Psychometrics Project

A self-esteem journey through gender identities

include_graphics(genderImagePath)

Introduction

We are going to perform a factor analysis on a dataset of answers to the Rosenberg Self-Esteem Scale carryed out in 2014, then check the results and compare them for three different gender identities stated in the questionnaire:

Females
Other
Males

There are 10 questions:

Q1. I feel that I am a person of worth, at least on an equal plane with others.
Q2. I feel that I have a number of good qualities.
Q3. All in all, I am inclined to feel that I am a failure.
Q4. I am able to do things as well as most other people.
Q5. I feel I do not have much to be proud of.
Q6. I take a positive attitude toward myself.
Q7. On the whole, I am satisfied with myself.
Q8. I wish I could have more respect for myself.
Q9. I certainly feel useless at times.
Q10. At times I think I am no good at all.

We start by checking the assumptions for the factor analysis

First

The nature of the variables

All the variables are Likert items.

Second

Normality

We start by plotting the variables

dataNormalities <- data %>% dplyr::select(-gender) 

dataNormalities <- dataNormalities %>%   pivot_longer(cols=everything() ,names_to = "VAR", values_to = "vals")

ggplot(dataNormalities,aes(x=vals))+geom_bar(fill="steelblue")+facet_wrap(~VAR) + theme_void()

It looks like the variables are generally normal , except for Q2 and Q4 that are more left-skewed than most , and Q10.

Let’s check with actual statistical tests:

lshap <- data %>% dplyr::select(-gender) %>% lapply(lillie.test)

lres <- sapply(lshap, `[`, c("statistic","p.value"))

lres <- t(lres)

datatable(lres)

The test outputs a strong rejection of the null hypothesis that the distributions are normal.

We’ll go on using methods that don’t assume normality of the distributions of our variables.

Third

KMO test ( since we can’t perform a Bartlett Sphericity test)

$KMO_j = \frac{\sum_{i\neq j}r_{ij}^2}{\sum_{i\neq j}r_{ij}^2+\sum_{i\neq j}u}$

where:

R = $r_{ij}$ is the correlation matrix, U = $u_{ij}$ is the partial covariance matrix

corrMatrix <- data %>% dplyr::select(-gender) %>% cor(method = ("spearman"))

corrplot(corrMatrix,type="upper",order="hclust")

psych::KMO(corrMatrix)

## Kaiser-Meyer-Olkin factor adequacy
## Call: psych::KMO(r = corrMatrix)
## Overall MSA =  0.92
## MSA for each item = 
##   Q1   Q2   Q3   Q4   Q5   Q6   Q7   Q8   Q9  Q10 
## 0.92 0.91 0.95 0.95 0.95 0.92 0.92 0.97 0.89 0.90

We expect an index of at least 0.6 to carry on the analysis , so in this case we are satisfied,as the 0.92 Measure of Sample Adequacy suggests that it is worthwhile to analyze this correlation matrix.

Exploratory factor analysis

We start by selecting the number of components/factors:

We do this with a parallel analysis scree plot

data %>% dplyr::select(-gender) %>% fa.parallel(fm="pa")

## Parallel analysis suggests that the number of factors =  4  and the number of components =  2

We choose 2 factors.

Principal axis analysis

res_PA<- data %>% dplyr::select(-gender) %>%  fa(nfactors = 2,n.obs=nrow(data),rotate="none", fm = 'pa')

