Chapter 4 HomeworkKey

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1 Chapter 4 Homework #1a. The data shown indicates this is a two way factorial design. You may also call this a 3 * 2 Factorial ANOVA. The first variable (happiness) has three level). The second variable (job satisfaction) has two levels. In a factorial design, there is a main effect for each IDV. Since this study has two IDV’s there will be two main effects. The main effect of happiness and the main effect of job satisfaction. There will be one hypothesis for each main effect. The hypothesis must include reference to the DV (hours worked per week). Research hypothesis 1: Are there differences in hours worked per week based on happiness? Research hypothesis 2: Is there a difference in job satisfaction based on hours worked per week? You may note that the hypothesis are written as questions – this is acceptable. A factorial design will include an interaction. An interaction is defined as the effects of one IDV variable on the DV differs based on the level of the other variable/variables. Research hypothesis 3: Does hours worked per change based on the interaction between job satisfaction and happiness? 1b. The best way to determine if there is an interaction is to examine the line graph. Excel is a good way to make the line graph. Enter the data into an excel spread sheet.

2 Highlight the data and the labels. Go to insert line graph:

3 A line graph will be generated: You can cut and paste that line graph to a word document: You can tell if there is an interaction if the lines cross on the chart. The lines cross (see the crossing of the blue and red line) there is an interaction. The interaction would most likely be significant. 35 36 37 38 39 40 41 42 43 44 very happy pretty happy not too happy very satisfied with job not very satisfied with job

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4 1c. You need to examine the marginal means to determine if there is main effect of a variable. The marginal means are simply the average of the cell means across the row or column. The marginal means for general happiness are 41.5, 42, and 39. Not too happy is probably different from very happy and pretty happy. If the means are different from each other – there is usually a significant main effect of that variable. If the means are similar – there is usually no significant effect. Please note, only one mean has to be different from the rest for there to be a main effect. 1d. The marginal means for happiness are 41 and 41. There is no difference between very satisfied and not very satisfied. 2. In this question, you are going to actually carry out the analysis from question 1 using SPSS. In the prior handout, you learned how to open a data set. Open the appropriate data set for this question! very satisfied with job not very satisfied with job very happy 43 40 41.5 pretty happy 42 42 42 not too happy 38 40 39 very satisfied with job not very satisfied with job very happy 43 40 pretty happy 42 42 not too happy 38 40 41 41

5 Within SPSS go to Analyze, General Linear Model, Univarite Click over the DV to the Dependent Variable (in this question it is Hrs1). Click over the IDV’s to Fixed Factor(s)(in this question (happy and satjob2). Click Model – Click Full Factorial (it is the default) – Click continue.

6 Click options, Check descriptive statistics, estimates of effect size, and homogeneity tests, click continue. Press Ok on the main dialogue window. Interpret your results! 2a. You should always interpret the interaction first in a factorial ANOVA! The interaction between happy and satjob2 is NOT significant. The probability associated with this result is .187. Since .187 is greater than .05 – it is NOT significant.

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7 2b. After you interpret the interaction, you are free to interpret the main effects! We again need to look at the probability associated with the result. Tests of Between-Subjects Effects Dependent Variable:Number of Hours Worked Last Week Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Corrected Model 1585.672 a 5 317.134 1.491 .190 .008 Intercept 700048.802 1 700048.802 3290.370 .000 .787 happy 679.432 2 339.716 1.597 .203 .004 satjob2 31.308 1 31.308 .147 .701 .000 happy * satjob2 715.617 2 357.809 1.682 .187 .004 Error 189566.344 891 212.757 Total 1757791.000 897 Corrected Total 191152.016 896 a. R Squared = .008 (Adjusted R Squared = .003) The probability associated with the main effect of happy is .203 the probability associated with the main effect of satjob2 is .701. Neither of these results is associated with a probability of less than .05. There no significant main effects in this analysis. 2c. Please see responses to question 1. I was incorrect in determining there was an interaction. I was correct in determining there was no main effect of job satisfaction, I was incorrect in determining there was a main effect of happiness. PLEASE NOTE SINCE THERE WAS NO FOLLOW-UP QUESTION ON THE BONFERRONI POST HOC TEST – SO I DID NOT RUN IT. 3a. Are there significant mean differences in salaries based on gender? Are there significant differences in salaries based on ethnicity? Is there an interaction in salaries between gender and ethnicity? 3b. For question 3b, you need to check your data for missing values, outliers, and normality. The steps for doing this if found in Chapter 3 of your textbook (reviewed here with this question!). Many of these steps are used to check to make sure the assumptions were met for this factorial ANOVA. An assumption is a condition that must be met in order to validly carry out a statistical analysis. In some instances when assumptions are not met – you should not carry out the analysis. The assumptions of the factorial ANOVA are: 1. The observations with each sample must be randomly sampled and must be independent of each other. This assumption is not tested statistically, you meet this assumption by using the appropriate methodology. The word “sample” actually means the independent groups formed in a factorial design. For example, a 2*2 Factorial ANOVA creates four independent groups. If this type of design had gender (male Vs. Female) and political party (republican Vs. Democrat). The issue of independence simply means that if you are in

