R für Einsteiger: Musterlösungen
Kapitel 18: Verfahren für die Testkonstruktion
Diese Musterlösung wurde erstellt von Peter Hähner (Ruhr-Universität Bochum).
(c) Luhmann: R für Einsteiger, 5. Aufl., Beltz, 2020
Vorbereitungen
Setzen Sie ein Arbeitsverzeichnis oder legen Sie ein entsprechendes R-Projekt an (Kap. 23).
Laden Sie dann die Datei erstis.RData.
load("erstis.RData")Laden Sie die benötigten Pakete (ggf. müssen Sie diese vorab noch installieren).
library(psych)
library(tidyverse)
library(GPArotation)Aufgabe 1
Führen Sie eine Itemanalyse für die Lebenszufriedenheits-Items (lz13 bis lz17) durch. Auf welches der fünf Items könnte man am ehesten verzichten?
# Variablen in Data Frame speichern
lezu <- select(erstis, lz13, lz14, lz15, lz16, lz17)
# Itemanalyse durchführen
psych::alpha(lezu)##
## Reliability analysis
## Call: psych::alpha(x = lezu)
##
## raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
## 0.86 0.86 0.84 0.56 6.3 0.017 4.9 1.1 0.53
##
## lower alpha upper 95% confidence boundaries
## 0.82 0.86 0.89
##
## Reliability if an item is dropped:
## raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r med.r
## lz13 0.80 0.81 0.77 0.51 4.2 0.023 0.0054 0.51
## lz14 0.83 0.83 0.80 0.56 5.0 0.021 0.0058 0.53
## lz15 0.82 0.82 0.79 0.54 4.7 0.022 0.0065 0.53
## lz16 0.83 0.84 0.81 0.57 5.3 0.021 0.0123 0.59
## lz17 0.85 0.86 0.82 0.60 6.0 0.017 0.0062 0.61
##
## Item statistics
## n raw.r std.r r.cor r.drop mean sd
## lz13 189 0.86 0.86 0.84 0.77 4.8 1.4
## lz14 189 0.79 0.80 0.74 0.67 5.0 1.3
## lz15 189 0.81 0.83 0.78 0.71 5.3 1.2
## lz16 189 0.79 0.78 0.70 0.65 5.0 1.5
## lz17 188 0.76 0.74 0.63 0.59 4.5 1.7
##
## Non missing response frequency for each item
## 1 2 3 4 5 6 7 miss
## lz13 0.02 0.05 0.10 0.18 0.28 0.32 0.05 0.01
## lz14 0.01 0.05 0.07 0.19 0.28 0.31 0.09 0.01
## lz15 0.01 0.03 0.06 0.13 0.20 0.46 0.11 0.01
## lz16 0.02 0.04 0.13 0.16 0.20 0.32 0.13 0.01
## lz17 0.04 0.12 0.10 0.24 0.19 0.19 0.12 0.02Am ehesten würde man auf das Item lz17 verzichten, da es die geringste Trennschärfe aufweist (r.drop = .59) und da Cronbachs Alpha durch das Weglassen dieses Items am wenigsten beeinträchtigt wird (raw_alpha = .85).
Aufgabe 2
Untersuchen Sie die faktorielle Struktur der Stimmungs-Items (stim1 bis stim12). Fassen Sie zunächst diese Items in einem neuen Data Frame zusammen, der keine fehlenden Werte enthält. Bestimmen Sie anschließend die Anzahl der zu extrahierenden Faktoren mit einer Parallelanalyse und führen Sie dann eine exploratorische Faktorenanalyse mit Maximum-Likelihood-Schätzung und obliquer Rotation durch.
# Variablen in Data Frame speichern und fehlende Werte entfernen
stimmung <- na.omit(select(erstis, contains("stim")))
# Parallelanalyse
fa.parallel(stimmung)
## Parallel analysis suggests that the number of factors = 4 and the number of components = 3Die Parallelanalyse favorisiert eine Lösung mit drei oder vier Faktoren.
Beide werden im Folgenden einmal dargestellt.
# 3-Faktorenlösung
faktoren.3 <- fa(stimmung, nfactors = 3,
fm = "ml",
rotate = "promax")
print(faktoren.3, digits = 2, cut = .3, sort = TRUE)## Factor Analysis using method = ml
