Exploratory Factor Analysis (EFA): Part 2

Relevant packages

• tidyverse
• psych
• GPArotation

Practical Issues with EFA

Data: pgpets.csv

A pet food company has conducted a questionnaire on the internet (\(n = 620\)) to examine whether owning a pet influences low mood. They asked 16 questions on a Likert scale (1-7, detailed below) followed by a simple Yes/No question concerning whether the respondent owned a pet.
There are lots of questions, and the researchers don’t really know much about the theory of mood disorders, but they think that they are likely picking up on multiple different types of “low mood”. They want to conduct a factor analysis to examine this, and then plan on investigating the group differences (pet owners vs not pet owners) on the factor scores.

The data is available at https://uoepsy.github.io/data/pgpets.csv

QuestionNumber	Over the last 2 weeks, how much have you had/have you been...
item1	Little interest or pleasure in doing things?
item2	Feeling down, depressed, or hopeless?
item3	Trouble falling or staying asleep, or sleeping too much?
item4	Feeling tired or having little energy?
item5	Poor appetite or overeating?
item6	Feeling bad about yourself - or that you are a failure or have let yourself or your family down?
item7	Reading the newspaper or watching television?
item8	Moving or speaking so slowly that other people could have noticed? Or the opposite - being so fidgety or restless that you have been moving around a lot more than usual?
item9	A lack of motivation to do anything at all?
item10	Feeling nervous, anxious or on edge?
item11	Not being able to stop or control worrying?
item12	Worrying too much about different things?
item13	Trouble relaxing?
item14	Being so restless that it is hard to sit still?
item15	Becoming easily annoyed or irritable?
item16	Feeling afraid as if something awful might happen?

Question 1

Read the data into R.
Create a new object in R that contains a subset the data. It should include all variables except for the participant ID and the variable corresponding to whether or not they have a pet (we’re going to come back to these later on).
In this new object, change the names of the columns to match the question number, rather than the question itself (see the data description above). This will be easier to work with.

Hint:
Check the output of the code paste0("item", 1:10). Consider this code in combination with another function: names(data). How could you combine the two codes to assign the new names to the current variable names?

Solution

library(tidyverse)
library(psych)

pgpets <- read_csv("https://uoepsy.github.io/data/pgpets.csv")

names(pgpets)

 [1] "little_interest_or_pleasure_in_doing_things"                                                                                                                          
 [2] "feeling_down_depressed_or_hopeless"                                                                                                                                   
 [3] "trouble_falling_or_staying_asleep_or_sleeping_too_much"                                                                                                               
 [4] "feeling_tired_or_having_little_energy"                                                                                                                                
 [5] "poor_appetite_or_overeating"                                                                                                                                          
 [6] "feeling_bad_about_yourself_or_that_you_are_a_failure_or_have_let_yourself_or_your_family_down"                                                                        
 [7] "trouble_concentrating_on_things_such_as_reading_the_newspaper_or_watching_television"                                                                                 
 [8] "moving_or_speaking_so_slowly_that_other_people_could_have_noticed_or_the_opposite_being_so_fidgety_or_restless_that_you_have_been_moving_around_a_lot_more_than_usual"
 [9] "a_lack_of_motivation_to_do_anything_at_all"                                                                                                                           
[10] "feeling_nervous_anxious_or_on_edge"                                                                                                                                   
[11] "not_being_able_to_stop_or_control_worrying"                                                                                                                           
[12] "worrying_too_much_about_different_things"                                                                                                                             
[13] "trouble_relaxing"                                                                                                                                                     
[14] "being_so_restless_that_it_is_hard_to_sit_still"                                                                                                                       
[15] "becoming_easily_annoyed_or_irritable"                                                                                                                                 
[16] "feeling_afraid_as_if_something_awful_might_happen"                                                                                                                    
[17] "do_you_own_a_pet"                                                                                                                                                     
[18] "ppt_id"                                                                                                                                                               

df <- pgpets %>% 
    select(-ppt_id, -do_you_own_a_pet)

names(df) <- paste0("item", 1:ncol(df))
head(df)

# A tibble: 6 × 16
  item1 item2 item3 item4 item5 item6 item7 item8 item9 item10 item11 item12
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>  <dbl>  <dbl>
1     5     5     5     4     6     7     4     3     6      4      4      5
2     4     4     3     5     3     5     2     4     4      3      3      5
3     5     5     5     5     3     7     5     5     6      5      5      5
4     4     5     5     6     4     5     4     6     7      6      6      4
5     2     1     3     3     3     3     5     4     2      3      3      3
6     6     4     6     5     5     4     3     5     6      5      4      4
# … with 4 more variables: item13 <dbl>, item14 <dbl>, item15 <dbl>,
#   item16 <dbl>

Question 2

Visualise the items (this might be the histograms of all marginal distributions, or a scatterplot matrix, or both).

