training<\/em><\/strong>.<\/p>\n\n\n\nImagine the new employees in the last exercise were randomly divided into a group that received no training, one that received training and one that received training and some team building exercises early on in their employment at the company. You need the file repeated measures mixed.sav (SPSS) or repeated measures mixed.xls (Excel).<\/p>\n\n\n\n
Conduct the two-way analysis for efficiency<\/em> and training<\/em> ignoring the column headed \u2018graduate\u2019 for now. Compare your results with those under \u2018Show Answer\u2019 below. Select Descriptives<\/strong> under Options<\/strong> to get the means and SDs for the efficiency<\/em> conditions. To get the means and SDs for the training groups on each efficiency level you can create a new variable which is the mean of the three efficiency measures then use Analyze\/Compare Means\/Means<\/strong>. E<\/u>M Means<\/strong> will give means but inappropriate SD estimates.<\/p>\n\n\n\nShow answer<\/summary>\nThere was a main effect for efficiency with the means rising from M = 43.2, SD = 7.33 at one month, through M = 45.8, SD = 7.68 at six months to M = 49.06, SD = 6.09 at twelve months. F2,90<\/sub> =1<\/em>2.<\/em>454, p<\/em> < .001, effect size (partial eta<\/em>2<\/sup>) = .217<\/p>\n\n\n\nThere was a main effect for training with means of M = 41.96, SD = 3.86 for the untrained group, M = 46.313, SD = 4.48 for the trained group and M = 49.813, SD = 4.30 for the trained and team building group. F<\/em>2,45<\/sub> = 13.879, p<\/em> < .001, effect size (partial eta<\/em>2<\/sup> ) = .382.<\/p>\n\n\n\nThe interaction was not significant. Sphericity was at an acceptable level (p<\/em> = .213). Levene\u2019s test for homogeneity of variance was not significant for any level of efficiency<\/em> so equality of variances was assumed.<\/p>\n<\/details>\n\n\n\nIf you\u2019re really feeling adventurous you could try the three-way mixed ANOVA that is produced by including the factor of graduate. This tells us whether the participant was a graduate or not. I have only provided brief details of results below but enough to let you see you\u2019ve performed the analysis correctly.<\/p>\n\n\n\nShow answer<\/summary>\nMain effect efficiency F<\/em>2,84<\/sub> = 13.034, p<\/em> <.001, eta2<\/sup> = .237
Main effect training F<\/em>2,42<\/sub> = 17.598, p<\/em> < .001, eta2<\/sup> = .456<\/p>\n\n\n\nMain effect graduate F1,42<\/sub> = 4.21, p<\/em> = 046, eta2<\/sup> = .091
Interaction efficiency x training not significant
Interaction efficiency x graduate not significant
Interaction training x graduate significant F<\/em>2,42<\/sub> = 5.424, p<\/em> = .008, eta2<\/sup> = .205
Three-way interaction efficiency x training x graduate approaching significance p<\/em> = .054<\/p>\n<\/details>\n\n\n\nExercise 24.3<\/h3>\n\n\n\n
Calculation of a two-way repeated measures ANOVA<\/strong><\/p>\n\n\n\n