standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and

Discussion in 'Calculator Requests' started by jaffar, Nov 14, 2016.

  1. jaffar

    jaffar New Member

    standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128
     

  2. math_celebrity

    math_celebrity Administrator Staff Member

    Standard Error (margin of Error) = Standard Deviation / sqrt(n)
    128 = 545/sqrt(n)

    Cross multiply:
    128sqrt(n) = 545

    Divide by 128
    sqrt(n) = 4.2578125

    Square both sides:
    n = 18.1289672852 But we need an integer, so the answer is 19
     

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