Solution

library(psych)
multi.hist(df)

The data have 6 variables, so we will create two plots each focusing on 8 at a time:

pairs.panels(df[, 1:8])

pairs.panels(df[, 9:16])

Question 3

Compute the correlation matrix for the items, and assess the suitability for factor analysis, using the Bartlett test and the Kaiser-Meyer-Olkin factor adequacy.

Solution

cordf = cor(df)

cortest.bartlett(cordf, n = nrow(df))

$chisq
[1] 3008

$p.value
[1] 0

$df
[1] 120

Bartlett’s test indicates that the correlation matrix is proportionally different from the identity matrix (\(\chi^2 (120) = 3008, p<.001\)), suggesting our correlations are significantly different from zero.

KMO(cordf)

Kaiser-Meyer-Olkin factor adequacy
Call: KMO(r = cordf)
Overall MSA =  0.9
MSA for each item = 
 item1  item2  item3  item4  item5  item6  item7  item8  item9 item10 item11 
  0.90   0.91   0.93   0.90   0.92   0.89   0.40   0.92   0.88   0.91   0.90 
item12 item13 item14 item15 item16 
  0.92   0.94   0.88   0.90   0.91 

The KMO for individual items are mainly “meritorious” or “marvelous”. The only item that may need further investigation is item 7, having an “unacceptable” KMO. The overall measure of sampling adequacy is 0.90, which suggests we have suitable data to perform a factor analysis

Question 4

Determine how many factors you will extract.

Solution

Using parallel analysis:

fa.parallel(df, fa = 'fa')

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

Using Velicer’s Minimum Average Partial:

VSS(df)

Very Simple Structure
Call: vss(x = x, n = n, rotate = rotate, diagonal = diagonal, fm = fm, 
    n.obs = n.obs, plot = plot, title = title, use = use, cor = cor)
VSS complexity 1 achieves a maximimum of 0.67  with  3  factors
VSS complexity 2 achieves a maximimum of 0.86  with  8  factors

The Velicer MAP achieves a minimum of 0.01  with  2  factors 
BIC achieves a minimum of  -477  with  2  factors
Sample Size adjusted BIC achieves a minimum of  -195  with  2  factors

Statistics by number of factors 
  vss1 vss2   map dof  chisq     prob sqresid  fit  RMSEA  BIC SABIC complex
1 0.66 0.00 0.035 104 1014.1 4.0e-149    12.1 0.66 0.1188  345   676     1.0
2 0.66 0.81 0.012  89   94.8  3.2e-01     6.8 0.81 0.0101 -477  -195     1.2
3 0.67 0.80 0.020  75   75.2  4.7e-01     6.5 0.82 0.0015 -407  -169     1.3
4 0.64 0.81 0.028  62   60.4  5.3e-01     5.9 0.83 0.0000 -338  -141     1.3
5 0.63 0.82 0.042  50   38.1  8.9e-01     5.2 0.85 0.0000 -283  -125     1.3
6 0.64 0.84 0.058  39   28.4  8.9e-01     4.4 0.88 0.0000 -222   -99     1.3
7 0.62 0.84 0.076  29   18.8  9.3e-01     4.0 0.89 0.0000 -168   -76     1.4
8 0.63 0.86 0.089  20    9.6  9.7e-01     3.7 0.89 0.0000 -119   -55     1.4
  eChisq   SRMR eCRMS eBIC
1 1880.5 0.1124 0.121 1212
2   73.0 0.0221 0.026 -499
3   55.9 0.0194 0.025 -426
4   40.9 0.0166 0.023 -358
5   25.1 0.0130 0.020 -296
6   16.1 0.0104 0.018 -235
7    9.2 0.0078 0.016 -177
8    4.4 0.0054 0.013 -124

Both parallel analysis and Velicer’s Minimum Average Partial suggested retaining 2 factors.

Question 5

Choosing an appropriate estimation method and rotation, perform a factor analysis to extract the desired number of factors (based on your answer to the previous question).

If you get an error, you may need to install the “GPArotation” package: install.packages("GPArotation").

