833 .000a

Residual 2312.367 379 6.101 1

Total 9627.741 385 a. Predictors: (Constant), Length, HP, Weight, Cylinders, WheelBase, Engine b. Dependent Variable: HiwayMPG

4. For all six predictors combined, R2 = .760, a value that is highly significant, F (6,

379) = 199.833, p < .001. The Coefficients table gives information for questions 5 and 6:

Coefficientsa

Unstandardized Coefficients

Standardized Coefficients Correlations

Model B Std. Error Beta t Sig. Zero-order Partial Part

(Constant) 28.154 2.431 11.580 .000 Engine -.970 .368 -.197 -2.639 .009 -.730 -.134 -.066

Cylinders .265 .189 .083 1.401 .162 -.643 .072 .035

HP -.012 .003 -.173 -4.192 .000 -.662 -.210 -.106

Weight -.006 .000 -.845 -16.361 .000 -.822 -.643 -.412

WheelBase .105 .042 .148 2.522 .012 -.470 .128 .063

1

Length .071 .020 .188 3.579 .000 -.375 .181 .090a. Dependent Variable: HiwayMPG 5. HiwayMPG = 28.154 -.970(Engine) +.265(Cylinders) -.012(HP) -.006(Weight)

+.105(WheelBase) +.071(Length)

64

6. The three variables that have the highest semipartial correlations with the dependent variable are Weight, HP, and Length.

Question 7 is answered from information shown in the table below:

Casewise Diagnosticsa

Case Number Make Std. Residual HiwayMPG Predicted Value Residual

13 Honda Civic HX 2dr

3.936 44.00 34.2776 9.72235

47 Toyota Echo 2dr manual

3.042 43.00 35.4870 7.51302

49 Toyota Echo 4dr 3.090 43.00 35.3663 7.63367 97 Volkswagen

Jetta GLS TDI 4dr

6.223 46.00 30.6282 15.37181

341 Chevrolet Tracker

-3.178 22.00 29.8490 -7.84896

a. Dependent Variable: HiwayMPG 7. Two cars could be considered outliers (their |std.residual|≥ 3.3). These are cars 13

and 97, the Honda Civic HX 2dr and the Volkswagen Jetta GLS TDI 4dr.

Exercise 11.3 1. The prediction equation is

Estimated enrollment = -9153.254 + .406 Hgrad + 4.275 Income + 450.125 Unemp. R2 for the equation is .962.

2. To summarize the errors in predicting enrollments using the formula, one could use the standard error of the estimate (which has the value 670.44). Alternatively, and perhaps more appropriate for the question being addressed here, one could use the standard error of the residuals that appears in the Residuals Statistics table (which has the value 633.51). The value is smaller than that of the standard error of the estimate because the denominator of its equation is N – 2 rather than N – k – 1 (in the case of the standard error of the estimate). The value 633.51 is a better answer to our question because the question does not call for generalizing from our sample data to a population value, which is the purpose of the standard error of the estimate. An even better answer to the question could be obtained by saving the unstandardized residuals (available by pressing the Save button on the Linear Regression Window) then transforming these saved scores to their absolute values with a Compute transformation and then reporting a summary measure of their central tendency available in the Explore procedure. The result is a mean absolute deviation of 487.91 (the median absolute deviation is 454.99). Thus, the mean number of students wrongly predicted using this model each year between 1961 and 1989 would be about 488.

3. The administrator’s suggestion could be examined using hierarchical regression in which the dependent variable, enrollment, is assessed in model 1 from the predictor

65

Year (it doesn’t matter if year is coded from 1 to 29 or if it is represented as 1961 to 1989). In model 2, the effect on R2 of introducing the other three variables is assessed. The tables below show the results of these steps:

Model Summaryc

.901a .812 .805 1438.03744 .812 116.375 1 27 .000

.986b .971 .967 594.29952 .160 44.695 3 24 .000

Model12

R R SquareAdjustedR Square

Std. Error ofthe Estimate

R SquareChange F Change df1 df2 Sig. F Change

Change Statistics

Predictors: (Constant), Year data obtained from 1961-1989 recoded as 1-29a.

Predictors: (Constant), Year data obtained from 1961-1989 recoded as 1-29, January unemployment rate (%) for New Mexico, Spring high schoolgraduates in New Mexico , Per capita income in Albuquerque (1961 dollars)

b.

Dependent Variable: Fall undergraduate enrollments, U of New Mexicoc.

ANOVAc

2.4E+008 1 240657781.4 116.375 .000a

55834696 27 2067951.6883.0E+008 282.9E+008 4 72003967.73 203.866 .000b

8476606 24 353191.9183.0E+008 28

RegressionResidualTotalRegressionResidualTotal

Model1

2

Sum ofSquares df Mean Square F Sig.

Predictors: (Constant), Year data obtained from 1961-1989 recoded as 1-29a.

Predictors: (Constant), Year data obtained from 1961-1989 recoded as 1-29,January unemployment rate (%) for New Mexico , Spring high schoolgraduates inNew Mexico , Per capita income in Albuquerque (1961 dollars)

b.

Dependent Variable: Fall undergraduate enrollments, U of New Mexicoc.

For model 2, R2 change is .16, a value that differs significantly from chance, F(3, 24) = 44.70, p < .001. Thus, including the variables Unemp, Hgrad, and Income significantly improves predicted enrollment over and above the linear effect of Year.

Exercise 11.5 1. The parameters of the equation for the males are shown in the table below:

Unstandardized Regression Coefficient Variable

-115.948 (Constant) 1.809 Chest diameter 1.805 Chest depth .961 Bitrochanteric diameter 1.228 Ankle diameter .198 Height .906 Knee diameter

For these data, R2 = .82, F(6, 240) = 177.11, p < .001.

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2. The parameters of the equation for the females are shown in the table below:

Unstandardized Regression Coefficient Variable

-111.257 (Constant) 2.899 Knee diameter 1.551 Chest depth 1.156 Chest diameter .199 Height .706 Bitrochanteric diameter

-1.731 Ankle diameter 1.947 Wrist diameter .366 Biiliac diameter, (pelvic breadth)

For these data, R2 = .81, F(8, 251) = 133.11, p < .001.

3. For males and females, the 4-term equations are:

Males Females

Variable

Unstandardized Regression Coefficient

Unstandardized Regression Coefficient

(Constant) -131.633 -110.809 Chest girth .527 .602 Hip girth .914 .774 Height .378 .276

For males, the standard error is 3.34. For females the standard error is 2.91. Note: Convert inches to cm by multiplying by 2.54; convert lbs to kg by multiplying by .45.

67

Chapter 12 Exercise 12.1 The question may be answered with a chi-square test of independence. The data can be represented by identifying 4 cells in a 2 x 2 table and weighting each by a frequency count as shown below:

Tables that show important output of the analysis are displayed below:

Still dating? * Writing condition Crosstabulation

Writing condition

Emotional Control Total

Count 34 22 56

Expected Count 28.7 27.3 56.0

Yes

Residual 5.3 -5.3 Count 10 20 30


No

Residual -5.3 5.3 Count 44 42 86

Still dating?

Total


Chi-Square Tests

Value df

Asymp. Sig. (2-sided)

Exact Sig. (2-sided)

Exact Sig. (1-sided)

Pearson Chi-Square 5.861a 1 .015 Continuity Correctionb 4.817 1 .028 Likelihood Ratio 5.943 1 .015 Fisher's Exact Test .023 .014

Linear-by-Linear Association 5.793 1 .016 N of Valid Cases 86 a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 14.65. b. Computed only for a 2x2 table

68

Symmetric Measures

Value Approx. Sig.

Phi .261 .015

Cramer's V .261 .015

Nominal by Nominal

N of Valid Cases 86 The analysis shows that emotional writing is significantly related to relationship stability, χ2 (1, N = 86) = 5.86, p < .025, φ = .26. The value of phi indicates a medium effect size for the relationship.

Exercise 12.3 The problem calls for a chi-square goodness-of-fit test. The theoretical values are the percents given for the U.S. in each speed category. Data entry is shown below:

The value labels for SpeedType are shown below:

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The data are first weighted by the frequency counts of the variable California:

Then a chi-square goodness-of-fit test is performed after entering the values of the U.S. percentages for each category of highway speed:

70

These tables were produced:

Speed category of road

Observed N Expected N Residual

<35 134 134.8 -.835.00 158 134.8 23.240.00 94 80.9 13.145.00 127 161.8 -34.850.00 56 45.0 11.055.00 330 341.6 -11.6Total 899

Test Statistics

Speed category of road

Chi-Square 16.701a

df 5Asymp. Sig. .005a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 45.0.

The distribution of fatalities as a function of speed limit differs significantly from that in the U.S. as a whole, χ2 (5, N = 899) = 16.70, p > .01, w = .14.

Exercise 12.5 The data can be read directly from the file HappyMarried.sav.

These are the variable names that are assigned to rows and columns.

