Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

Confidence intervals - what would the required sample size be?

Update:

Assume that you have a 95% confidence intervaI for a popuIation mean based on a sample size of 60. lf you wish to have a confidence intervaI of the same confidence IeveI but with a Iength which is one fifth of the one you aIready have, then what would the required sampIe size be?

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  • Alan
    Lv 7
    6 months ago
    Favorite Answer

    CI_low = mean - Z_critical* standard deviation/sqrt(N)

    CI_High = mean + Z_critical*standard deviation/sqrt(N)

    length of Interval = 2*Z_critical*standard deviation/sqrt(N)

    original length = 2*Z_critical*standard deviation/sqrt(60)

    so 2 , Z_critical, and standard deviation don't change

    2*Z_critical*standard deviation is a constant

    (1/5)*k/sqrt(60) = k/sqrt(new size)

    1/5sqrt(60) = 1/sqrt(new_size)

    multiply both side by sqrt(new_size)

    sqrt(new_size)/5*sqrt(60) = 1

    multiply both sides by 5*sqrt(60)

    sqrt(New_size) = 5*sqrt(60)

    square both sides

    New_size = 25*60 = 1500

    sample size = 1500

    ----- checking

    old size =k/sqrt(60) = k /0.129099445

    new_size = k/sqrt(1500) = k /0.025819889

    new_size/old size= sqrt(60)/sqrt(1500) = sqrt(1)/sqrt(25) = 1/5

    standard deviation/sqrt(60)

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  • Anonymous
    6 months ago

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