Solution

We probably shouldn’t use MLE as an estimation method here, because we’re going to be treating what looks like some sort of Likert data (ordinal, responses of 1-7) as if it is continuous.
Let’s use minres here.

Finally, do we have any reason to think that the factors we are going to extract are orthogonal? Given what the questions look to be evaluating, it’s hard to see distinctly unrelated constructs within the 16 items.

pgmod <- fa(df, nfactors = 2, rotate = "oblimin", fm = "minres")
pgmod

Factor Analysis using method =  minres
Call: fa(r = df, nfactors = 2, rotate = "oblimin", fm = "minres")
Standardized loadings (pattern matrix) based upon correlation matrix
         MR1   MR2     h2   u2 com
item1   0.45 -0.02 0.1947 0.81 1.0
item2   0.54 -0.05 0.2685 0.73 1.0
item3   0.50  0.44 0.6071 0.39 2.0
item4   0.75  0.04 0.5871 0.41 1.0
item5   0.44 -0.12 0.1678 0.83 1.1
item6   0.81 -0.02 0.6464 0.35 1.0
item7   0.01  0.03 0.0012 1.00 1.4
item8   0.57 -0.02 0.3112 0.69 1.0
item9   0.84 -0.03 0.6921 0.31 1.0
item10  0.01  0.68 0.4669 0.53 1.0
item11 -0.05  0.63 0.3778 0.62 1.0
item12  0.05  0.41 0.1887 0.81 1.0
item13  0.07  0.51 0.2933 0.71 1.0
item14 -0.03  0.77 0.5818 0.42 1.0
item15 -0.04  0.65 0.4033 0.60 1.0
item16  0.01  0.59 0.3579 0.64 1.0

                       MR1  MR2
SS loadings           3.23 2.92
Proportion Var        0.20 0.18
Cumulative Var        0.20 0.38
Proportion Explained  0.53 0.47
Cumulative Proportion 0.53 1.00

 With factor correlations of 
     MR1  MR2
MR1 1.00 0.39
MR2 0.39 1.00

Mean item complexity =  1.1
Test of the hypothesis that 2 factors are sufficient.

The degrees of freedom for the null model are  120  and the objective function was  4.91 with Chi Square of  3008
The degrees of freedom for the model are 89  and the objective function was  0.16 

The root mean square of the residuals (RMSR) is  0.02 
The df corrected root mean square of the residuals is  0.03 

The harmonic number of observations is  620 with the empirical chi square  73  with prob <  0.89 
The total number of observations was  620  with Likelihood Chi Square =  94.8  with prob <  0.32 

Tucker Lewis Index of factoring reliability =  0.997
RMSEA index =  0.01  and the 90 % confidence intervals are  0 0.025
BIC =  -477
Fit based upon off diagonal values = 0.99
Measures of factor score adequacy             
                                                   MR1  MR2
Correlation of (regression) scores with factors   0.94 0.92
Multiple R square of scores with factors          0.88 0.85
Minimum correlation of possible factor scores     0.77 0.69

DETAILS

If we think about a factor analysis being a set of regressions (convention in factor analysis is to use \( {\lambda_{,}} \) instead of \(\beta\)), then we can think of a given item being the manifestation of some latent factors, plus a bit of randomness (or ‘stray causes’):

\[\begin{aligned} \text{Item}_{1} &= {\lambda_{1,1}} \cdot \text{Factor}_{1} + {\lambda_{2,1}} \cdot \text{Factor}_{2} + \u{1} \\ \text{Item}_{2} &= {\lambda_{1,2}} \cdot \text{Factor}_{1} + {\lambda_{2,2}} \cdot \text{Factor}_{2} + \u{2} \\ &\vdots \\ \text{Item}_{16} &= {\lambda_{1,16}} \cdot \text{Factor}_{1} + {\lambda_{2,16}} \cdot \text{Factor}_{2} + \u{16} \end{aligned}\]

As you can see from the above, the 16 different items all stem from the same two factors (\( \text{Factor}_{1} , \text{Factor}_{2} \)), plus some item-specific errors (\(\u{1}, \dots, \u{16}\)). The \( {\lambda_{,}} \) terms are called factor loadings, or loadings in short

Communality is sum of the squared factor loadings for each item.

Intuitively, for each row, the two \( {\lambda_{,}} \)s tell us how much each item depends on the two factors shared by the 16 items. The sum of the squared loadings tells us how much of one item’s information is due to the shared factors.