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These are the tables that are produced:

General Happiness * Marital Status Crosstabulation

Marital Status

Married Widowed Divorced Separated Never married Total

Count 264 3 32 5 71 375

Expected Count 195.3 11.9 58.1 14.7 95.0 375.0

Very happy

Residual 68.7 -8.9 -26.1 -9.7 -24.0 Count 316 24 119 25 188 672

Expected Count 350.0 21.3 104.2 26.3 170.2 672.0

Pretty happy

Residual -34.0 2.7 14.8 -1.3 17.8 Count 45 11 35 17 45 153

Expected Count 79.7 4.8 23.7 6.0 38.8 153.0

Not too happy

Residual -34.7 6.2 11.3 11.0 6.2 Count 625 38 186 47 304 1200

General Happiness

Total

Expected Count 625.0 38.0 186.0 47.0 304.0 1200.0

Chi-Square Tests

Value df

Asymp. Sig. (2-sided)

Pearson Chi-Square 112.180a 8 .000 Likelihood Ratio 110.902 8 .000 Linear-by-Linear Association 52.891 1 .000 N of Valid Cases 1200 a. 1 cells (6.7%) have expected count less than 5. The minimum expected count is 4.85.

Symmetric Measures

Value Approx. Sig.

Phi .306 .000

Cramer's V .216 .000

Nominal by Nominal

N of Valid Cases 1200 The analysis shows that marital status is significantly related to general happiness, χ2 (8, N = 1,200) = 112.18, p < .001, w = .31. The value of w indicates a medium effect size for the relationship.

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Chapter 13 Exercise 13.1 1. Here is the Recode window that can produce the needed transformations:

2. The Explore procedure indicates that the distribution of the variable mem1 may not be normal for the low and high suggestibility group due to the presence of too many low scores. Histograms and Q-Q plots are shown below for each of these groups:

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3. The new variable mem2 is created using the Compute Variable procedure using the formula mem2 = mem1 * mem1. The Explore procedure for the new variable mem2 shows the assumption of normality is better met.

4. The results of the t-test for independent samples is shown below:

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Group Statistics

group N Mean Std. Deviation Std. Error Mean

1.00 32 .3422 .23731 .04195 mem2

2.00 22 .4364 .23165 .04939

Independent Samples Test

.000 .990 -1.447 52 .154 -.09418 .06510 -.22480 .03645

-1.453 46.009 .153 -.09418 .06480 -.22461 .03626

Equal variancesassumedEqual variancesnot assumed

mem2F Sig.

Levene's Test forEquality of Variances

t df Sig. (2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the

Difference

t-test for Equality of Means

5. The outcome of the t-test can be summarized as follows:

The corrected recognition scores of the low suggestibility participants (those having objective scores on the Barber Suggestibility Scale of 0.0-2.5) and high suggestibility participants (those with objective scores on the Barber Suggestibility Scale of 5.50-8.00) were transformed to their squared values to better meet the assumption that scores in each sample were normally distributed. A t-test for independent samples was used to determine whether the mean squared corrected recognition score of the low suggestibility participants (M = .34, SD = .24) differed from that of the high suggestibility participants (M = .44, SD = .23). The two means were found to not differ significantly, t (52) = 1.45, p > .10. Thus, it appears that the recognition ability of high and low suggestible participants, as defined and measured in this study, did not differ significantly.

Exercise 13.3 1. Output of a single sample t-test performed on the variable MPG and using the

reference value 55 is shown below: One-Sample Statistics

74 47.8743 5.14521 .59812MPGN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

-11.913 73 .000 -7.12568 -8.3177 -5.9336MPGt df Sig. (2-tailed)

MeanDifference Lower Upper


Difference

Test Value = 55

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Using Explore, a histogram and normal Q-Q plot was obtained to check the assumption that the variable MPG was approximately normally distributed:

MPG60.0055.0050.0045.0040.0035.00

Freq

uenc

y

12.5

10.0

7.5

5.0

2.5

0.0

Histogram

Mean =47.87�Std. Dev. =5.145�

N =74

Observed Value65605550454035

Expe

cted

Nor

mal

3

2

1

0

-1

-2

-3

Normal Q-Q Plot of MPG

Here is how the outcome of the test can be described:

The mean combined city/highway mileage of 74 2006 Prius owners was found to be 47.87 (SD = 5.15). From an informal analysis of the variable using a histogram and normal Q-Q plot no serious threats to the assumption of normality were apparent. A single-sample t-test revealed that the mean mpg differed significantly from 55, the combined EPA city/highway mpg estimate, t (73) = 11.91, p = .001. The effect size as measured by d was 1.39, a value that can be considered large. Thus, the actual combined city/highway mileage of the 2006 Prius is significantly different the EPA estimate of 55mpg.

2. The actual percent of highway driving differs significantly from the value 45 used by the EPA. Output of the single sample t-test is shown below:

One-Sample Statistics

69 58.30 23.963 2.885HiwayN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

4.612 68 .000 13.304 7.55 19.06Hiwayt df Sig. (2-tailed)



Difference

Test Value = 45

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Exercise 13.5 1. The variable Dfry was created using the following Compute statement:

2. Data for Pennsylvania (state = 0) were selected using the Select Cases: If window:

A single sample t-test performed on the variable Dfry using the reference value 0:

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The following output was produced:


71 .0192 .06212 .00737DfryN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

2.598 70 .011 .01915 .0045 .0339Dfryt df Sig. (2-tailed)



Difference

Test Value = 0

There was a significant increase in the price of fries. 3. A similar analysis was performed for the restaurants in New Jersey:


295 .0168 .07004 .00408DfryN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

4.131 294 .000 .01685 .0088 .0249Dfryt df Sig. (2-tailed)



Difference

Test Value = 0

Again, there was a significant increase in the price of fries.

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4. A t-test for independent samples was performed for the restaurants in Pennsylvania and New Jersey to compare the value of the variable Dfry.

Group Statistics

71 .0192 .06212 .00737295 .0168 .07004 .00408

StatePennsylvaniaNew Jersey

DfryN Mean Std. Deviation

Std. ErrorMean

Independent Samples Test

1.147 .285 .254 364 .799 .00231 .00907 -.01552 .02014

.274 116.784 .785 .00231 .00843 -.01438 .01899

Equal variancesassumedEqual variancesnot assumed

DfryF Sig.

Levene's Testfor Equality of

Variances

t df Sig. (2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper


Difference

t-test for Equality of Means

The mean value of Dfry did not differ for the two states, t < 1. This may indicate that

the increase in the price of a small fry that appeared related to an increase in minimum wage in New Jersey was due to some other, perhaps more general, cause (and not the increase in minimum wage).

79

Chapter 14 Exercise 14.1 1. The data were labeled and values entered for the variable Major:

2. Cases were selected using the IF option from the Select Cases window:

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3. A one-way ANOVA was run using the following windows:

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Here is the output:

Descriptives Expected grade in class

95% Confidence Interval for Mean

N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum

criminal justice 17 3.029 .5818 .1411 2.730 3.329 2.0 4.0exercise science 11 3.709 .3562 .1074 3.470 3.948 3.0 4.0nursing 25 3.608 .4051 .0810 3.441 3.775 2.5 4.0psychology 12 3.500 .2828 .0816 3.320 3.680 3.0 3.9Total 65 3.454 .4985 .0618 3.330 3.577 2.0 4.0

Test of Homogeneity of Variances

Expected grade in class

Levene Statistic df1 df2 Sig.

1.953 3 61 .131

ANOVA

Expected grade in class

Sum of Squares df Mean Square F Sig.

Between Groups 4.399 3 1.466 7.776 .000Within Groups 11.503 61 .189 Total 15.902 64

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4. The results of the analysis are summarized below:

Differences among the mean expected grades of students in four majors in college (see Table 14.2 below) were assessed with a one-way ANOVA. A Levene test of homogeneity of variance conducted prior to the ANOVA did not indicate the assumption of homogeneity of variance was significantly violated (p > .10). The ANOVA revealed the differences among the mean expected grade of the four majors was significant, F (3, 61) = 7.77, p < .001, η2 = .28.

Table 14.2. Descriptive statistics for ANOVA example.

Major N Mean Standard Deviation

Criminal Justice 17 3.03 0.58 Exercise Science 11 3.71 0.36 Nursing 25 3.61 0.41 Psychology 12 3.50 0.28

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5. This question can be addressed with a Dunnett test (or a Tukey test). A Dunnett test is appropriate because the same group is being contrasted to every other individual group:

The table shown below gives the outcome of the test:

Multiple Comparisons Expected grade in class Dunnett t (2-sided)a

95% Confidence Interval (I) Major in college

(J) Major in college

Mean Difference (I-J) Std. Error Sig. Lower Bound Upper Bound

exercise science criminal justice .6797* .1680 .000 .274 1.086nursing criminal justice .5786* .1365 .000 .249 .908psychology criminal justice .4706* .1637 .015 .075 .866a. Dunnett t-tests treat one group as a control, and compare all other groups against it. *. The mean difference is significant at the 0.05 level.

The analysis shows that the mean expected grade of the criminal justice majors differs significantly from that of every other individual major in the study (all ps < .025).

