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Homework Help: Confidence Level for Proportion

  1. Mar 13, 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.

    A) 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.

    B) The Company would estimate at 3.0 % near the proportion of axle shafts located outside the standards , 19 times out of 20. How many additional axle shafts should you consider it ? You must take into account the results obtained in A) .

    2. Relevant equations
    (ME)/(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) = Zalpha/2

    3. The attempt at a solution

    Known variable:
    N = 1800
    [15%,25%] ---> P= (15+25)/2 =20% or 0.20
    ME = (25-15)/2 = 5% or 0.05

    (ME)/(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) = Zalpha/2
    (0.05)/(sqrt((0.2*(1-0.2))/n)*sqrt(1-(150/1800)) = Zalpha/2
    Zalpha/2 = 1.59901

    normcdf(-1.59901,1.59901) = 0.89 or 89%

    This is correct answer for A) checked in book.

    Now for part B)
    ME = 3% = 0.03
    19 times out of 20= 95% = 1-alpha

    [p+invnorm(0.025)*(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) , p+invnorm(1-0.025)*(sqrt((P*(1-P))/n)*sqrt(1-(n/N))]
    [0.2+invnorm(0.025)*(sqrt((0.5*(1-0.5))/150)*sqrt(1-(150/1900)) , 0.2+invnorm(1-0.025)*(sqrt((0.5*(1-0.5))/150)*sqrt(1-(150/1800))]


    solve(invnorm(1-0.025)*(sqrt((0.2766*(1-0.2766))/n)*sqrt(1-(n/1800))=0.03) for n
    n = 579.225

    1800-580 = 1220 addition needs.

    I am doing something wrong for the part B).
    Please point me to the right direction, maybe my P that I have is not correct.
  2. jcsd
  3. Mar 14, 2016 #2
    anyone can help me :(
  4. Mar 14, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I cannot understand what part B is is asking, because I cannot understand what it is saying. Are you asking for the sample size that would allow the firm to estimate (with probability >= 95%) the percentage of items outside the standard to an accuracy of within ± 3 percentage points--that is, an interval (p - 3/100, p + 3/100)--or are you asking for an estimate accurate to within a 3% error--that is, an interval (0.97*p, 1.03*p)? These are two very different concepts.
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