The communality is a bit like the \(R^2\) (the proportion of variance of an item that is explained by the factor structure). And the standardised loadings are a bit like the \(R^2\) for an item onto a specific factor (i.e. the proportion of variance of an item that is explained by factor X).

The Uniqueness of each item is simply \(1 - \text{communality}\).
This is the leftover bit; the variance in each item that is left unexplained by the latent factors (this could be specific variance, or it could be error variance. the one thing we know is, it’s not common variance).

Side note: this is what sets Factor Analysis apart from PCA, which is the linear combination of total variance (including error) in all our items. FA allows some of the variance to be shared by the underlying factors, and considers the remainder to be unique to the individual items (or, in another, error in how each item measures the construct).

The Complexity of an item corresponds to how well an item reflects a single underlying construct. Specifically, it is \({(\sum \lambda_i^2)^2}/{\sum \lambda_i^4}\), where \(\lambda_i\) is the loading on to the \(i^{th}\) factor. It will be equal to 1 for an item which loads only on one factor, and 2 if it loads evenly on to two factors, and so on.

In R, we will often see these estimats under specific columns:

• h2 = item communality
  
• u = item uniqueness
  
• com = item complexity

Question 6

Using fa.sort(), examine the loadings of each item onto the factors, along with communalities, uniqueness and complexity scores.

• Do all factors load on 3+ items at a salient level?
• Do all items have at least one loading above a salient cut off?
• Are there any Heywood cases (communalities or standardised loadings \(\geq1\))?
• Should we perhaps remove some complex items?
• Does the solution account for an acceptable level of variance?
  
• Is the factor structure (items that load on to each factor) coherent, and does it make theoretical sense?

Solution

fa.sort(pgmod)

Factor Analysis using method =  minres
Call: fa(r = df, nfactors = 2, rotate = "oblimin", fm = "minres")
Standardized loadings (pattern matrix) based upon correlation matrix
         MR1   MR2     h2   u2 com
item9   0.84 -0.03 0.6921 0.31 1.0
item6   0.81 -0.02 0.6464 0.35 1.0
item4   0.75  0.04 0.5871 0.41 1.0
item8   0.57 -0.02 0.3112 0.69 1.0
item2   0.54 -0.05 0.2685 0.73 1.0
item3   0.50  0.44 0.6071 0.39 2.0
item1   0.45 -0.02 0.1947 0.81 1.0
item5   0.44 -0.12 0.1678 0.83 1.1
item14 -0.03  0.77 0.5818 0.42 1.0
item10  0.01  0.68 0.4669 0.53 1.0
item15 -0.04  0.65 0.4033 0.60 1.0
item11 -0.05  0.63 0.3778 0.62 1.0
item16  0.01  0.59 0.3579 0.64 1.0
item13  0.07  0.51 0.2933 0.71 1.0
item12  0.05  0.41 0.1887 0.81 1.0
item7   0.01  0.03 0.0012 1.00 1.4

                       MR1  MR2
SS loadings           3.23 2.92
Proportion Var        0.20 0.18
Cumulative Var        0.20 0.38
Proportion Explained  0.53 0.47
Cumulative Proportion 0.53 1.00

 With factor correlations of 
     MR1  MR2
MR1 1.00 0.39
MR2 0.39 1.00

Mean item complexity =  1.1
Test of the hypothesis that 2 factors are sufficient.

The degrees of freedom for the null model are  120  and the objective function was  4.91 with Chi Square of  3008
The degrees of freedom for the model are 89  and the objective function was  0.16 

The root mean square of the residuals (RMSR) is  0.02 
The df corrected root mean square of the residuals is  0.03 

The harmonic number of observations is  620 with the empirical chi square  73  with prob <  0.89 
The total number of observations was  620  with Likelihood Chi Square =  94.8  with prob <  0.32 

Tucker Lewis Index of factoring reliability =  0.997
RMSEA index =  0.01  and the 90 % confidence intervals are  0 0.025
BIC =  -477
Fit based upon off diagonal values = 0.99
Measures of factor score adequacy             
                                                   MR1  MR2
Correlation of (regression) scores with factors   0.94 0.92
Multiple R square of scores with factors          0.88 0.85
Minimum correlation of possible factor scores     0.77 0.69

• Both factors load on 3+ items at a salient level.
• All except item7 has loadings \(>0.3\).
• There are no Heywood cases.
• Item 3 loads quite highly on both factors (high complexity)
• The solution explains 38% of the variance.
  
• The coherence of the factor structure will require us to look back at the questions themselves. We’ll not do this right now.