6. The outcome of the one-way ANOVA performed on the dependent variable

Number is shown below:

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Descriptives

Number of correct answers

18 8.0000 2.14202 .50488 6.9348 9.0652 3.00 11.0011 8.1818 1.32802 .40041 7.2896 9.0740 6.00 10.0027 7.9259 2.60068 .50050 6.8971 8.9547 2.00 12.0012 8.0000 1.53741 .44381 7.0232 8.9768 6.00 10.0068 8.0000 2.10897 .25575 7.4895 8.5105 2.00 12.00

criminal justiceexercise sciencenursingpsychologyTotal

N Mean Std. Deviation Std. Error Lower Bound Upper Bound

95% Confidence Interval forMean

Minimum Maximum



2.955 3 64 .039

LeveneStatistic df1 df2 Sig.

ANOVA


.512 3 .171 .037 .991297.488 64 4.648298.000 67

Between GroupsWithin GroupsTotal

Sum ofSquares df Mean Square F Sig.

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The results show that the mean number of correct answers on the statistical concepts test did not differ significantly among the four majors, F < 1.

Exercise 14.3 1. The data are entered in two columns as shown below (for a subset of the cases):

Suitable variable names and value labels are created:

The analysis is requested with the following windows:

86

This output is produced:

Descriptives

Nwords

95% Confidence Interval for Mean

N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum

High School 7 3.8571 1.57359 .59476 2.4018 5.3125 2.00 6.00Young Adult 7 3.7143 1.38013 .52164 2.4379 4.9907 2.00 6.00Middle Aged 7 6.2857 1.11270 .42056 5.2566 7.3148 5.00 8.00Retired 7 7.4286 1.71825 .64944 5.8395 9.0177 5.00 10.00Total 28 5.3214 2.12661 .40189 4.4968 6.1460 2.00 10.00


Nwords

Levene Statistic df1 df2 Sig.

.493 3 24 .690

ANOVA

Nwords

87

Sum of Squares df Mean Square F Sig.

Between Groups 70.679 3 23.560 10.994 .000Within Groups 51.429 24 2.143 Total 122.107 27

1. The Levene test of homogeneity of variance is not significant, so the assumption of

homogeneity of variance is not rejected. The highly significant F value indicates that the mean number of words spelled correctly is significantly affected by age.

2. Testing if high school students correctly spell a significantly different number of words correctly than every other individual age group calls for a Dunnett test (a Tukey HSD test would also be acceptable). The window that requests this test from the Post Hoc tests is shown below:

88

The data produced by this test is shown next:

Multiple Comparisons Nwords Dunnett t (2-sided)a

95% Confidence Interval (I) Age Category

(J) Age Category

Mean Difference (I-J) Std. Error Sig. Lower Bound Upper Bound

Young Adult High School -.14286 .78246 .996 -2.1042 1.8185Middle Aged High School 2.42857* .78246 .013 .4672 4.3899Retired High School 3.57143* .78246 .000 1.6101 5.5328a. Dunnett t-tests treat one group as a control, and compare all other groups against it. *. The mean difference is significant at the 0.05 level. The test reveals that the mean number of words spelled correctly by the high school students differs significantly from that of the middle aged (p < .025) and the retired groups (p < .001).

89

Chapter 15 Exercise 15.1 1. The data were coded into SPSS by creating three variables:

2. The 2-way ANOVA was run from the General Linear Model using the windows

shown below:

90

The output is shown below:

Between-Subjects Factors

Value Label N

1.00 a1 10Factor A

2.00 a2 101.00 b1 10Factor B 2.00 b2 10

91


Dependent Variable:Dependent variable

Factor A Factor B Mean Std. Deviation N

b1 6.2000 1.30384 5

b2 16.0000 2.34521 5

a1

Total 11.1000 5.46606 10b1 11.8000 1.30384 5b2 9.0000 1.58114 5

a2

Total 10.4000 2.01108 10b1 9.0000 3.19722 10b2 12.5000 4.14327 10

Total

Total 10.7500 4.02460 20

Levene's Test of Equality of Error Variancesa


F df1 df2 Sig.

2.228 3 16 .124Tests the null hypothesis that the error variance of the dependent variable is equal across groups. a. Design: Intercept + A + B + A * B

Tests of Between-Subjects Effects


Source Type III Sum of

Squares df Mean Square F Sig. Partial Eta Squared

Corrected Model 262.150a 3 87.383 30.661 .000 .852Intercept 2311.250 1 2311.250 810.965 .000 .981A 2.450 1 2.450 .860 .368 .051B 61.250 1 61.250 21.491 .000 .573A * B 198.450 1 198.450 69.632 .000 .813Error 45.600 16 2.850 Total 2619.000 20 Corrected Total 307.750 19 a. R Squared = .852 (Adjusted R Squared = .824)

92

The plot was edited to look like this:

The results can be described like this:

The effect of factor A (a1, a2) and factor B (b1, b2) on mean values of the dependent variable was examined using a two-way analysis of variance (ANOVA) for independent groups. Descriptive statistics for the dependent variable as a function of levels of the two factors are shown in Table 15.4. The ANOVA revealed the interaction of factors A and B was significant, F (1, 16) = 69.63, p < .001, partial η2 = .81. In addition, there was a significant main effect of factor B, F (1, 16) = 21.49, p < .001, partial η2 = .57. The main effect of factor A was not significant, F < 1.

93

Table 15.4. Descriptive statistics for the dependent variable number of words recalled as a function of cue condition and cognitive load.

Mean Dependent Variable Values as a Function of Factors A and B Factor A Factor B Mean Std. Deviation N

b1 6.2000 1.30384 5 b2 16.0000 2.34521 5

a1

Total 11.1000 5.46606 10 b1 11.8000 1.30384 5 b2 9.0000 1.58114 5

a2

Total 10.4000 2.01108 10 b1 9.0000 3.19722 10 b2 12.5000 4.14327 10

Total

Total 10.7500 4.02460 20

Exercise 15.3 The arrows show the means compared for each contrast type:

Deviation last

b1 b2 1 2 3 4 Grand

mean

b1 b2 1 2 3 4 Grand

mean

b1 b2 1 2 3 4 Grand

mean

Simple first

b1 b2 1 2 3 4

b1 b2 1 2 3 4

b1 b2 1 2 3 4

Difference none

b1 b2 1 2 3 4

b1 b2 1 2 3 4

b1 b2 1 2 3 4

Helmert none

b1 b2 1 2 3 4 n

b1 b2 1 2 3 4

b1 b2 1 2 3 4

Repeated none

b1 b2 1 2 3 4

b1 b2 1 2 3 4

b1 b2 1 2 3 4

94

Chapter 16 Exercise 16.1 A one-way within-subjects ANOVA was performed to answer the questions in parts 1 and 2 using the variables Oct2004, Oct2005, and Oct2006 as the three variables. The following selections were made to implement the analysis:

Any variable name can be specified here. Press Add

then Define.

95

Transfer these variables by pressing this button.

96

The output produced is shown below:

Within-Subjects Factors Measure:MEASURE_1

time Dependent

Variable

1 Oct2004 2 Oct2005 3 Oct2006


Mean Std. Deviation N

Number of reports of mishandled baggage per 1000 passengers in 2004

4.4794 2.78175 17


5.3800 2.69124 17


8.4818 4.99308 17

Multivariate Testsb

Effect Value F Hypothesis df Error df Sig. Partial Eta Squared

Pillai's Trace .761 23.837a 2.000 15.000 .000 .761

Wilks' Lambda .239 23.837a 2.000 15.000 .000 .761

Hotelling's Trace 3.178 23.837a 2.000 15.000 .000 .761

time

Roy's Largest Root 3.178 23.837a 2.000 15.000 .000 .761a. Exact statistic b. Design: Intercept Within Subjects Design: time

Transfer this factor to the Horizontal Axis then press Add.

97

Mauchly's Test of Sphericityb

Measure:MEASURE_1 Epsilona Within

Subjects Effect Mauchly's W

Approx. Chi-Square df Sig.

Greenhouse-Geisser Huynh-Feldt Lower-bound

time .452 11.898 2 .003 .646 .679 .500Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. a. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table. b. Design: Intercept Within Subjects Design: time

Tests of Within-Subjects Effects

Measure:MEASURE_1



Sphericity Assumed 149.888 2 74.944 32.309 .000 .669

Greenhouse-Geisser 149.888 1.292 115.985 32.309 .000 .669

Huynh-Feldt 149.888 1.358 110.395 32.309 .000 .669

time

Lower-bound 149.888 1.000 149.888 32.309 .000 .669Sphericity Assumed 74.227 32 2.320 Greenhouse-Geisser 74.227 20.677 3.590 Huynh-Feldt 74.227 21.724 3.417

Error(time)

Lower-bound 74.227 16.000 4.639

Tests of Within-Subjects Contrasts

Measure:MEASURE_1

Source time Type III Sum of


Linear 136.160 1 136.160 44.615 .000 .736time Quadratic 13.728 1 13.728 8.649 .010 .351Linear 48.831 16 3.052 Error(time)

Quadratic 25.396 16 1.587

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Tests of Between-Subjects Effects Measure:MEASURE_1 Transformed Variable:Average



Intercept 1906.260 1 1906.260 54.044 .000 .772Error 564.362 16 35.273

1. Because Mauchly’s test of sphericity is significant, the effect of time could be

assessed using either a univariate F based on corrected degrees of freedom (the Huynh-Feldt,{ XE "Huynh-Feldt correction" } Greenhouse-Geisser{ XE "Greenhouse-Geisser correction" }, or lower-bound procedure) or a multivariate statistic. Both approaches indicate the effect of time was highly significant.

