Answers to

--- Practice problems ---

Return to the questions

  1. By definition, the normal range includes the central 95% of results on normal people.

  2. The 95% confidence interval
    = mean - 2 SD to mean + 2 SD
    = 80 - 2x2 to 80 + 2x2
    = 76 - 84 mg/dl

  3. C.V. = 100 x ( SD / mean ) %
    = 100 x (2/80)%
    = 2.5 %

  4. The 95% confidence interval
    = mean - 2 SD to mean + 2 SD

      ( SD = C.V. x mean / 100

    = 2.5%x250 mg/dl/100% = 6.25 mg/dl )

    The 95% confidence interval
    = 250 - 2x6.25 to 250 + 2x6.25
    = 238 - 262 mg/dl

  5. sensitivity = 98/100 = 98 %

  6. specificity = 198/200 = 99 %

  7. PVP = TP / ( TP + FP )

        TP = prevalence x sensitivity

    = 0.01x0.98 = 0.0098
        FP = ( 1 - prevalence ) x ( 1 - specificity )
    = (1 - 0.01)x(1 - 0.99) = 0.99x0.01 = 0.0099

    PVP = 0.0098/( 0.0098 + 0.0099 ) = 50 %

  8. PVN = TN / ( TN + FN )

        TN = ( 1 - prevalence ) x ( specificity)

    = ( 1 - 0.01 ) x (0.99) = 0.9801
        FN = prevalence x ( 1 - sensitivity )
    = 0.01 x ( 1 - .98 ) = 0.01 x 0.02 = 0.0002

    PVN = 0.9801 / ( 0.9801 + 0.0002 ) = 99.98 %