Question 7

If you think any items should be removed, do so now, and perform the factor analysis once more.

Solution

We’re going to remove that problematic item 7 (which seems to be about “reading the newspaper and watching tv”), as well as item 3 (which seems to be about sleepiness, and relates to both factors quite highly).

df2 <- df %>% 
    select(-item7,-item3)
cortest.bartlett(cor(df2), n = nrow(df2))

$chisq
[1] 2516

$p.value
[1] 0

$df
[1] 91

KMO(cor(df2))

Kaiser-Meyer-Olkin factor adequacy
Call: KMO(r = cor(df2))
Overall MSA =  0.88
MSA for each item = 
 item1  item2  item4  item5  item6  item8  item9 item10 item11 item12 item13 
  0.90   0.92   0.88   0.91   0.87   0.91   0.85   0.89   0.89   0.91   0.92 
item14 item15 item16 
  0.86   0.89   0.90 

fa.parallel(df2, fa = 'fa')

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

VSS(df2)

Very Simple Structure
Call: vss(x = x, n = n, rotate = rotate, diagonal = diagonal, fm = fm, 
    n.obs = n.obs, plot = plot, title = title, use = use, cor = cor)
VSS complexity 1 achieves a maximimum of 0.73  with  3  factors
VSS complexity 2 achieves a maximimum of 0.83  with  8  factors

The Velicer MAP achieves a minimum of 0.01  with  2  factors 
BIC achieves a minimum of  -342  with  2  factors
Sample Size adjusted BIC achieves a minimum of  -139  with  2  factors

Statistics by number of factors 
  vss1 vss2   map dof chisq     prob sqresid  fit RMSEA  BIC SABIC complex
1 0.62 0.00 0.046  77 983.7 2.8e-157    10.9 0.62 0.138  489   733     1.0
2 0.73 0.81 0.015  64  69.6  2.9e-01     5.6 0.81 0.012 -342  -139     1.1
3 0.73 0.80 0.024  52  51.3  5.0e-01     5.3 0.81 0.000 -283  -118     1.2
4 0.71 0.81 0.039  41  34.4  7.6e-01     4.7 0.84 0.000 -229   -99     1.2
5 0.67 0.81 0.056  31  21.9  8.9e-01     4.3 0.85 0.000 -177   -79     1.3
6 0.64 0.81 0.075  22  13.5  9.2e-01     3.9 0.86 0.000 -128   -58     1.3
7 0.71 0.82 0.100  14   4.9  9.9e-01     4.0 0.86 0.000  -85   -41     1.4
8 0.64 0.83 0.149   7   2.3  9.4e-01     3.4 0.88 0.000  -43   -20     1.5
   eChisq   SRMR eCRMS eBIC
1 1857.56 0.1283 0.139 1362
2   50.68 0.0212 0.025 -361
3   34.68 0.0175 0.023 -300
4   21.55 0.0138 0.021 -242
5   12.82 0.0107 0.018 -186
6    6.65 0.0077 0.016 -135
7    2.50 0.0047 0.012  -88
8    0.88 0.0028 0.010  -44

pgmod2 <- fa(df2, nfactors = 2, rotate = "oblimin", fm = "minres")
pgmod2$loadings

Loadings:
       MR1    MR2   
item1   0.438       
item2   0.531       
item4   0.743       
item5   0.431 -0.113
item6   0.806       
item8   0.564       
item9   0.837       
item10         0.676
item11         0.629
item12         0.408
item13         0.509
item14         0.766
item15         0.639
item16         0.593

                 MR1   MR2
SS loadings    2.894 2.644
Proportion Var 0.207 0.189
Cumulative Var 0.207 0.396

fa.sort(pgmod2)

Factor Analysis using method =  minres
Call: fa(r = df2, nfactors = 2, rotate = "oblimin", fm = "minres")
Standardized loadings (pattern matrix) based upon correlation matrix
         MR1   MR2   h2   u2 com
item9   0.84 -0.01 0.70 0.30 1.0
item6   0.81  0.00 0.65 0.35 1.0
item4   0.74  0.06 0.59 0.41 1.0
item8   0.56 -0.01 0.31 0.69 1.0
item2   0.53 -0.03 0.27 0.73 1.0
item1   0.44 -0.02 0.19 0.81 1.0
item5   0.43 -0.11 0.16 0.84 1.1
item14 -0.02  0.77 0.58 0.42 1.0
item10  0.02  0.68 0.47 0.53 1.0
item15 -0.02  0.64 0.40 0.60 1.0
item11 -0.03  0.63 0.38 0.62 1.0
item16  0.03  0.59 0.36 0.64 1.0
item13  0.08  0.51 0.29 0.71 1.0
item12  0.06  0.41 0.19 0.81 1.0