2. The linear effect of time is highly significant, F (1, 16) = 44.62, p < .001, partial η2 = .74.

Exercise 16.3 The data should be analyzed as a mixed two-way ANOVA. The between-subjects factor is given the name condition (new, standard) and the within-subjects factor is called display (time1, time2, time3). The question calls for analysis of the main effect of the

99

between-subjects factor, condition. The selections used to analyze the data are shown below:

Press Add then Define.

Transfer these highlighted variables by pressing this button.

Use condition as the between-subjects factor.

100

Output produced for the analysis is shown next:

Within-Subjects FactorsMeasure:MEASURE_1

display Dependent

Variable

1 time1 2 time2 3 time3

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Between-Subjects Factors

Value Label N

1 new 60 processing condition

2 control 60


processing condition Mean Std. Deviation N

new 232.15 82.854 60

control 247.53 100.981 60

time1

Total 239.84 92.298 120 new 238.02 123.236 60 control 436.68 147.955 60

time2

Total 337.35 168.325 120 new 401.78 128.641 60 control 299.27 131.827 60

time3

Total 350.53 139.536 120

Multivariate Testsb

Effect Value F Hypothesis df Error df Sig. Partial Eta Squared

Pillai's Trace .514 61.875a 2.000 117.000 .000 .514

Wilks' Lambda .486 61.875a 2.000 117.000 .000 .514

Hotelling's Trace 1.058 61.875a 2.000 117.000 .000 .514

display

Roy's Largest Root 1.058 61.875a 2.000 117.000 .000 .514Pillai's Trace .585 82.530a 2.000 117.000 .000 .585Wilks' Lambda .415 82.530a 2.000 117.000 .000 .585Hotelling's Trace 1.411 82.530a 2.000 117.000 .000 .585

display * condition

Roy's Largest Root 1.411 82.530a 2.000 117.000 .000 .585a. Exact statistic b. Design: Intercept + condition Within Subjects Design: display

Mauchly's Test of Sphericityb

Measure:MEASURE_1

Epsilona Within Subjects Effect Mauchly's W

Approx. Chi-Square df Sig.

Greenhouse-Geisser Huynh-Feldt Lower-bound

display .984 1.889 2 .389 .984 1.000 .500Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. a. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table. b. Design: Intercept + condition Within Subjects Design: display

102

Tests of Within-Subjects Effects

Measure:MEASURE_1

Source Type III Sum of Squares df Mean Square F Sig.

Partial Eta Squared

Sphericity Assumed 877290.239 2 438645.119 56.781 .000 .325

Greenhouse-Geisser

877290.239 1.968 445669.175 56.781 .000 .325

Huynh-Feldt 877290.239 2.000 438645.119 56.781 .000 .325

display

Lower-bound 877290.239 1.000 877290.239 56.781 .000 .325Sphericity Assumed 1382045.906 2 691022.953 89.451 .000 .431Greenhouse-Geisser

1382045.906 1.968 702088.353 89.451 .000 .431

Huynh-Feldt 1382045.906 2.000 691022.953 89.451 .000 .431

display * condition

Lower-bound 1382045.906 1.000 1382045.906 89.451 .000 .431Sphericity Assumed 1823135.856 236 7725.152 Greenhouse-Geisser

1823135.856 232.280 7848.855

Huynh-Feldt 1823135.856 236.000 7725.152

Error(display)

Lower-bound 1823135.856 118.000 15450.304

Tests of Within-Subjects Contrasts

Measure:MEASURE_1

Source display Type III Sum of Squares df Mean Square F Sig.

Partial Eta Squared

Linear 735048.017 1 735048.017 108.920 .000 .480display Quadratic 142242.222 1 142242.222 16.346 .000 .122Linear 208506.150 1 208506.150 30.897 .000 .208display *

condition Quadratic 1173539.756 1 1173539.756 134.862 .000 .533Linear 796321.833 118 6748.490 Error(display)

Quadratic 1026814.022 118 8701.814

Tests of Between-Subjects Effects Measure:MEASURE_1 Transformed Variable:Average



Intercept 3.443E7 1 3.443E7 1204.435 .000 .911condition 124396.844 1 124396.844 4.352 .039 .036Error 3372790.611 118 28582.971

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Estimated Marginal Means processing condition

Measure:MEASURE_1

95% Confidence Interval processing condition Mean Std. Error Lower Bound Upper Bound

new 290.650 12.601 265.696 315.604 control 327.828 12.601 302.874 352.782

From the table labeled Tests of Between-subjects Effects the effect of condition is found to be significant, F (1, 118) = 4.35, p < .05, partial η2 = .04.

104

Chapter 17 Exercise 17.1 To test this question, a binomial test could be performed. Here, the number of hurricanes is 7 and the number of non-hurricanes is 20. These data could be entered in the SPSS Data Editor in the following way:

Then, the values of the numeric variable StormType would be weighted by values of the variable Number.

The binomial test would be run using the test proportion is .248.

105

The following table indicates the proportion of hurricanes vs. non-hurricanes during 2005 did not differ from the historical proportion.


N Mean Std. Deviation Minimum Maximum

StormType 27 1.7407 .44658 1.00 2.00

Binomial Test

Category N Observed Prop. Test Prop.

Asymp. Sig. (1-tailed)

Group 1 hurricane 7 .259 .248 .520a

Group 2 not hurricane 20 .741 StormType

Total 27 1.000 a. Based on Z Approximation.

A description of the outcome is shown below:

Historically, since records were begun in 1944, the proportion of Atlantic storms classified as hurricanes during the hurricane season has been .248. A binomial test was used to determine whether the number of storms classified as hurricanes during the 2005 hurricane season differed significantly from this value (Nhurricane = 7, Nnot hurricane = 20). The test revealed that the proportion of hurricanes during 2005 did not differ significantly from the historical proportion of .248, p > .25.

Exercise 17.3 The question may be addressed with a Mann-Whitney U test. The data can be read from the file HappyMarried.sav. The two groups are defined by selecting the values 1 and 5 for the variable marital.

106

These tables are produced:

Ranks

Marital Status N Mean Rank Sum of Ranks

Married 625 431.69 269804.50

Never married 304 533.49 162180.50

General Happiness

Total 929

Test Statisticsa

General Happiness

Mann-Whitney U 74179.500Wilcoxon W 269804.500Z -6.095Asymp. Sig. (2-tailed) .000a. Grouping Variable: Marital Status

The results could be summarized as follows:

Using a Mann-Whitney U test, the two distributions of happiness ratings (mean rankmarried = 431.69, mean ranknever married = 533.49) were found to differ significantly, U = 74179.5, z = 6.10, p < .001, A = .61. The A value .61 corresponds to a small to medium effect (Vargha and Delaney, 2000). Note that a lower mean happiness rank corresponds to greater judged happiness.

Exercise 17.5 The question can be addressed with a Mann-Whitney U test.

107

The following tables are produced:

Ranks

processing condition N Mean Rank Sum of Ranks

new 22 14.64 322.00

control 22 30.36 668.00

time

Total 44

Test Statisticsa

time

Mann-Whitney U 69.000Wilcoxon W 322.000Z -4.061Asymp. Sig. (2-tailed) .000a. Grouping Variable: processing condition

The table labeled Ranks indicates that reading times in the control condition are greater than those in the new condition. The z value is large and its probability is low, indicating the distribution of reading times for the new and control instruction conditions differ significantly. The effect size, A is

86.22

223

22668

, =⎟⎠⎞

⎜⎝⎛ −

=newcontrolA

108

The results could be summarized as follows: Using a Mann-Whitney U test, the two distributions of ratings (mean ranknew = 14.64, mean rankcontrol = 30.36) were found to differ significantly, U = 69, z = 4.06, p < .001, A = .86. The A value .86 corresponds to a large effect (Vargha and Delaney, 2000).

Exercise 17.7 Because the data are paired, that is, each participant provides a reading time for the first and the second page, the data may be analyzed using a Wilcoxon signed ranks test. The window that requests the analysis is shown below:

The output produced consists of the following tables:

Ranks

N Mean Rank Sum of Ranks

Negative Ranks 5a 13.00 65.00

Positive Ranks 55b 32.09 1765.00

Ties 0c

time2 - time1

Total 60 a. time2 < time1 b. time2 > time1 c. time2 = time1

Test Statisticsb

time2 - time1

Z -6.257a

Asymp. Sig. (2-tailed) .000a. Based on negative ranks. b. Wilcoxon Signed Ranks Test

109

The analysis reveals that there is a significant difference between the median reading times for the two pages. The median reading time for each page mentioned in the summary shown below was obtained using the Explore procedure.

The median reading time for page one was 22.45 seconds and that for page two was 42.65 seconds. This difference, tested with a Wilcoxon signed ranks test was significant, z = 6.26, p < .001. For this test, the smaller sum of ranks was 65.00, N = 5, and there were no tied ranks. The A value for the effect was .92, which corresponds to a large effect (Vargha and Delaney, 2000).

