Exam 1

  1. Metainformation

    Tag Value
    fileDescriptive-statistics_eur-descriptive-108-en_eur-descriptive-108-en
    nameeur-descriptive-108-en
    sectionDescriptive statistics/Summary Statistics/Bivariate statistics/Covariance
    typenum
    solution0.738
    tolerance0.01
    TypeCalculate
    ProgramCalculater
    LanguageEnglish
    LevelStatistical Literacy

    Question

    Below you find the scores of six students on the course 1.1 exam (social psychology) and the course 1.3 exam (statistics). Calculate the covariance between the exam scores of course 1.1 and course 1.3.

    Students Exam Course 1.1 Exam Course 1.3
    1 7.5 6.8
    2 6.2 5.4
    3 5.2 7.6
    4 8.2 7.6
    5 9.2 8.4
    6 6.9 4.7

    Solution

        |Students |Exam Course 1.1 |Exam Course 1.3 |Product
    deviation 1.1 – deviation 1.3
    1 7.5 6.8 (7.5 ‐ 7.2)(6.8 ‐ 6.75)
    2 6.2 5.4 (6.2 ‐ 7.2)(5.4 ‐ 6.75)
    3 5.2 7.6 (5.2 ‐ 7.2)(7.6 ‐ 6.75)
    4 8.2 7.6 (8.2 ‐ 7.2)(7.6 ‐ 6.75)
    5 9.2 8.4 (9.2 ‐ 7.2)(8.4 ‐ 6.75)
    6 6.9 4.7 (6.9 ‐ 7.2)(4.7 ‐ 6.75)
    Sum 43.2 40.5 4.43
    Mean 7.2 6.75 0.738

    The mean is calculated by summing the scores on the exam and dividing this sum by the number of participants:

    Mean Exam 1.1 = 43.206=7.20\frac{43.20}{6} = 7.20

    Mean Exam 1.3 = 40.506=6.75\frac{40.50}{6} = 6.75

    The covariance is calculated by summing the product of the deviation scores and divide this by N: Covariance Exam 1.1, Exam 1.3: CovE1.1,E1.34.436=0.738Cov_{E1.1, E1.3} \frac{4.43}{6}=0.738