8 one group – you cannot be in the other. For example, a male republican CANNOT at the same time be a female democrat. 2. The distribution of scores on the DV must be normal in the population from which the data were sampled. Each of the groups formed in a factorial design has its own distribution. Those distributions must be normal (skew and kurtosis!). 3. The distribution of scores on the DV must have equal variance. Each of the groups formed in a factorial design has its own distribution. Those distributions must have equal variances. In the example of Gender and Political Party there will be four separate distributions of scores. Each distribution must be normal – and the variances for the distributions must be equal. We use SPSS to check for assumptions 2 and 3! First we start off with the frequency tables for each IV and the DV. I did not include the frequency of the DV – the printout was too long. The frequency tables of Sex and Minority classification do not indicate any problems with missing data (there is no missing data). Nor are there any problems with splits. For a split to be a problem – you would look for 90 percent of the cases falling into one category. In this example, you have a 54 and 45 split for sex and a 78 and 21 split for minority – both are OK. Political Party Republican Democrat Gender Male Male Republcans Male Democrats Female Female Republicans Female Democrats

9 Next we check for outlier! We need to use the explore procedure in SPSS. Go to Analyze, Descriptives, Explore. Move the DV to the Dependent List (SALNOW) – Move the IDV’s to the Factor List (Sex and Minority).

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10 Click statistics – check outliers. Click continue. Click Plots – Check Boxplots:Factors levels together and Stem-and-Leaf. Click continue Click OK on the main dialogue window. I am going to use the Stem-and-Leaf plots to examine extreme values. There will be four stem and leaf plots (this is a 2*2 factorial ANOVA – creating four independent group!).

11 CURRENT SALARY Stem-and-Leaf Plot for sex= MALES Frequency Stem & Leaf 1.00 0 . 7 18.00 0 . 888888889999999999 64.00 1 . 0000000000000000000000000000000000000011111111111111111111111111 60.00 1 . 222222222222222222222222222222222222222222222333333333333333 22.00 1 . 4444444444445555555555 16.00 1 . 6666666666667777 11.00 1 . 88888889999 9.00 2 . 000001111 10.00 2 . 2222222333 8.00 2 . 44444445 14.00 2 . 66666666777777 6.00 2 . 888999 5.00 3 . 00111 4.00 3 . 2233 10.00 Extremes (>=34500) Stem width: 10000 Each leaf: 1 case(s) CURRENT SALARY Stem-and-Leaf Plot for sex= FEMALES Frequency Stem & Leaf 5.00 6 . 33444 12.00 6 . 566777788999 4.00 7 . 0223 11.00 7 . 56688888999 17.00 8 . 00111222333334444 23.00 8 . 55555666777777888889999 24.00 9 . 000000111111222233333444 28.00 9 . 5566666666677777777778899999 14.00 10 . 00000022223344 13.00 10 . 5555566668999 12.00 11 . 011111124444 12.00 11 . 566667777899 3.00 12 . 012 4.00 12 . 5566 6.00 13 . 001233 7.00 13 . 5578889 4.00 14 . 0224 2.00 14 . 68 3.00 15 . 014 1.00 15 . 5 11.00 Extremes (>=16320) Stem width: 1000 Each leaf: 1 case(s)