110

Chapter 18 Exercise 18.1 These are the screens that were used to analyze the data:

111

Here is the output produced:

112


53.18 24.379 84

67.71 10.751 84

62.80 9.432 84

73.35 23.308 84

1.999 1.1132 84

51.607 38.3961 84

3730.95 4719.432 84

29.119 11.8499 84

9.60 4.728 84

3.6016 2.11883 84

3.958 1.8857 84

17.61 15.930 8478.48 20.434 8466.87 28.551 84

People living in cities (%)Average female lifeexpectancyAverage male lifeexpectancyPeople who read (%)Population increase (%per year))Infant mortality (deathsper 1000 live births)Gross domestic product /capitaBirth rate per 1000 peopleDeath rate per 1000peopleBirth to death ratioFertility: average numberof kidscropgrowMales who read (%)Females who read (%)

Mean Std. Deviation Analysis N

Correlation Matrix

1.000 .749 .731 .620 -.254 -.725 .591 -.596 -.588 .174 -.565 -.213 .590 .615

.749 1.000 .979 .837 -.470 -.955 .575 -.838 -.793 .121 -.815 .062 .775 .817

.731 .979 1.000 .768 -.370 -.924 .564 -.766 -.839 .211 -.750 .039 .715 .744

.620 .837 .768 1.000 -.624 -.880 .452 -.842 -.550 -.112 -.837 .144 .947 .973

-.254 -.470 -.370 -.624 1.000 .489 -.366 .806 -.007 .743 .792 -.445 -.614 -.633

-.725 -.955 -.924 -.880 .489 1.000 -.586 .838 .727 -.089 .807 -.097 -.808 -.843

.591 .575 .564 .452 -.366 -.586 1.000 -.586 -.263 -.131 -.505 .020 .418 .430

-.596 -.838 -.766 -.842 .806 .838 -.586 1.000 .471 .320 .972 -.262 -.791 -.832

-.588 -.793 -.839 -.550 -.007 .727 -.263 .471 1.000 -.551 .494 .175 -.487 -.512

.174 .121 .211 -.112 .743 -.089 -.131 .320 -.551 1.000 .296 -.395 -.144 -.138

-.565 -.815 -.750 -.837 .792 .807 -.505 .972 .494 .296 1.000 -.219 -.793 -.837

-.213 .062 .039 .144 -.445 -.097 .020 -.262 .175 -.395 -.219 1.000 .178 .173.590 .775 .715 .947 -.614 -.808 .418 -.791 -.487 -.144 -.793 .178 1.000 .964.615 .817 .744 .973 -.633 -.843 .430 -.832 -.512 -.138 -.837 .173 .964 1.000


Correlation

People livingin cities (%)

Averagefemale lifeexpectancy

Averagemale life

expectancyPeople who

read (%)

Populationincrease (%per year))

Infantmortality(deaths

per 1000live births)

Grossdomesticproduct /

capitaBirth rate per1000 people

Death rate per1000 people

Birth todeath ratio

Fertility:averagenumberof kids cropgrow

Males whoread (%)

Females whoread (%)

KMO and Bartlett's Test

.865

2010.83491

.000

Kaiser-Meyer-Olkin Measure of SamplingAdequacy.

Approx. Chi-SquaredfSig.

Bartlett's Test ofSphericity

113

Communalities

1.000 .671

1.000 .953

1.000 .915

1.000 .877

1.000 .927

1.000 .932

1.000 .373

1.000 .932

1.000 .839

1.000 .866

1.000 .894

1.000 .4061.000 .8091.000 .861


Initial Extraction

Extraction Method: Principal Component Analysis.

Total Variance Explained

8.658 61.841 61.841 8.658 61.841 61.841 8.458 60.413 60.4132.597 18.550 80.391 2.597 18.550 80.391 2.797 19.978 80.391.929 6.637 87.027.707 5.049 92.077.432 3.082 95.159.267 1.907 97.066.131 .933 97.999.095 .676 98.675.057 .410 99.084.046 .326 99.411.036 .260 99.671.020 .145 99.816.019 .133 99.949.007 .051 100.000

Component1234567891011121314

Total % of Variance Cumulative % Total % of Variance Cumulative % Total % of Variance Cumulative %Initial Eigenvalues Extraction Sums of Squared Loadings Rotation Sums of Squared Loadings

Extraction Method: Principal Component Analysis.

114

Component Number1413121110987654321

Eige

nval

ue10

8

6

4

2

0

Scree Plot

Component Matrixa

-.951 -.168

.951 .222

.934 -.064-.932 .253.923 -.099

-.916 .235

.903 .315

.893 -.108

.741 .350

-.657 -.638

.611 .001

-.088 .927

-.650 .711

.150 -.619

Infant mortality (deathsper 1000 live births)Average female lifeexpectancyPeople who read (%)Birth rate per 1000 peopleFemales who read (%)Fertility: average numberof kidsAverage male lifeexpectancyMales who read (%)People living in cities (%)Death rate per 1000peopleGross domestic product /capitaBirth to death ratioPopulation increase (%per year))cropgrow

1 2Component

Extraction Method: Principal Component Analysis.2 components extracted.a.

115

Rotated Component Matrixa

.975 .046

-.965 .008

.945 .145

.907 -.232

.889 -.265-.870 .418.859 -.268

-.858 .397

.792 .209

-.762 -.508

.601 -.110

.082 .927

-.510 .817

.035 -.636

Average female lifeexpectancyInfant mortality (deathsper 1000 live births)Average male lifeexpectancyPeople who read (%)Females who read (%)Birth rate per 1000 peopleMales who read (%)Fertility: average numberof kidsPeople living in cities (%)Death rate per 1000peopleGross domestic product /capitaBirth to death ratioPopulation increase (%per year))cropgrow

1 2Component

Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Rotation converged in 3 iterations.a.

Component Transformation Matrix

.983 -.182

.182 .983

Component12

1 2

Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Component 11.00.50.0-0.5-1.0

Com

pone

nt 2

1.0

0.5

0.0

-0.5

-1.0

lit_fema

lit_male

cropgrow

fertilty

b_to_d

death_rt

birth_rt

gdp_cap

babymort

pop_incr

literacy

lifeexpm

lifeexpf

urban

Component Plot in Rotated Space

116

Component Score Coefficient Matrix

.109 .117

.124 .064

.125 .100

.102 -.044

-.024 .283

-.120 -.044

.069 -.013

-.088 .115

-.119 -.228

.055 .353

-.088 .108

-.026 -.238.094 -.059.098 -.057


1 2Component

Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization. Component Scores.

Here are the answers to the questions: 1. Although there were 109 countries in the data set, due to missing data in some of the

variables, there were 84 countries used in the analysis. This number is obtained from the table of descriptive statistics in the column labeled N.

2. The KMO statistic has the value .87. Values above .50 are considered acceptable; a value above .80 is considered meritorious.

3. The variables with the lowest communalities were gross domestic product/ capita (.373) and cropgrow (.406).

4. Two principal components were retained. 5. Principal component 1 had an eigenvalue of 8.66 and principal component 2 had an

eigenvalue of 2.60. 6. The two retained principal components accounted for about .80 of the total variance. 7. The three variables loading highest on each component are shown in the table below:

Component 1 Loading Component 2 Loading lifeexpf .98 b_to_d .93

babymort -.97 pop_incr .82 lifeexpm ,95 cropgrow -.64

8. Component 1 may be how developed a country is in terms of its infrastructure and ability to provide healthcare to its inhabitants. Component 2 may be rate of development and growth.

117

Exercise 18.3 These are the screens that were used to analyze the data:

118


119


Mean Std. Deviation Analysis N

q1 1.98 1.086 84q2 2.48 1.246 84q3 4.01 1.035 84q4 2.30 1.220 84q5 2.43 1.254 84q6 3.42 1.282 84q7 3.80 1.062 84q8 2.62 1.270 84q9 2.27 1.274 84q10 1.42 .853 84q11 2.13 1.315 84q12 1.95 1.171 84q13 2.69 1.481 84q14 3.75 .903 84q15 2.14 1.214 84

Correlation Matrix

q1 q2 q3 q4 q5 q6 q7 q8 q9 q10 q11 q12 q13 q14 q15

q1 1.000 .409 -.396 .424 .229 -.097 -.119 .343 .405 .284 .112 .520 .085 -.043 -.116

q2 .409 1.000 -.135 .341 .230 -.178 -.099 .580 .334 .264 -.097 .585 -.076 -.075 -.300

q3 -.396 -.135 1.000 -.003 -.152 .450 .287 -.235 -.240 .035 -.072 -.099 .105 .312 -.136

q4 .424 .341 -.003 1.000 .514 -.180 -.083 .510 .652 .053 .103 .415 .112 -.008 .003

q5 .229 .230 -.152 .514 1.000 -.225 -.169 .626 .808 .034 .104 .375 .066 -.298 .070

q6 -.097 -.178 .450 -.180 -.225 1.000 .399 -.308 -.226 .005 -.004 -.123 .069 .476 -.008

q7 -.119 -.099 .287 -.083 -.169 .399 1.000 -.344 -.163 -.265 -.145 -.182 -.033 .524 -.099

q8 .343 .580 -.235 .510 .626 -.308 -.344 1.000 .654 .093 .102 .595 .033 -.221 -.019

q9 .405 .334 -.240 .652 .808 -.226 -.163 .654 1.000 .082 .137 .388 .192 -.233 .029

q10 .284 .264 .035 .053 .034 .005 -.265 .093 .082 1.000 .187 .334 .189 -.113 .105

q11 .112 -.097 -.072 .103 .104 -.004 -.145 .102 .137 .187 1.000 .208 .392 -.043 .645

q12 .520 .585 -.099 .415 .375 -.123 -.182 .595 .388 .334 .208 1.000 .033 -.011 -.122

q13 .085 -.076 .105 .112 .066 .069 -.033 .033 .192 .189 .392 .033 1.000 .005 .394

q14 -.043 -.075 .312 -.008 -.298 .476 .524 -.221 -.233 -.113 -.043 -.011 .005 1.000 -.077