res_PA

## Factor Analysis using method =  pa
## Call: fa(r = ., nfactors = 2, n.obs = nrow(data), rotate = "none", 
##     fm = "pa")
## Standardized loadings (pattern matrix) based upon correlation matrix
##       PA1  PA2   h2   u2 com
## Q1   0.75 0.33 0.67 0.33 1.4
## Q2   0.68 0.37 0.60 0.40 1.5
## Q3  -0.75 0.13 0.58 0.42 1.1
## Q4   0.61 0.26 0.43 0.57 1.3
## Q5  -0.71 0.10 0.51 0.49 1.0
## Q6   0.78 0.11 0.62 0.38 1.0
## Q7   0.76 0.08 0.58 0.42 1.0
## Q8  -0.55 0.21 0.35 0.65 1.3
## Q9  -0.72 0.36 0.65 0.35 1.5
## Q10 -0.77 0.33 0.71 0.29 1.3
## 
##                        PA1  PA2
## SS loadings           5.06 0.64
## Proportion Var        0.51 0.06
## Cumulative Var        0.51 0.57
## Proportion Explained  0.89 0.11
## Cumulative Proportion 0.89 1.00
## 
## Mean item complexity =  1.3
## Test of the hypothesis that 2 factors are sufficient.
## 
## The degrees of freedom for the null model are  45  and the objective function was  5.31 with Chi Square of  254553.3
## The degrees of freedom for the model are 26  and the objective function was  0.22 
## 
## The root mean square of the residuals (RMSR) is  0.03 
## The df corrected root mean square of the residuals is  0.04 
## 
## The harmonic number of observations is  47974 with the empirical chi square  3674.65  with prob <  0 
## The total number of observations was  47974  with Likelihood Chi Square =  10568.66  with prob <  0 
## 
## Tucker Lewis Index of factoring reliability =  0.928
## RMSEA index =  0.092  and the 90 % confidence intervals are  0.09 0.093
## BIC =  10288.42
## Fit based upon off diagonal values = 1
## Measures of factor score adequacy             
##                                                    PA1  PA2
## Correlation of (regression) scores with factors   0.96 0.79
## Multiple R square of scores with factors          0.93 0.63
## Minimum correlation of possible factor scores     0.85 0.25

We can see that:

from the loadings - partial correlation coefficients of factors to items -

$a_{ia}=\sqrt\lambda_a-u_{ia}$

the first factor explains most of the variables, and we have items like Q2,Q4,Q8,Q9,Q10 that have cross-loading across factors, this is reflected in the complexities as seen below.

from the complexities - how much an item reflects each construct. Computed with Hoffman’s index:

$Comp=\frac{\sum\lambda^2_i}{\sum\lambda^4_i}$

they are still kind of high , with mean item complexity = 1.3 .

from the communalities - % of item variance explained by the extracted factors , basically the equivalent of a regression ’s $R^2$

$Comm_i= \sum_{a=1}^m a^2_{ia}$

we can consider as acceptable the communalities from 0.2 on , since we are considering an acceptable factor loading as 0.4. This is because , approximately:

$Item variance = FL^2$

In our case they are already much higher than that.

Uniqueness , as a result , will be $1 - communality$.

lastly , about the correlation between factors:

cor(res_PA$scores)

##              PA1          PA2
## PA1  1.000000000 -0.005227977
## PA2 -0.005227977  1.000000000

No high correlation between factors is found. In general , if the correlations are >0.85 the factors are not distinct from each other, thus they overlap , leading to multicollinearity, so they should be combined (Brown,2015).

Let’s check the path diagram

fa.diagram(res_PA)

As we stated before , all the variables are explained by the first factor only , so we’ll need to dive deeper and look for a rotation.

We go for an oblique rotation:

The rotation of choice is oblimin (Fabrigar & Wegener, 2012):

$R=P\phi P+U^2$

as we leave room for variables to be correlated , ’cause we know that with those 10 items we are trying to explain just one construct.

res_PA_Oblimin<- data %>% dplyr::select(-gender) %>%  fa(nfactors = 2,n.obs=nrow(data),rotate="oblimin", fm = 'pa')

res_PA_Oblimin

## Factor Analysis using method =  pa
## Call: fa(r = ., nfactors = 2, n.obs = nrow(data), rotate = "oblimin", 
##     fm = "pa")
## Standardized loadings (pattern matrix) based upon correlation matrix
##       PA1   PA2   h2   u2 com
## Q1   0.00  0.82 0.67 0.33 1.0
## Q2   0.08  0.82 0.60 0.40 1.0
## Q3   0.58 -0.23 0.58 0.42 1.3
## Q4  -0.02  0.65 0.43 0.57 1.0
## Q5   0.52 -0.25 0.51 0.49 1.4
## Q6  -0.30  0.55 0.62 0.38 1.6
## Q7  -0.32  0.51 0.58 0.42 1.7
## Q8   0.57 -0.03 0.35 0.65 1.0
## Q9   0.86  0.08 0.65 0.35 1.0
## Q10  0.84  0.00 0.71 0.29 1.0
## 
##                        PA1  PA2
## SS loadings           2.91 2.78
## Proportion Var        0.29 0.28
## Cumulative Var        0.29 0.57
## Proportion Explained  0.51 0.49
## Cumulative Proportion 0.51 1.00
## 
##  With factor correlations of 
##       PA1   PA2
## PA1  1.00 -0.68
## PA2 -0.68  1.00
## 
## Mean item complexity =  1.2
## Test of the hypothesis that 2 factors are sufficient.
## 
## The degrees of freedom for the null model are  45  and the objective function was  5.31 with Chi Square of  254553.3
## The degrees of freedom for the model are 26  and the objective function was  0.22 
## 
## The root mean square of the residuals (RMSR) is  0.03 
## The df corrected root mean square of the residuals is  0.04 
## 
## The harmonic number of observations is  47974 with the empirical chi square  3674.65  with prob <  0 
## The total number of observations was  47974  with Likelihood Chi Square =  10568.66  with prob <  0 
## 
## Tucker Lewis Index of factoring reliability =  0.928
## RMSEA index =  0.092  and the 90 % confidence intervals are  0.09 0.093
## BIC =  10288.42
## Fit based upon off diagonal values = 1
## Measures of factor score adequacy             
##                                                    PA1  PA2
## Correlation of (regression) scores with factors   0.94 0.94
## Multiple R square of scores with factors          0.89 0.87
## Minimum correlation of possible factor scores     0.77 0.75

