Exam 1

  1. Metainformation

    Tag Value
    fileReliability_eur-reliability-112-en_eur-reliability-112-en
    nameeur-reliability-112-en
    sectionReliability/Analysis/Cronbach's alpha
    typenum
    solution0.896
    tolerance0
    TypeCalculate
    ProgramCalculator
    LanguageEnglish
    LevelStatistical Literacy

    Question

    Twenty parallel items are removed from an 80 items questionnaire that had an alpha of .92. What will be the new alpha? Use the formula below:

    Rxx−revised=n×RXX−original1+(n−1)RXX−originalR_{xx-revised} = \frac{n \times R_{XX-original}}{1+(n-1) R_{XX-original}}

    Note that n = lengthening factor. Original number of items x lengthening factor = new number of items.


    Solution

    In order to calculate the new alpha we need the Spearman‐Brown formula. In this case n is 6080=.75\frac{60}{80}=.75 So: Rxx−revised=.75×.921+(.75−1).92=.896R_{xx-revised} = \frac{.75\times.92}{1+(.75-1).92} = .896.