Correlation

q15 -.116 -.300 -.136 .003 .070 -.008 -.099 -.019 .029 .105 .645 -.122 .394 -.077 1.000

KMO and Bartlett's Test

Kaiser-Meyer-Olkin Measure of Sampling Adequacy. .666 Approx. Chi-Square 553.949 df 105

Bartlett's Test of Sphericity

Sig. .000

120

Communalities

Initial Extraction

q1 1.000 .754q2 1.000 .685q3 1.000 .877q4 1.000 .653q5 1.000 .799q6 1.000 .613q7 1.000 .702q8 1.000 .735q9 1.000 .840q10 1.000 .735q11 1.000 .731q12 1.000 .721q13 1.000 .532q14 1.000 .728q15 1.000 .786Extraction Method: Principal Component Analysis.

Total Variance Explained

Initial Eigenvalues Extraction Sums of Squared Loadings Rotation Sums of Squared Loadings Component Total % of Variance Cumulative % Total % of Variance Cumulative % Total % of Variance Cumulative %

1 4.372 29.144 29.144 4.372 29.144 29.144 3.168 21.117 21.1172 2.137 14.248 43.392 2.137 14.248 43.392 2.155 14.365 35.4823 1.925 12.834 56.225 1.925 12.834 56.225 2.111 14.076 49.5584 1.424 9.492 65.718 1.424 9.492 65.718 2.078 13.852 63.4105 1.034 6.895 72.613 1.034 6.895 72.613 1.380 9.203 72.6136 .757 5.050 77.663 7 .650 4.335 81.997 8 .575 3.833 85.830 9 .525 3.501 89.332 10 .442 2.950 92.281 11 .396 2.643 94.924 12 .267 1.781 96.705 13 .205 1.364 98.069 14 .173 1.151 99.220 15 .117 .780 100.000 Extraction Method: Principal Component Analysis.

121

Component Matrixa

Component

1 2 3 4 5

q8 .834 -.116 .021 -.120 .107 q9 .806 .009 .146 -.410 .024 q5 .725 .025 .055 -.490 .172 q12 .697 -.198 .295 .330 -.002 q4 .667 -.118 .319 -.304 .000 q1 .609 -.144 .179 .330 -.470 q2 .608 -.446 .098 .324 .042 q15 .020 .856 .155 -.091 -.141 q11 .213 .742 .320 .104 -.149 q13 .125 .570 .425 .009 .105 q14 -.379 -.280 .647 -.004 -.295 q6 -.444 -.112 .631 .039 .062 q7 -.433 -.309 .497 -.276 -.311 q10 .290 .179 .195 .694 .316 q3 -.395 -.151 .512 -.089 .655 Extraction Method: Principal Component Analysis. a. 5 components extracted.

122

Rotated Component Matrixa

Component

1 2 3 4 5

q9 .888 .100 .111 -.121 -.118 q5 .864 -.019 .068 -.218 .013 q4 .774 .196 .057 .098 -.051 q8 .735 .339 -.044 -.257 -.112 q10 -.135 .743 .266 -.226 .210 q12 .431 .722 .005 .013 -.117 q2 .357 .669 -.309 -.045 -.110 q1 .287 .589 .042 .125 -.554 q15 -.019 -.201 .852 -.098 -.097 q11 .065 .118 .836 -.021 -.116 q13 .123 .092 .687 .055 .183 q14 -.129 .046 -.013 .840 .056 q7 -.043 -.238 -.112 .794 .030 q6 -.196 .044 .102 .645 .383 q3 -.052 -.006 -.032 .326 .876 Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization. a. Rotation converged in 6 iterations.

123

Component Score Coefficient Matrix

Component

1 2 3 4 5

q1 -.041 .249 .034 .200 -.434 q2 .015 .308 -.153 .014 .002 q3 .096 .065 -.025 .011 .689 q4 .286 -.035 .017 .122 .031 q5 .348 -.173 .007 -.060 .141 q6 -.006 .091 .067 .277 .188 q7 .088 -.121 -.024 .427 -.141 q8 .213 .047 -.042 -.069 .053 q9 .328 -.117 .034 .023 .018 q10 -.194 .462 .116 -.159 .249 q11 -.026 .048 .401 .047 -.104 q12 .033 .325 -.002 .065 -.012 q13 .043 .043 .323 .032 .148 q14 .004 .060 .027 .448 -.120 q15 -.005 -.118 .406 -.007 -.103 Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Component Score Covariance Matrix

Component 1 2 3 4 5

1 1.000 .000 .000 .000 .000 2 .000 1.000 .000 .000 .000 3 .000 .000 1.000 .000 .000 4 .000 .000 .000 1.000 .000 5 .000 .000 .000 .000 1.000 Extraction Method: Principal Component Analysis. Rotation Method: Varimax with Kaiser Normalization.

Of most importance to the question being addressed in this problem is the data shown in the Rotated Component Matrix table. The values in the table are loadings of the individual questions on the five retained components. Component 1 corresponds most strongly to questions 9, 5, 4, and 8. The text of these questions is shown below:

9. Some boys/girls push and shove other kids. How often do you do this?

5. Some boys/girls hit other kids. How often do you do this? How often do you do this?

4. When they are mad at someone, some boys/girls get back at the kid by not letting the kid be in their group anymore. How often do you do this?

8. Some boys/girls yell at other and call them mean names. How often

124

do you do this? These four questions all seem to have aggression in common. Component 2 corresponds most strongly to questions 10, 12, 2, and 1. The text of these questions is shown below:

10. Some boys/girls tell their friends that they will stop liking them unless the friends do what they say. How often do you tell friends this?

12. Some boys/girls try to keep other from liking a kid by saying mean things about the kid. How often do you do this?

2. Some boys/girls try to keep certain people from being in their group when it is time to play or do an activity. How often do you do this?

1. Some boys/girls tell lies (fibs/make up stories) about a kid so that others won’t like that kid anymore. How often do you do this?

These questions all appear to have relational aggression in common. Component 3 corresponds most strongly to questions 15, 11, and 12. The text of these questions is shown below:

15. Some boys/girls have a lot of kids who like to play with them. How often do the other kids like to play with you?

11. Some boys/girls have a lot of friends. How often do you have a lot of friends?

13. Some boys/girls wish that they had more friends. How often do you feel this way?

These questions appear to deal with social connections. Component 4 corresponds most strongly to questions 14, 7, and 6. The text of these questions is shown below:

14. Some boys/girls say or do nice things for other kids. How often do you do this?

7. Some boys/girls help out other kids when they need it .How often do you do this?

6. Some boys/girls let other know that they care about them. How often do you do this?

These questions appear to have positive social behavior in common. Component 5 corresponds most strongly just to question 3. The text of the question is shown below:

3. Some boys/girls try to cheer up other kids who feel upset or sad. The question appears to also have positive social behavior, and seems to logically belong to component 4. However, it may differ from the questions comprising component 4 in that it may have an element of humor in it.

125

Chapter 19 Exercise 19.1 These are the screens that show the requests made for the analysis:

126


Group Statistics

Valid N (listwise) own Mean Std. Deviation Unweighted Weighted

iq 95.0000 4.08248 4 4.000 1.00

yrsed 11.7500 2.06155 4 4.000 iq 105.0000 4.08248 4 4.000 2.00 yrsed 17.0000 2.58199 4 4.000 iq 103.2500 4.57347 4 4.000 3.00 yrsed 23.0000 4.24264 4 4.000 iq 101.0833 5.96137 12 12.000 Total yrsed 17.2500 5.56164 12 12.000

Tests of Equality of Group Means

Wilks' Lambda F df1 df2 Sig.

iq .416 6.309 2 9 .019 yrsed .255 13.150 2 9 .002

Covariance Matrices

own iq yrsed

iq 16.667 3.3331.00

yrsed 3.333 4.250iq 16.667 6.6672.00 yrsed 6.667 6.667iq 20.917 18.0003.00 yrsed 18.000 18.000

127

Analysis 1 Box's Test of Equality of Covariance Matrices

Log Determinants

own Rank Log Determinant

1.00 2 4.0902.00 2 4.2003.00 2 3.961Pooled within-groups 2 4.468The ranks and natural logarithms of determinants printed are those of the group covariance matrices.

Test Results

Box's M 3.462Approx. .390df1 6df2 2018.769

F

Sig. .886Tests null hypothesis of equal population covariance matrices.