## Call: fa(r = stimmung, nfactors = 3, rotate = "promax", fm = "ml")
## Standardized loadings (pattern matrix) based upon correlation matrix
## item ML3 ML1 ML2 h2 u2 com
## stim3 3 0.91 0.65 0.35 1.1
## stim9 9 0.80 0.60 0.40 1.1
## stim6 6 -0.62 0.42 0.58 1.3
## stim12 12 -0.53 0.50 0.50 1.4
## stim11 11 0.39 -0.34 0.43 0.57 2.0
## stim8 8 0.98 0.75 0.25 1.1
## stim1 1 0.79 0.60 0.40 1.1
## stim10 10 0.42 -0.38 0.36 0.64 2.2
## stim7 7 0.92 0.70 0.30 1.1
## stim5 5 0.68 0.43 0.57 1.1
## stim2 2 -0.59 0.55 0.45 1.3
## stim4 4 0.38 0.62 2.9
##
## ML3 ML1 ML2
## SS loadings 2.43 2.02 1.91
## Proportion Var 0.20 0.17 0.16
## Cumulative Var 0.20 0.37 0.53
## Proportion Explained 0.38 0.32 0.30
## Cumulative Proportion 0.38 0.70 1.00
##
## With factor correlations of
## ML3 ML1 ML2
## ML3 1.00 -0.60 0.43
## ML1 -0.60 1.00 -0.46
## ML2 0.43 -0.46 1.00
##
## Mean item complexity = 1.5
## Test of the hypothesis that 3 factors are sufficient.
##
## The degrees of freedom for the null model are 66 and the objective function was 4.82 with Chi Square of 873.13
## The degrees of freedom for the model are 33 and the objective function was 0.42
##
## The root mean square of the residuals (RMSR) is 0.04
## The df corrected root mean square of the residuals is 0.06
##
## The harmonic number of observations is 187 with the empirical chi square 41.2 with prob < 0.15
## The total number of observations was 187 with Likelihood Chi Square = 74.44 with prob < 4.9e-05
##
## Tucker Lewis Index of factoring reliability = 0.896
## RMSEA index = 0.082 and the 90 % confidence intervals are 0.057 0.107
## BIC = -98.18
## Fit based upon off diagonal values = 0.99
## Measures of factor score adequacy
## ML3 ML1 ML2
## Correlation of (regression) scores with factors 0.93 0.93 0.91
## Multiple R square of scores with factors 0.86 0.87 0.83
## Minimum correlation of possible factor scores 0.72 0.74 0.67# 4-Faktorenlösung
faktoren.4 <- fa(stimmung,
nfactors = 4,
fm = "ml",
rotate = "promax")
print(faktoren.4, digits = 2, cut = .3, sort = TRUE)## Factor Analysis using method = ml
## Call: fa(r = stimmung, nfactors = 4, rotate = "promax", fm = "ml")
## Standardized loadings (pattern matrix) based upon correlation matrix
## item ML3 ML2 ML1 ML4 h2 u2 com
## stim8 8 0.90 0.70 0.30 1.1
## stim1 1 0.90 0.67 0.33 1.1
## stim4 4 -0.40 0.43 0.57 2.9
## stim11 11 -0.40 0.41 0.59 1.9
## stim7 7 0.92 0.69 0.31 1.1
## stim5 5 0.68 0.43 0.57 1.1
## stim2 2 -0.60 0.56 0.44 1.3
## stim10 10 0.35 -0.39 0.37 0.63 2.7
## stim3 3 0.96 0.80 0.20 1.0
## stim9 9 0.60 0.55 0.45 1.3
## stim12 12 0.90 0.83 0.17 1.0
## stim6 6 0.53 0.47 0.53 1.7
##
## ML3 ML2 ML1 ML4
## SS loadings 1.99 1.90 1.70 1.32
## Proportion Var 0.17 0.16 0.14 0.11
## Cumulative Var 0.17 0.32 0.47 0.58
## Proportion Explained 0.29 0.28 0.25 0.19
## Cumulative Proportion 0.29 0.56 0.81 1.00
##
## With factor correlations of
## ML3 ML2 ML1 ML4
## ML3 1.00 -0.50 -0.50 0.59
## ML2 -0.50 1.00 0.41 -0.34
## ML1 -0.50 0.41 1.00 -0.60
## ML4 0.59 -0.34 -0.60 1.00
##
## Mean item complexity = 1.5
## Test of the hypothesis that 4 factors are sufficient.
##
## The degrees of freedom for the null model are 66 and the objective function was 4.82 with Chi Square of 873.13
## The degrees of freedom for the model are 24 and the objective function was 0.23
##
## The root mean square of the residuals (RMSR) is 0.03
## The df corrected root mean square of the residuals is 0.05
##
## The harmonic number of observations is 187 with the empirical chi square 22.07 with prob < 0.57
## The total number of observations was 187 with Likelihood Chi Square = 40.62 with prob < 0.018
##
## Tucker Lewis Index of factoring reliability = 0.942
## RMSEA index = 0.061 and the 90 % confidence intervals are 0.025 0.093
## BIC = -84.92
## Fit based upon off diagonal values = 0.99
## Measures of factor score adequacy
## ML3 ML2 ML1 ML4
## Correlation of (regression) scores with factors 0.93 0.91 0.93 0.93
## Multiple R square of scores with factors 0.86 0.84 0.87 0.86
## Minimum correlation of possible factor scores 0.73 0.67 0.74 0.72In diesem Fall würde man sich vermutlich für die 4-Faktorenlösung entscheiden, da die Struktur der Faktorladungen einfacher zu interpretieren ist. Es gibt zum Beispiel weniger Items, die auf zwei Faktoren laden.