As we can see from the results:

from the loadings , we don’t have as much cross-loading across factors as before, this is reflected in the complexities as seen below.
from the complexities , the mean lowered , so now the mean item complexity = 1.2 .
from the communalities , just like before ,we can consider as acceptable the communalities from 0.2 on , since we are considering an acceptable factor loading as 0.4 on : $ Item variance = FL^2$ In our case they are already much higher than that.
lastly , about the correlation between factors:

res_PA_Oblimin$score.cor

##            [,1]       [,2]
## [1,]  1.0000000 -0.6994701
## [2,] -0.6994701  1.0000000

These correlations reflect some interesting things : * the two factors are explaining two opposite things. * we have a clue that the two factors are explaining the same construct as expected - even though of course we still can’t be sure of the construct we’re measuring , that’s validity’s job -.

In general though , interpreting the results , we should drop Q3,Q5,Q6 and Q7 , as they have a pretty high complexity:

res_PA_Oblimin<- data %>% dplyr::select(-c(gender,Q3,Q5,Q6,Q7)) %>% fa(nfactors = 2,n.obs=nrow(mydata),rotate="oblimin", fm ='pa')

res_PA_Oblimin

## Factor Analysis using method =  pa
## Call: fa(r = ., nfactors = 2, n.obs = nrow(mydata), rotate = "oblimin", 
##     fm = "pa")
## Standardized loadings (pattern matrix) based upon correlation matrix
##       PA1   PA2   h2   u2 com
## Q1  -0.04  0.80 0.69 0.31   1
## Q2   0.05  0.83 0.64 0.36   1
## Q4  -0.05  0.63 0.44 0.56   1
## Q8   0.54 -0.05 0.33 0.67   1
## Q9   0.86  0.04 0.70 0.30   1
## Q10  0.84 -0.04 0.75 0.25   1
## 
##                        PA1  PA2
## SS loadings           1.78 1.76
## Proportion Var        0.30 0.29
## Cumulative Var        0.30 0.59
## Proportion Explained  0.50 0.50
## Cumulative Proportion 0.50 1.00
## 
##  With factor correlations of 
##       PA1   PA2
## PA1  1.00 -0.63
## PA2 -0.63  1.00
## 
## Mean item complexity =  1
## Test of the hypothesis that 2 factors are sufficient.
## 
## The degrees of freedom for the null model are  15  and the objective function was  2.42 with Chi Square of  115976.2
## The degrees of freedom for the model are 4  and the objective function was  0 
## 
## The root mean square of the residuals (RMSR) is  0.01 
## The df corrected root mean square of the residuals is  0.01 
## 
## The harmonic number of observations is  47974 with the empirical chi square  69.01  with prob <  3.7e-14 
## The total number of observations was  47974  with Likelihood Chi Square =  194.76  with prob <  5e-41 
## 
## Tucker Lewis Index of factoring reliability =  0.994
## RMSEA index =  0.032  and the 90 % confidence intervals are  0.028 0.035
## BIC =  151.65
## Fit based upon off diagonal values = 1
## Measures of factor score adequacy             
##                                                    PA1  PA2
## Correlation of (regression) scores with factors   0.93 0.92
## Multiple R square of scores with factors          0.86 0.84
## Minimum correlation of possible factor scores     0.72 0.68

Now we can see that the mean item complexity is 1 .

The loadings lead us to do this separation:

Q1,Q2,Q4 belong to PA2
Q8,Q9,Q10 belong to PA1

Let’s carve an interpretation out of these question groups:

PA2: Evenness with others (neither inferiority complex , nor superiority complex) OR positive attitude towards self :

Q1. I feel that I am a person of worth, at least on an equal plane with others.
Q2. I feel that I have a number of good qualities.
Q4. I am able to do things as well as most other people.