Summary of Canonical Discriminant Functions

Eigenvalues

Function Eigenvalue % of Variance Cumulative %

Canonical Correlation

1 3.259a 74.6 74.6 .875 2 1.109a 25.4 100.0 .725 a. First 2 canonical discriminant functions were used in the analysis.

Wilks' Lambda

Test of Function(s) Wilks' Lambda Chi-square df Sig.

1 through 2 .111 18.661 4 .001 2 .474 6.343 1 .012

Standardized Canonical Discriminant Function

Coefficients

Function

1 2

iq -.560 1.298yrsed 1.314 -.522

128

Structure Matrix

Function

1 2

yrsed .918* .396iq .369 .929*

Pooled within-groups correlations between discriminating variables and standardized canonical discriminant functions Variables ordered by absolute size of correlation within function. *. Largest absolute correlation between each variable and any discriminant function

Functions at Group Centroids

Function own 1 2

1.00 -1.527 -.9332.00 -.622 1.2383.00 2.148 -.305Unstandardized canonical discriminant functions evaluated at group means

Classification Statistics

Classification Processing Summary

Processed 12Missing or out-of-range group codes

0Excluded

At least one missing discriminating variable

0

Used in Output 12

Prior Probabilities for Groups

Cases Used in Analysis own Prior Unweighted Weighted

1.00 .333 4 4.0002.00 .333 4 4.0003.00 .333 4 4.000Total 1.000 12 12.000

129

Classification Function Coefficients

own

1.00 2.00 3.00

iq 9.244 9.788 8.952yrsed -7.732 -7.714 -6.282(Constant) -394.775 -449.389 -391.004Fisher's linear discriminant functions

Classification Resultsb,c

Predicted Group Membership

own 1.00 2.00 3.00 Total

1.00 4 0 0 4

2.00 0 4 0 4

Count

3.00 0 0 4 4

1.00 100.0 .0 .0 100.0

2.00 .0 100.0 .0 100.0

Original

%

3.00 .0 .0 100.0 100.0 1.00 3 1 0 4 2.00 1 3 0 4

Count

3.00 0 0 4 4 1.00 75.0 25.0 .0 100.0 2.00 25.0 75.0 .0 100.0

Cross-validateda

%

3.00 .0 .0 100.0 100.0 a. Cross validation is done only for those cases in the analysis. In cross validation, each case is classified by the functions derived from all cases other than that case. b. 100.0% of original grouped cases correctly classified. c. 83.3% of cross-validated grouped cases correctly classified.

Answers: 1. The Fisher classification functions are

775.394732.7244.91 −×−×= YrsEdIQG 389.449714.7788.92 −×−×= YrsEdIQG 004.391282.6952.83 −×−×= YrsEdIQG

2. Based on the leave-one-out procedure, 83.3% of the cases will be correctly

classified. 3. For case 1:

775.39426732.7109244.91 −×−×=G = 411.78 389.44926714.7109788.92 −×−×=G = 416.94 004.39126282.6109952.83 −×−×=G = 421.43

The largest of the 3 values corresponds to G3, so the case is classified as category 3, a renter.

These are the functions used to answer question 1.

130

For case 2

775.39412732.799244.91 −×−×=G = 427.60 389.44912714.799788.92 −×−×=G = 427.06 004.39112282.699952.83 −×−×=G = 419.86

The largest of the 3 values corresponds to G1, so the case is classified as category 1, a non-owner.

Exercise 19.3 These are the screens that show the requests made for the analysis:

131

Selected output is shown below:

Analysis Case Processing Summary

Unweighted Cases N Percent

Valid 507 100.0 Missing or out-of-range group codes

0 .0

At least one missing discriminating variable

0 .0

Both missing or out-of-range group codes and at least one missing discriminating variable

0 .0

Excluded

Total 0 .0 Total 507 100.0

132

Group Statistics

36.5031 1.77922 260 260.000 41.2413 2.08716 247 247.000

31.4615 2.04918 260 260.000 32.5267 1.86513 247 247.000

12.3669 .83637 260 260.000 14.4571 .88254 247 247.0009.8742 .66163 260 260.000 11.2462 .63590 247 247.000

27.5815 2.30748 260 260.000 28.0915 2.06710 247 247.000

17.7246 1.83206 260 260.000 20.8065 2.14363 247 247.00026.0973 1.81881 260 260.000 29.9490 2.08311 247 247.00018.0969 1.18660 260 260.000 19.5619 1.07136 247 247.00013.0265 .86606 260 260.000 14.7441 .94424 247 247.000

100.3038 6.47060 260 260.000 116.5016 6.49802 247 247.00086.0600 6.17041 260 260.000 100.9899 7.20902 247 247.00069.8035 7.58775 260 260.000 84.5332 8.78224 247 247.00083.7458 9.96163 260 260.000 87.6623 8.38488 247 247.00095.6527 6.94073 260 260.000 97.7632 6.22804 247 247.00057.1958 4.63600 260 260.000 56.4980 4.24667 247 247.00028.0973 2.70948 260 260.000 34.4036 2.98204 247 247.00023.7604 1.68225 260 260.000 28.2405 1.77932 247 247.00035.2600 2.57808 260 260.000 37.1955 2.27300 247 247.00035.0062 2.61313 260 260.000 37.2069 2.64514 247 247.00021.2058 1.43882 260 260.000 23.1591 1.72909 247 247.00015.0592 .84941 260 260.000 17.1903 .90800 247 247.00060.6004 9.61570 260 260.000 78.1445 10.51289 247 247.000

164.8723 6.54460 260 260.000 177.7453 7.18363 247 247.000

Biacromial diameter (incm)Bitrochanteric diameter(in cm)Elbow diameter (in cm)Wrist diameter (in cm)Biiliac diameter, orpelvic breadth (in cm)Chest depth (in cm)Chest diameter (in cm)Knee diameter (in cm)Ankle diameter (in cm)Shoulder girth (in cm)Chest girth (in cm)Waist girth (in cm)Navel girth (in cm)Hip girth (in cm)Thigh girth (in cm)Bicep girth (in cm)Forearm girth (in cm)Knee girth (in cm)Calf girth (in cm)Ankle girth (in cm)Wrist girth (in cm)Weight (in kg)Height (in cm)

Mean Std. Deviation Unweighted WeightedValid N (listwise)


0 male

Gender; 1 for males and 0 for females


Wilks' Lambda F df1 df2 Sig.

Biacromial diameter (in cm) .399 759.223 1 505 .000Biiliac diameter, or pelvic breadth (in cm)

.987 6.845 1 505 .009

Bitrochanteric diameter (in cm) .931 37.347 1 505 .000Chest depth (in cm) .624 303.814 1 505 .000Chest diameter (in cm) .506 493.162 1 505 .000Elbow diameter (in cm) .403 749.653 1 505 .000Wrist diameter (in cm) .472 565.636 1 505 .000Knee diameter (in cm) .704 212.183 1 505 .000Ankle diameter (in cm) .525 456.265 1 505 .000Shoulder girth (in cm) .390 790.480 1 505 .000Chest girth (in cm) .445 629.622 1 505 .000Waist girth (in cm) .552 409.577 1 505 .000Navel girth (in cm) .957 22.821 1 505 .000Hip girth (in cm) .975 12.939 1 505 .000Thigh girth (in cm) .994 3.114 1 505 .078Bicep girth (in cm) .448 622.151 1 505 .000Forearm girth (in cm) .373 849.251 1 505 .000Knee girth (in cm) .863 80.083 1 505 .000Calf girth (in cm) .850 88.775 1 505 .000Ankle girth (in cm) .725 191.928 1 505 .000Wrist girth (in cm) .404 745.468 1 505 .000Weight (in kg) .567 385.030 1 505 .000Height (in cm) .531 445.610 1 505 .000

133

Analysis 1

Box's Test of Equality of Covariance Matrices

Log Determinants

23 16.97123 21.59823 20.724

Gender; 1 for malesand 0 for females0malePooled within-groups

RankLog

Determinant

The ranks and natural logarithms of determinantsprinted are those of the group covariance matrices.

Test Results

756.7302.613

276771799.8

.000

Box's MApprox.df1df2Sig.

F

Tests null hypothesis of equal population covariance matrices.

Summary of Canonical Discriminant Functions

Eigenvalues

7.128a 100.0 100.0 .936Function1

Eigenvalue % of Variance Cumulative %CanonicalCorrelation

First 1 canonical discriminant functions were used in theanalysis.

a.

Wilks' Lambda

.123 1034.041 23 .000Test of Function(s)1

Wilks'Lambda Chi-square df Sig.