                       MR1  MR2
SS loadings           2.89 2.64
Proportion Var        0.21 0.19
Cumulative Var        0.21 0.40
Proportion Explained  0.52 0.48
Cumulative Proportion 0.52 1.00

 With factor correlations of 
     MR1  MR2
MR1 1.00 0.35
MR2 0.35 1.00

Mean item complexity =  1
Test of the hypothesis that 2 factors are sufficient.

The degrees of freedom for the null model are  91  and the objective function was  4.1 with Chi Square of  2516
The degrees of freedom for the model are 64  and the objective function was  0.11 

The root mean square of the residuals (RMSR) is  0.02 
The df corrected root mean square of the residuals is  0.03 

The harmonic number of observations is  620 with the empirical chi square  50.7  with prob <  0.89 
The total number of observations was  620  with Likelihood Chi Square =  69.6  with prob <  0.29 

Tucker Lewis Index of factoring reliability =  0.997
RMSEA index =  0.012  and the 90 % confidence intervals are  0 0.028
BIC =  -342
Fit based upon off diagonal values = 0.99
Measures of factor score adequacy             
                                                   MR1  MR2
Correlation of (regression) scores with factors   0.93 0.91
Multiple R square of scores with factors          0.87 0.83
Minimum correlation of possible factor scores     0.75 0.66

Reliability

Measurement Error & Reliability

You will often find research that foregoes the measurement model by taking a scale score (i.e., the sum or mean of a set of likert-type questions). For instance, think back to our exercises on path mediation, where “Health Locus of Control” was measured as the “average score on a set of items relating to perceived control over ones own health”. In doing so, we make the assumption that these variables provide measurements of the underlying latent construct without error.
In fact, if we think about it a little, in our simple regression framework \(y = \beta_0 + \beta_1x_1 + ... + \beta_kx_k + \varepsilon\) all our predictor variables \(x_1\) to \(x_k\) are assumed to be measured without error - it is only our outcome \(y\) that our model considers to have some randomness included (the \(\varepsilon\)). This seems less problematic if our variables are representing something that is quite easily and reliably measured (time, weight, height, age, etc.), but it seems inappropriate when we are concerned with something more abstract. For instance, two people both scoring 11 on the Generalised Anxiety Disorder 7 (GAD-7) scale does not necessarily mean that they have identical levels of anxiety.

The inconsistency with which an observed variable reflects the underlying construct that we consider it to be measuring is termed the reliability.

Reliability: A silly example

Suppose I’m trying to weigh my dog. I have a set of scales, and I put him on the scales. He weighs in at 13.53kg. I immediately do it again, and the scales this time say 13.41kg. I do it again. 13.51kg, and again, 13.60kg. What is happening? Is Dougal’s weight (Dougal is the dog, by the way) randomly fluctuating by 100g? Or are my scales just a bit inconsistent, and my observations contain measurement error?

I take him to the vets, where they have a much better set of weighing scales, and I do the same thing (measure him 4 times). The weights are 13.47, 13.49, 13.48, 13.48.
The scales at the vets are clearly more reliable. We still don’t know Dougal’s true weight, but we are better informed to estimate it if we go on the measurements from the scales at the vet.¹

Another way to think about reliability is to consider the idea that more error means less reliability:

\[\text{observations = truth + error}\]

There are different types of reliability:

• test re-test reliability: correlation between values over repeated measurements.
• alternate-form reliability: correlation between scores on different forms/versions of a test (we might want different versions to avoid practice effects).
• Inter-rater reliability: correlation between values obtained from different raters.
• split-half reliability: correlation between scores of two equally sized subsets of items.

The form of reliability we are going to be most concerned with here is known as Internal Consistency. This is the extent to which items within a scale are correlated with one another. There are two main measures of this:

alpha and omega

Cronbach’s \(\alpha\) ranges from 0 to 1 (higher is better). You can get this using the alpha() function from the psych package. The formula is:

\[ \begin{aligned} \text{Cronbach's } \alpha = \frac{n \cdot \overline{cov(ij)}}{\overline{\sigma^2_i} + (n-1) \cdot \overline{cov(ij)}} \\ \\ \text{where}: \qquad & n = \text{number of items} \\ & \overline{cov(ij)} = \text{average covariance between item-pairs} \\ & \overline{\sigma^2_i} = \text{average item variance} \\ \end{aligned} \]

McDonald’s Omega (\(\omega\)) is substantially more complicated, but avoids the limitation that Cronbach’s alpha which assumes that all items are equally related to the construct. You can get it using the omega() function from the psych package. If you want more info about it then the help docs (?omega()) are a good place to start.

