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Estimation - Student-t distribution - Confidence Level

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data

    A company produces batches of 1800 axle shafts . These are tested to determine the proportion of those with too rough surface according to the standards set in the industry.

    The quality control department of a sample of 150 axle shafts in a lot and concluded that there are between 15 % and 25 % of the axle shafts in this set that are outside the norms . Calculate the confidence level associated with this estimate.



    2. Relevant equations
    tn-1,sigma/2=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
    (1-sigma)=tcdf(-tn-1,sigma/2,+tn-1,sigma/2,n-1)

    3. The attempt at a solution
    Known variable:
    N = 1800
    n=150
    [15%,25%] ---> average= (15+25)/2 =20% or 0.20
    ME = (25-15)/2 = 5% or 0.05

    Unknown variable:
    (1-sigma)
    Average of N
    Standard Deviation of n


    I am stuck here, I cannot continue without knowing SD.
     
  2. jcsd
  3. Mar 6, 2016 #2

    RUber

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    Homework Helper

    This is a proportion. So, standard deviation of a proportion is
    ##\sqrt{p(1-p)}.##
     
  4. Mar 6, 2016 #3
    Hello,

    I dont understand, what would be p ?
     
  5. Mar 6, 2016 #4
    do I still use Student -t formula ?
     
  6. Mar 6, 2016 #5
    I did that:
    p=(15%+25%)/2 =0.2
    s= √p(1−p) = √0.2(1−0.2) = 0.4
    t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
    t = 0.05/((0.4/sqrt(150))* (sqrt(1-(150/1800)))
    (1-sigma)=tcdf(-t,+t,n-1)
    (1-sigma) = 1

    It does not work :(
    anyone please help me.
     
  7. Mar 7, 2016 #6

    RUber

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    Homework Helper

    I am not sure what you last line is doing.
    Normally, you would take your calculate t standard error of the proportion...in this case:
    t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N))) = .031,
    And compare that to your range ( +/- 5% ) .
    This gives you an idea of what your ##t_{\frac{\alpha}{2}, df}## should be equal to.
     
  8. Mar 7, 2016 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    Google "confidence interval for proportion".
     
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