134

Standardized Canonical Discriminant Function Coefficients

.293

-.100

.153-.052

-.109

.134-.061.093.246.172

-.1651.335-.746-.237-.502.424.506

-.078-.039.005

-.093-.633.414

Biacromial diameter (incm)Bitrochanteric diameter(in cm)Elbow diameter (in cm)Wrist diameter (in cm)Biiliac diameter, orpelvic breadth (in cm)Chest depth (in cm)Chest diameter (in cm)Knee diameter (in cm)Ankle diameter (in cm)Shoulder girth (in cm)Chest girth (in cm)Waist girth (in cm)Navel girth (in cm)Hip girth (in cm)Thigh girth (in cm)Bicep girth (in cm)Forearm girth (in cm)Knee girth (in cm)Calf girth (in cm)Ankle girth (in cm)Wrist girth (in cm)Weight (in kg)Height (in cm)

1Function

Structure Matrix

.486

.469

.459

.456

.455

.418

.416

.396

.370

.356

.352

.337

.327

.291

.243

.231

.157

.149

.102

.080

.060

.044

-.029

Forearm girth (in cm)Shoulder girth (in cm)Biacromial diameter (incm)Elbow diameter (in cm)Wrist girth (in cm)Chest girth (in cm)Bicep girth (in cm)Wrist diameter (in cm)Chest diameter (in cm)Ankle diameter (in cm)Height (in cm)Waist girth (in cm)Weight (in kg)Chest depth (in cm)Knee diameter (in cm)Ankle girth (in cm)Calf girth (in cm)Knee girth (in cm)Bitrochanteric diameter(in cm)Navel girth (in cm)Hip girth (in cm)Biiliac diameter, orpelvic breadth (in cm)Thigh girth (in cm)

1Function

135

Canonical Discriminant Function Coefficients

.151

-.051

.178-.080

-.050

.067-.031.082.272.027

-.025.163

-.081-.036-.113.149.292

-.032-.015.003

-.106-.063.060

-21.313

Biacromial diameter (incm)Bitrochanteric diameter(in cm)Elbow diameter (in cm)Wrist diameter (in cm)Biiliac diameter, orpelvic breadth (in cm)Chest depth (in cm)Chest diameter (in cm)Knee diameter (in cm)Ankle diameter (in cm)Shoulder girth (in cm)Chest girth (in cm)Waist girth (in cm)Navel girth (in cm)Hip girth (in cm)Thigh girth (in cm)Bicep girth (in cm)Forearm girth (in cm)Knee girth (in cm)Calf girth (in cm)Ankle girth (in cm)Wrist girth (in cm)Weight (in kg)Height (in cm)(Constant)

1Function

Unstandardized coefficients


-2.5972.734

Gender; 1 for malesand 0 for females0male

1Function

Unstandardized canonical discriminantfunctions evaluated at group means

Classification Statistics

Prior Probabilities for Groups

.500 260 260.000

.500 247 247.0001.000 507 507.000

Gender; 1 for malesand 0 for females0maleTotal

Prior Unweighted WeightedCases Used in Analysis

136

Classification Function Coefficients

2.868 3.675

-2.550 -2.823

-2.398 -1.4496.674 6.248

3.074 2.810

6.570 6.9292.548 2.381

14.742 15.1814.709 6.1593.124 3.2665.147 5.0159.361 10.230

-1.247 -1.6787.220 7.0299.264 8.663-.947 -.152

17.273 18.8313.708 3.5389.740 9.660

.004 .022-.797 -1.361

-27.972 -28.30711.390 11.712

-2105.098 -2219.080

Biacromial diameter (incm)Bitrochanteric diameter(in cm)Elbow diameter (in cm)Wrist diameter (in cm)Biiliac diameter, orpelvic breadth (in cm)Chest depth (in cm)Chest diameter (in cm)Knee diameter (in cm)Ankle diameter (in cm)Shoulder girth (in cm)Chest girth (in cm)Waist girth (in cm)Navel girth (in cm)Hip girth (in cm)Thigh girth (in cm)Bicep girth (in cm)Forearm girth (in cm)Knee girth (in cm)Calf girth (in cm)Ankle girth (in cm)Wrist girth (in cm)Weight (in kg)Height (in cm)(Constant)

0 male

Gender; 1 for malesand 0 for females

Fisher's linear discriminant functions


Predicted Group Membership

Gender; 1 for males and 0 for females 0 female 1 male Total

0 female 257 3 260 Count

1 male 5 242 247

0 female 98.8 1.2 100.0

Original

%

1 male 2.0 98.0 100.0 0 female 257 3 260 Count 1 male 6 241 247 0 female 98.8 1.2 100.0

Cross-validateda

% 1 male 2.4 97.6 100.0

a. Cross validation is done only for those cases in the analysis. In cross validation, each case is classified by the functions derived from all cases other than that case. b. 98.4% of original grouped cases correctly classified. c. 98.2% of cross-validated grouped cases correctly classified.

137

Answers: 1. Yes. The function has a Wilks’ lambda = .12, χ2 (23, N = 507) = 1,034.04, p < .001. 2. The discriminant function successfully predicted computer gender of 98% of the

cases in the sample. 3. The two variables that were most strongly correlated with the discriminant function

were forearm girth and shoulder girth. Based on the standardized dircriminant function coefficients, waist girth and navel girth made the greatest independent contributions to discriminating between males and females.

To answer question 4, a discriminant analysis was conducted using the predictors forearm girth, shoulder girth, waist girth and navel girth. Selected output is shown below:

Analysis Case Processing Summary

507 100.0

0 .0

0 .0

0 .0

0 .0507 100.0

Unweighted CasesValid

Missing or out-of-rangegroup codesAt least one missingdiscriminating variableBoth missing orout-of-range group codesand at least one missingdiscriminating variableTotal

Excluded

Total

N Percent

Group Statistics

23.7604 1.68225 260 260.000100.3038 6.47060 260 260.000

69.8035 7.58775 260 260.00083.7458 9.96163 260 260.00028.2405 1.77932 247 247.000

116.5016 6.49802 247 247.00084.5332 8.78224 247 247.00087.6623 8.38488 247 247.00025.9430 2.83058 507 507.000

108.1951 10.37483 507 507.00076.9795 11.01269 507 507.00085.6538 9.42413 507 507.000

Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)

Gender; 1 for malesand 0 for females0

male

Total


138


.373 849.251 1 505 .000

.390 790.480 1 505 .000

.552 409.577 1 505 .000

.957 22.821 1 505 .000

Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)

Wilks'Lambda F df1 df2 Sig.

Covariance Matrices

2.830 8.134 9.034 10.7188.134 41.869 35.636 39.5389.034 35.636 57.574 63.132

10.718 39.538 63.132 99.2343.166 8.132 6.573 6.3598.132 42.224 33.152 29.3746.573 33.152 77.128 64.9526.359 29.374 64.952 70.306

Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)

Gender; 1 for malesand 0 for females0

male

Forearmgirth (in cm)

Shouldergirth (in cm)

Waist girth(in cm)

Navel girth(in cm)

Box's Test of Equality of Covariance Matrices

Log Determinants

4 10.5064 10.8774 10.880

Gender; 1 for malesand 0 for females0malePooled within-groups

RankLog

Determinant

The ranks and natural logarithms of determinantsprinted are those of the group covariance matrices.

Test Results

97.8359.700

101211974

.000

Box's MApprox.df1df2Sig.

F

Tests null hypothesis of equal population covariance matrices.

139

Summary of Canonical Discriminant Functions Eigenvalues

3.309a 100.0 100.0 .876Function1

Eigenvalue % of Variance Cumulative %CanonicalCorrelation

First 1 canonical discriminant functions were used in theanalysis.

a.

Wilks' Lambda

.232 734.696 4 .000Test of Function(s)1

Wilks'Lambda Chi-square df Sig.

Standardized Canonical Discriminant Function Coefficients

.588

.3241.019

-1.249


1Function

Structure Matrix

.713

.688

.495

.117


1Function

Pooled within-groups correlations between discriminatingvariables and standardized canonical discriminant functions Variables ordered by absolute size of correlation within function.

Canonical Discriminant Function Coefficients

.340

.050

.124-.135

-12.190

Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)(Constant)

1Function

Unstandardized coefficients

140


-1.7691.863

Gender; 1 for malesand 0 for females0male

1Function

Unstandardized canonical discriminantfunctions evaluated at group means

Classification Statistics Classification Function Coefficients

3.168 4.4012.081 2.263-.793 -.341.415 -.077

-132.390 -176.832

Forearm girth (in cm)Shoulder girth (in cm)Waist girth (in cm)Navel girth (in cm)(Constant)

0 male

Gender; 1 for malesand 0 for females

Fisher's linear discriminant functions


249 11 26010 237 247

95.8 4.2 100.04.0 96.0 100.0249 11 26011 236 247

95.8 4.2 100.04.5 95.5 100.0

Gender; 1 for malesand 0 for females0male0male0male0male

Count

%

Count

%

Original

Cross-validateda

0 male

Predicted GroupMembership

Total

Cross validation is done only for those cases in the analysis. In cross validation,each case is classified by the functions derived from all cases other than thatcase.

a.

95.9% of original grouped cases correctly classified.b.

95.7% of cross-validated grouped cases correctly classified.c.

4. Yes. The function has a Wilks’ lambda = .23, χ2 (4, N = 507) = 734.70, p < .001.

The discriminant function successfully predicted computer gender of 96% of the cases in the sample. The Fisher functions could be used to predict a person’s gender based on values of the four variables. The functions are:

39.132415.793.081.2168.3 −+−+= navelwaistshoulderforearmGenderfemale 832.176077.341.263.2401.4 −−−+= navelwaistshoulderforearmGendermale .

A person is given the gender classification of the function that has the higher value.

A Visual Approach to SPSS for Windows

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