yes, you may proceed further and create your variable.

You can create a new variable. But it is necessary to ensure that these items are components of the same factor or variable. Therefore, I believe that you should first analyze the dimensionality of the items (parallel analysis, for example) and complement the evaluation with a factor analysis. The factor analysis generates weight for the items, which facilitates the calculation of the weighted average. Free software like R does such procedures with ease.

Jonhatan Magno Norte da Silva Doesn't the acceptable Cronbach Alpha inform you that the items are all under one component? How is it different from Factor Analysis?

I have never read in any specialized literature that Cronbach's alpha greater than 0.70 indicates that all items form a latent factor or trait.

In a nutshell, Cronbach's alpha is a measure of reliability that indicates how much the items correlate in the mean. The fact that the items correlate well (alpha> 0.70, for example), does not guarantee that they form or are part of the same factor.

Factor analysis is a family of methods and techniques that use correlations (interrelations) to define a set of dimensions to which the items in a questionnaire belong. Thus, if all items have a good factor load and commonality in the same dimension, there is evidence that all items measure the same factor. However, if a part of the items has a good factor load and commonality in dimension 1 and another part of the items has a good factor load and commonality in dimension 2, for example, there are indications that the questionnaire measures two factors, that is, it is two-dimensional. And so on for 3, 4, ..., n dimensions.

There are different ways to define the name of the dimension formed by the items. I, in particular, like to analyze the semantic part of the items. If all the items talk about cooperation from the boss, co-operation of coworkers, help from top management in tasks, for example, we could say that the items measure 'work support'.

Hope this helps.

Jonhatan Magno Norte da Silva Let's assume that I want to identify items under your example of 'work support'. I have conducted interviews to derive all the items that I believe constitute the dimension of 'work support'. Are you saying that I have to conduct a factor analysis to determine if the items I have identified are of one dimension?

Dimensionality can be observed by some techniques, including factor analysis (for your exploratory case, I believe). I always recommend Factorial analysis as it is a simple and robust method of simultaneously observing the validity and dimensionality of the measuring instrument. But you can get evidence of unidimensionality through the analysis of the parallel analysis graph too, for example. On the internet you will find tutorials and videos showing how to easily perform parallel analysis on software such as R.

Jonhatan Magno Norte da Silva I was reading up about factor analysis and came across the topics of extraction and rotation. What would be a simple way to decide which extraction method to use (Principle Component Analysis or Principle Axis Factoring), and which rotation to use (Varimax)?

The second part is about validity as you mentioned - so can I assume that factor analysis provides the validity for my construct and cronbach alpha provides the reliability?

Also, I am using SPSS, if that is important. Thank you!

Yes, the analysis of the main components and varimax rotation are techniques that can be used to solve your problem. You are on the right track.

When the alpha value and the factor analysis present satisfactory parameters, it provides strong indications of validity and reliability for the Instrument. In particular, I only consider the Instrument to be totally valid and reliable when replicated by other researchers, in addition to being reassessed by means of confirmatory factor analysis, structural equation model or item response theory, among other psychometric techniques.

But, as I mentioned, the signs of validity and reliability are very strong when Factor Analysis parameters and the alpha value are satisfactory.

Regarding the use of SPSS no problem. However, when publishing their findings in scientific journals, editors may ask you for information about your license, since it is SPSS is paid software. Therefore, free (free) software such as R, Python and Julia are growing in popularity among researchers.

Jonhatan Magno Norte da Silva If my factor analysis show that all my items belong under one factor/construct, does that mean that I can compute a new variable to represent that construct by averaging all items? This would be a simple average, but how may I calculate a weighted average based on the factor loadings of the items?

Some software calculates as factorial loads, commonality of the items and the score of each respondent.

However, the score of each respondent can be calculated by a weighted sum as well. To do so, you must add the multiplication of each answer by the factorial load of the item. For example, let's say that the respondent answered alternative 2 for item 1, 4 for item 2, 4 for item 3, 4 for item 4, and 1 for item 5. Let's also say that the factorial loads of the items are 0.552, 0.480, 0.710, 0.350 and 0.450, respectively for items 1, 2, 3, 4 and 5. Therefore, the respondent's score will be equal to (2 * 0.552) + (4 * 0.480) + (4 * 0.710) + (4 * 0.350) + (1 * 0.450).

Note that in this example the item with the highest load is more important in calculating the score. This procedure is useful, showing that not all items are equally important in calculating the latent factor or trait.

Jonhatan Magno Norte da Silva so I have 6 items measured on a 5-point likert scale. The component scores of the items are: 0.848, 0.84, 0.792, 0.741, 0.624, 0.375

For a respondent that has rated 4 in all 6 items, I got a score of 16.88. Is this correct? Does this mean that the respondent rated 16.88/30 as a summary of all 6 items?

You see. Someone who answered 4 for all items will have a score equal to: (0.848 × 4) + (0.84 × 4) + (0.792 × 4) + (0.741 × 4) + (0.624 × 4) + (0.375 × 4) = 16.88

And the result will be 16.88

You don't need to divide by any value.

To classify respondents, you can calculate the mean or median of the score of all respondents (benchmark).

So, let's say that the average of all respondents is equal to 16.5, then we can say that the respondent with a score equal to 16.88 had a high exposure to the factor assessed by the 6 items, because his score is higher than the average of the scores.

A respondent with a score equal to 9.1, for example, has a low exposure to the factor assessed by the 6 items.

Jonhatan Magno Norte da Silva I have named the factor representing the 6 items as "aspects of customer satisfaction at theme parks". The 5-point scale is the respondent's likelihood to include items 1 to 6 when gauging their satisfaction at a theme park.

So does a score higher than the mean indicate that the respondent gauges their satisfaction more closely with the factor?

If my objective is to simply identify the aspects of customer satisfaction at theme parks, is the calculation of the weighted average necessary, or can I just report the component scores?

In this case, you should calculate the score of each respondent as we discussed earlier. Then, calculate the average or median of all scores. Then you will analyze how many are above average or median (satisfied with the water park) and how many are below average or median (not satisfied with the water park).

By stating in your paper the number and percentage of respondents satisfied and not satisfied with the water park, you will have indications of the level of satisfaction of the users of the water park. The average talvel is not necessary, but the weighted sum will help you in the classification of users (satisfied and not satisfied).

Jonhatan Magno Norte da Silva Yes I understand. However, if I am only interested in reporting the aspects (i.e. the 6 items), I could just report the factor loadings of each item, right? What is the minimum factor loading that is recommended in factor analysis before I should remove an item from the factor?