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CI help

  1. Jul 31, 2012 #1
    Your first assignment at El Smello Perfume Co. is to determine the proper setting for filling the bottles. The company wishes to keep profits high by using as little perfume as possible, but a government regulation states that no more than 2.5% of the bottles may contain less than the stated volume on the bottle. If the bottles you are filling are stated to contain 100 mL, what is the minimum volume you should use for filling if the uncertainty in the bottle wall thickness is ± 1.2 mL and the uncertainty in the fill level is ± 1.6 mL. (Hint: assume an infinite number of samples where the CI is X ± st and t= 1.96 for 95% confidence and 2.58 for 99% confidence).

    heres my attempt.
    CI=x-st for the lower bound since it asks for minimum V?
    set CI = 100 and find X
    s not sure if its just 1.2 + 1.6 or sqrt(1.2^2+1.6^2)
    t not sure if i just use 95% or interpolate for 97.5

    let me know whats the correct way? thanks
     
  2. jcsd
  3. Jul 31, 2012 #2

    chiro

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    Science Advisor

    Hey blehxpo and welcome to the forums.

    You had the right idea with 1.6^2 + 1.2^2: the reason for using this is because you are incorporating two kinds of variances: the wall thickness variance and the fill variance. If we assume that these are the only sources of variation, then we use the fact that Var[X+Y] = Var[X] + Var[Y] (if X and Y are independent, which we assume) and the reason you square is because we assume that that the uncertainties correspond to standard deviations and if this is the case we square each to get the variance.

    So now we use the fact the normal distribution is symmetric around the mean (and we need to find our new mean). We know that P(X - 1.96*SQRT(1.6^2 + 1.2^2) < 100) = 0.025 (this is given by the regulation). Now we can use P(X < 100 + 1.96*SQRT(1.6^2 + 1.2^2)) = 0.025. Now to turn into standard normal we subtract the population mean mu and divide by the total standard deviation which is SQRT(1.2^2 + 1.6^2) giving us:

    P((X - mu)/SQRT(1.6^2 + 1.2^2) < (100 - mu)/SQRT(1.2^2 + 1.6^2)) = P(Z < (100-mu)/SQRT(1.6^2 + 1.2^2)) = 0.025. We know that (100-mu)/SQRT(1.6^2 + 1.2^2) = -1.96, so we solve for mu which gives us:

    100 - mu = -1.96*SQRT(1.2^2 + 1.6^2) which gives us
    mu = 100 + 1.96*SQRT(1.2^2 + 1.6^2)

    The above assumes that the uncertainty for the fill and the bottle width is also normal: if it's not, we can't use this.

    You can see that the mean has increased and you would expect this since we need to have a lower failure of error which means we need to raise the filling level until we get a 2.5 probability under the area of the curve in the < 100 area. Also the other thing is that the normal allows you to get negative fill rates, but this is probably just an exercise to get you used to statistics.
     
  4. Jul 31, 2012 #3
    Thanks for the help and detailed explanation.
    Thanks!
     
  5. Jul 31, 2012 #4

    Stephen Tashi

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    Science Advisor

    Is this problem assuming that volume = fill level + wall thickness? What shape bottle has that property?
     
  6. Aug 8, 2012 #5
    Hi i found this question that i cant the answer in the way u said is incorrect
    can u help me in finding the answer ?
    here is two question
    1. for the coin that flirt 10000 times with 4950 heads what is the 90% confidence interval for the probeblity of coin landing on heads

    2. what is the 95% of confidence interval of the following data ?
    0.79
    0.70
    0.73
    0.66
    0.65
    0.70
    0.74
    0.81
    0.71
    0.70

    tnx for your help
     
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