Question 8

Using the relevant function, obtain alpha for the set of items in each factor from your final factor analysis model of low mood.

Solution

The first factor has high loadings for items 1 to 9 (we have excluded 3 and 7).

names(df2)

 [1] "item1"  "item2"  "item4"  "item5"  "item6"  "item8"  "item9"  "item10"
 [9] "item11" "item12" "item13" "item14" "item15" "item16"

alpha(select(df2, item1:item9))

Reliability analysis   
Call: alpha(x = select(df2, item1:item9))

  raw_alpha std.alpha G6(smc) average_r S/N  ase mean   sd median_r
      0.82      0.81     0.8      0.37 4.2 0.01    4 0.98     0.35

    95% confidence boundaries 
         lower alpha upper
Feldt     0.79  0.82  0.84
Duhachek  0.80  0.82  0.84

 Reliability if an item is dropped:
      raw_alpha std.alpha G6(smc) average_r S/N alpha se var.r med.r
item1      0.82      0.81    0.80      0.41 4.2    0.010 0.022  0.43
item2      0.81      0.79    0.78      0.39 3.9    0.011 0.024  0.35
item4      0.77      0.76    0.75      0.35 3.2    0.013 0.016  0.31
item5      0.82      0.81    0.80      0.42 4.3    0.010 0.020  0.43
item6      0.76      0.75    0.74      0.34 3.0    0.013 0.015  0.30
item8      0.80      0.79    0.78      0.38 3.8    0.011 0.023  0.35
item9      0.76      0.75    0.73      0.33 3.0    0.014 0.014  0.30

 Item statistics 
        n raw.r std.r r.cor r.drop mean  sd
item1 620  0.54  0.57  0.44   0.40  4.0 1.1
item2 620  0.61  0.63  0.52   0.48  4.0 1.2
item4 620  0.79  0.77  0.75   0.68  4.0 1.5
item5 620  0.50  0.54  0.40   0.36  4.0 1.1
item6 620  0.83  0.80  0.79   0.72  4.1 1.7
item8 620  0.64  0.65  0.56   0.51  4.0 1.2
item9 620  0.85  0.82  0.82   0.74  3.9 1.8

Non missing response frequency for each item
         1    2    3    4    5    6    7 miss
item1 0.01 0.07 0.26 0.33 0.23 0.08 0.01    0
item2 0.02 0.09 0.22 0.34 0.23 0.09 0.02    0
item4 0.06 0.10 0.20 0.28 0.20 0.11 0.06    0
item5 0.01 0.07 0.24 0.36 0.24 0.06 0.01    0
item6 0.07 0.12 0.17 0.22 0.21 0.13 0.08    0
item8 0.02 0.09 0.23 0.30 0.27 0.07 0.02    0
item9 0.13 0.12 0.15 0.20 0.17 0.14 0.10    0

alpha(select(df2, item10:item16))

Reliability analysis   
Call: alpha(x = select(df2, item10:item16))

  raw_alpha std.alpha G6(smc) average_r S/N   ase mean   sd median_r
      0.81       0.8    0.78      0.37 4.1 0.012    4 0.89     0.37

    95% confidence boundaries 
         lower alpha upper
Feldt     0.78  0.81  0.83
Duhachek  0.78  0.81  0.83

 Reliability if an item is dropped:
       raw_alpha std.alpha G6(smc) average_r S/N alpha se  var.r med.r
item10      0.77      0.77    0.74      0.35 3.3    0.014 0.0072  0.35
item11      0.78      0.78    0.75      0.37 3.5    0.013 0.0085  0.37
item12      0.81      0.80    0.78      0.41 4.1    0.012 0.0035  0.41
item13      0.79      0.79    0.76      0.38 3.7    0.013 0.0077  0.41
item14      0.76      0.75    0.72      0.34 3.1    0.015 0.0045  0.35
item15      0.78      0.77    0.75      0.36 3.4    0.013 0.0072  0.35
item16      0.78      0.78    0.75      0.37 3.5    0.013 0.0077  0.37

