Exam 1

  1. Metainformation

    Tag Value
    fileReliability_eur-reliability-111-en_eur-reliability-111-en
    nameeur-reliability-111-en
    sectionReliability/Analysis/Cronbach's alpha
    typenum
    solution30
    tolerance0
    TypeCalculate
    ProgramCalculator
    LanguageEnglish
    LevelStatistical Literacy

    Question

    A test consisting of 10 items has an alpha estimate of reliability of .50. How many parallel items have to be added to get an alpha estimate of reliability of .80?

    Use n=Rxx−revised(1−Rxx−original)Rxx−original(1−Rxx−revised)n =\frac{R_{xx-revised}(1-R_{xx-original})}{R_{xx-original}(1-R_{xx-revised})} in which n is the factor with which the original number of items has to be multiplied


    Solution

    In order to calculate the number of items that have to be added we need the Spearman Brown formula: n=Rxx−revised(1−Rxx−original)Rxx−original(1−Rxx−revised)n = \frac{R_{xx-revised}(1-R_{xx-original})}{R_{xx-original}(1-R_{xx-revised})} in which n is the factor with which the original number of items has to be multiplied. So in our case: n=.8(1−.5).5(1−.8)=.4.1=4n = \frac{.8(1-.5)}{.5(1-.8)} = \frac{.4}{.1} = 4. The number of items that has to be added is 40‐10=3040 ‐ 10 = 30.