12 CURRENT SALARY Stem-and-Leaf Plot for minority= WHITE Frequency Stem & Leaf 12.00 6 . 334446677899 10.00 7 . 0225888999 38.00 8 . 00112233334445555566677777778888889999 46.00 9 . 0000000111112223333334455666666666777777778899 42.00 10 . 000000112222333334555555566667788899999999 37.00 11 . 0011111122344444445566666777777778999 36.00 12 . 000001111223333333333444556666667789 26.00 13 . 00012233345555555788888999 13.00 14 . 0112222444468 13.00 15 . 0011144556789 13.00 16 . 0000113346899 5.00 17 . 23449 8.00 18 . 00124479 3.00 19 . 025 6.00 20 . 024558 6.00 21 . 026799 9.00 22 . 000236678 3.00 23 . 257 7.00 24 . 0012577 1.00 25 . 0 36.00 Extremes (>=26000) Stem width: 1000 Each leaf: 1 case(s) CURRENT SALARY Stem-and-Leaf Plot for minority= NONWHITE Frequency Stem & Leaf 5.00 6 . 57789 6.00 7 . 366888 10.00 8 . 1234557899 16.00 9 . 0012446667777999 23.00 10 . 00022345555666666668899 13.00 11 . 1111122335689 16.00 12 . 0011233333345677 2.00 13 . 38 5.00 14 . 01124 1.00 15 . 3 1.00 16 . 0 6.00 Extremes (>=17580) Stem width: 1000 Each leaf: 1 case(s) How to use the stem-and-leaf plots. Look for the plot that allows you to eliminate the majority of the outliers! In this example there are 10 males that make greater than or equal to (GE) 34,000 per year, 11 females that make GE 16,320 per year, 36 Whites that make GE 26,000 per year, and 6 nonwhites that make GE 17580 per year. I would eliminate cases that make 26,000 per year or more on the DV.

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13 How to eliminate extreme values on the DV within SPSS. PLEASE NOTE, THE HOMEWORK DOES NOT REQUIRE YOU TO DO THIS! IT IS SHOWN HERE BECAUSE IT MAY BE REQUIRED FOR EXAM PURPOSES. Within SPSS go to transform, recode into different variables. Click over the variable you want to re-create (Salnow in this example).

14 Give the variable a new name – the label is not required (but good to use). Click old and new values. Under Old Value – Click range value through HIGHEST. Type in 26,000. Under New Value click System-missing. Click Add.

15 Do not click continue yet! Under Old Value click – All other values. Under New Value – Click Copy old Values – Click Add. Click continue. On the main dialogue window (shown below) click change – then Click OK.

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16 Back to the homework…. Once you have examined outliers, you need to examine normality. You could examine normality by looking at the histogram or Komogorov Smirnov (based on what was requested – you have neither of these in your current output). So just refer to the descriptive and check for skewness (which can indicate a violation in normality). The data is substantially positively skewed (especially for nonwhites). You are looking for values that deviate from 0 (COULD BE POSTIVE OR NEGATIVE). It is + indicating a log transformation!

17 3c. Allows you to skip the transformation and go straight to the analysis. Please note, steps for transformation were reviewed in chapter 3 homework. Steps for carrying out the analysis were reviewed in question 2 of this homework assignment! Under options I want to ask for Homgeneity Tests. This is the test for equal variances! Look at and interpret Levene’s test According to Levene’s test. The assumption for equal variances across groups has NOT been met. The probability associated with this result is .000. ,000 is less than .05 – therefore, the finding is significant! However, with larger samples sizes the factorial analysis is robust to this violation of the assumption . Robust is the means an analysis can be carried out when an assumption is violated.

18 As with all factorial designs, you need to first interpret the interaction. If the interaction was significant, you would not interpret the main effects. The probability associated with the interaction between sex and minority status on salary is .306 (.306 is greater than .05 and therefore not significant). glyph1197ext, interpret the main effects. There is a significant main effect of sex and minority. The probability associated with both is .000 (each less than .05). The effect size associated with sex is close to medium (.177). The effect size associated with minority (.060) is considered small. 3d. Please see writing up results on page 79 of your textbook (4 th edition).

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19 Effect size conventions. ANOVA Effect Size of effect f % of variance small .1 1 medium .25 6 large .4 14

Chapter 4 HomeworkKey

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