 Item statistics 
         n raw.r std.r r.cor r.drop mean  sd
item10 620  0.74  0.73  0.67   0.61  4.0 1.4
item11 620  0.69  0.69  0.61   0.55  4.0 1.3
item12 620  0.54  0.56  0.43   0.38  4.0 1.2
item13 620  0.62  0.63  0.53   0.48  3.9 1.2
item14 620  0.79  0.77  0.74   0.66  3.9 1.5
item15 620  0.69  0.69  0.62   0.56  4.0 1.3
item16 620  0.67  0.67  0.60   0.54  4.0 1.2

Non missing response frequency for each item
          1    2    3    4    5    6    7 miss
item10 0.04 0.10 0.22 0.27 0.24 0.08 0.04    0
item11 0.04 0.08 0.19 0.33 0.24 0.10 0.03    0
item12 0.01 0.09 0.24 0.32 0.25 0.08 0.01    0
item13 0.03 0.08 0.25 0.35 0.19 0.10 0.01    0
item14 0.06 0.11 0.22 0.27 0.20 0.10 0.05    0
item15 0.03 0.10 0.24 0.31 0.20 0.10 0.03    0
item16 0.02 0.09 0.23 0.30 0.25 0.08 0.01    0

Cronbach’s \(\alpha\) of 0.82 and 0.81 for the two factors and suggests high internal consistency.

Replicability

Question 9

Split the dataset in half, and assess the replicability of your factor structure of low mood by examining the factor congruence in scores on each subset of the data.

Hint: see the lectures!

Solution

library(tidyverse)
set1 <- df2 %>%
  sample_frac(.5) # randomly select one half
set2 <- anti_join(df2, set1) # select the non-matching cases

res1 <- fa(set1, nfactors = 2, rotate = "oblimin")
res2 <- fa(set2, nfactors = 2, rotate = "oblimin")
fa.congruence(res1, res2) 

     MR1   MR2
MR1 0.98 -0.01
MR2 0.03  1.00

We can see there is a high level of factor congruence across the two subsets of data.

Factor Scores

Question 10

Extract the factor scores for each factor of low-mood from your model, and attach them to original dataset (the one which has information on pet ownership).
Then, conduct a \(t\)-test to examine whether the pet-owners differ from non-pet-owners in their levels of each factor of low mood.

As a bonus, can you come up with some description of the two factors? (you will have to look back to the question numbers and questions right at the start).

QuestionNumber	Over the last 2 weeks, how much have you had/have you been...
item1	Little interest or pleasure in doing things?
item2	Feeling down, depressed, or hopeless?
item3	Trouble falling or staying asleep, or sleeping too much?
item4	Feeling tired or having little energy?
item5	Poor appetite or overeating?
item6	Feeling bad about yourself - or that you are a failure or have let yourself or your family down?
item7	Reading the newspaper or watching television?
item8	Moving or speaking so slowly that other people could have noticed? Or the opposite - being so fidgety or restless that you have been moving around a lot more than usual?
item9	A lack of motivation to do anything at all?
item10	Feeling nervous, anxious or on edge?
item11	Not being able to stop or control worrying?
item12	Worrying too much about different things?
item13	Trouble relaxing?
item14	Being so restless that it is hard to sit still?
item15	Becoming easily annoyed or irritable?
item16	Feeling afraid as if something awful might happen?

Solution

pgpets <- 
  pgpets %>% 
    mutate(
      depression = pgmod2$scores[,1],
      anxiety = pgmod2$scores[,2],
      pet = factor(do_you_own_a_pet)
    )

t.test(pgpets$depression ~ pgpets$pet)

    Welch Two Sample t-test

data:  pgpets$depression by pgpets$pet
t = 1, df = 600, p-value = 0.3
alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
95 percent confidence interval:
 -0.0691  0.2273
sample estimates:
mean in group 0 mean in group 1 
         0.0375         -0.0416 

ggplot(pgpets, aes(x=pet, y=depression))+
  geom_boxplot()

t.test(pgpets$anxiety~pgpets$pet)

    Welch Two Sample t-test

data:  pgpets$anxiety by pgpets$pet
t = 4, df = 597, p-value = 0.00004
alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
95 percent confidence interval:
 0.159 0.444
sample estimates:
mean in group 0 mean in group 1 
          0.143          -0.158 

ggplot(pgpets, aes(x=pet, y=anxiety))+
  geom_boxplot()

Footnotes

1. Of course this all assuming that the scales aren’t completely miscalibrated