PA1: Self Disheartening (in italian: avvilirsi) OR negative attitude towards self:

Q8. I wish I could have more respect for myself.
Q9. I certainly feel useless at times.
Q10. At times I think I am no good at all.

The communalities are very good as most exceed the value of 0.6 , so they exceed the usual researchers’ threshold of 0.5, even though we have Q4 and Q8 that are respectively 0.4 and 0.3, but they respect out fixed minimum of 0.2 anyway.

Lastly,let’s check the factor correlations:

cor(res_PA_Oblimin$scores)

##            PA1        PA2
## PA1  1.0000000 -0.7133108
## PA2 -0.7133108  1.0000000

The correlations are still negative and significant as we expected , so what we told before is still valid.

Let’s see the diagram:

fa.diagram(res_PA_Oblimin)

This diagram accounts for everything we’ve said before.

So, in conclusion, going into the actual analysis of the self-esteem differences through gender , we will use these results.

Let’s assess the reliability:

To do that , we are going to compute Cronbach’s alpha, as it indicates the internal consistency reliability.

Let’s start with the first factor , the one that accounts for the negative attitude towards self.

positiveSelf = c('Q8','Q9','Q10')
negativeSelf = c('Q1','Q2','Q4')

relPositive = psych::alpha(data[,positiveSelf])
relNegative = psych::alpha(data[,negativeSelf])

print(relPositive, digits = 3 )

## 
## Reliability analysis   
## Call: psych::alpha(x = data[, positiveSelf])
## 
##   raw_alpha std.alpha G6(smc) average_r  S/N     ase mean    sd median_r
##      0.796     0.795   0.744     0.564 3.87 0.00161 2.68 0.861    0.493
## 
##  lower alpha upper     95% confidence boundaries
## 0.793 0.796 0.8 
## 
##  Reliability if an item is dropped:
##     raw_alpha std.alpha G6(smc) average_r  S/N alpha se var.r med.r
## Q8      0.838     0.840   0.724     0.724 5.24  0.00147    NA 0.724
## Q9      0.658     0.660   0.493     0.493 1.94  0.00310    NA 0.493
## Q10     0.643     0.643   0.474     0.474 1.80  0.00326    NA 0.474
## 
##  Item statistics 
##         n raw.r std.r r.cor r.drop mean    sd
## Q8  47974 0.767 0.779 0.569  0.521 2.68 0.969
## Q9  47974 0.872 0.870 0.798  0.701 2.77 1.007
## Q10 47974 0.887 0.878 0.812  0.711 2.57 1.084
## 
## Non missing response frequency for each item
##         0     1     2     3     4 miss
## Q8  0.004 0.138 0.240 0.405 0.212    0
## Q9  0.005 0.139 0.197 0.393 0.265    0
## Q10 0.004 0.217 0.219 0.325 0.235    0

Cronbach’s alpha = 0.79

print(relNegative, digits = 3 )

## 
## Reliability analysis   
## Call: psych::alpha(x = data[, negativeSelf])
## 
##   raw_alpha std.alpha G6(smc) average_r  S/N     ase mean    sd median_r
##      0.804     0.804   0.739     0.578 4.11 0.00155    3 0.715    0.546
## 
##  lower alpha upper     95% confidence boundaries
## 0.801 0.804 0.807 
## 
##  Reliability if an item is dropped:
##    raw_alpha std.alpha G6(smc) average_r  S/N alpha se var.r med.r
## Q1     0.690     0.690   0.527     0.527 2.22  0.00283    NA 0.527
## Q2     0.706     0.706   0.546     0.546 2.40  0.00268    NA 0.546
## Q4     0.795     0.796   0.661     0.661 3.90  0.00187    NA 0.661
## 
##  Item statistics 
##        n raw.r std.r r.cor r.drop mean    sd
## Q1 47974 0.873 0.868 0.777  0.690 3.00 0.876
## Q2 47974 0.858 0.860 0.761  0.677 3.09 0.824
## Q4 47974 0.812 0.815 0.649  0.589 2.91 0.829
## 
## Non missing response frequency for each item
##        0     1     2     3     4 miss
## Q1 0.002 0.063 0.180 0.438 0.317    0
## Q2 0.007 0.043 0.131 0.495 0.324    0
## Q4 0.005 0.051 0.213 0.494 0.237    0

Cronbach’s alpha = 0.80

We have good values as they are both greater than 0.7.

Let’s analyze the different genders:

dataGenders <- cbind(res_PA_Oblimin$scores,data)

dataFemales <- dataGenders %>% dplyr::filter(gender == 2) %>% dplyr::select(c(PA2,PA1))

dataOther <- dataGenders %>% dplyr::filter(gender == 3) %>% dplyr::select(c(PA2,PA1))

dataMales <- dataGenders %>% dplyr::filter(gender == 1) %>% dplyr::select(c(PA2,PA1))

# mean(dataFemales$PA2)

Rather than going into the analysis with some hypothesis on how the genders will behave towards their self-esteems , maybe it’d be better to let the numbers speak , then draw conclusions.

We remark the meanings of the factors:

PA2: Evenness with others PA1: Self Disheartening

The mean of factors’scores for females are:

PA2: -0.08 -0.08
PA1: 0.06 0.08

The mean of factors’scores for ‘other genders’ are:

PA2: -0.51 -0.51
PA1: 0.26 0.46

The mean of factors’scores for males are:

PA2: 0.15 0.16
PA1: -0.1 -0.1

Females compared to Males:

PA2: Females have considerably lower scores for the first factor on average.
PA1: Females have considerably higher scores for the second factor on average.

While females have more of an inferiority complex towards the world compared to males, they also tend to dishearten more . So , the data reflects a well-spread stereotype: at least on the surface , men are psychologically “stronger” than men.

Basically our data is playing The Cure for us:

include_graphics(cureImagePath)

Females compared to ‘other’ genders:

PA2: Females have lower scores for the first factor on average.
PA1: Females have higher scores for the second factor on average.

The interpretation is the same as before: While females have more of an inferiority complex towards the world compared to ‘other genders’, they also tend to dishearten more .

Males compared to ‘other’ genders:

PA2: Males have considerably higher scores for the second factor on average.
PA1: Males have considerably lower scores for the second factor on average.

Basically males don’t suffer from inferiority complex towards the world as much as other genders. Males also tend to dishearten less .

include_graphics(mentalPath)

In conclusion , it seems like cisgender males are the ones with a better overall self-esteem and , infering from that , better mental well-being than other genders.

But…why?

Regarding ‘other’ genders : This may be due to the fact that cisgender males are on average treated in a better way in this historical moment compared to transgender people for example, that are still seeking acceptance from the world.

Regarding women : only recently are starting to pave their way through the western world - in 1945 women voted for the first time in Italy - while we are aware that in the eastern world it’s not the same at all.

Let’s check whether or not those results are statistically significant:

We’ll perform some Welch t-tests , as neither the sample sizes , nor the variances are equal for the different gender samples.

t.test(dataFemales$PA2,dataOther$PA2,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataFemales$PA2 and dataOther$PA2
## t = 9.4844, df = 543.2, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.3385766 0.5154575
## sample estimates:
##   mean of x   mean of y 
## -0.08497743 -0.51199447

t.test(dataFemales$PA1,dataOther$PA1,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataFemales$PA1 and dataOther$PA1
## t = -9.4941, df = 547.31, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.4594393 -0.3019161
## sample estimates:
##  mean of x  mean of y 
## 0.07753153 0.45820926

t.test(dataFemales$PA2,dataMales$PA2,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataFemales$PA2 and dataMales$PA2
## t = -28.149, df = 38264, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.2578549 -0.2242835
## sample estimates:
##   mean of x   mean of y 
## -0.08497743  0.15609178

t.test(dataFemales$PA1,dataMales$PA1,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataFemales$PA1 and dataMales$PA1
## t = 24.647, df = 36990, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1997876 0.2343080
## sample estimates:
##   mean of x   mean of y 
##  0.07753153 -0.13951625

t.test(dataOther$PA2,dataMales$PA2,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataOther$PA2 and dataMales$PA2
## t = -14.78, df = 551.91, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.7568762 -0.5792964
## sample estimates:
##  mean of x  mean of y 
## -0.5119945  0.1560918

t.test(dataOther$PA1,dataMales$PA1,alternative = "two.sided", var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  dataOther$PA1 and dataMales$PA1
## t = 14.814, df = 561.21, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.5184721 0.6769789
## sample estimates:
##  mean of x  mean of y 
##  0.4582093 -0.1395163

We reject the null hypothesis for all our t-tests: the differences in means are statically significant for each one of our comparisons.

Return back to introduction

A work by Antonio Taurisano

hedislimanners@protonmail.com