Statistics with Probability

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  • FinalReview 6
  • #12964

    Ezra Halleck

    6. A manufacturer wants to know if an experimental version of its child toothpaste works better to prevent tooth decay than the existing formula. For children using its existing formula, cavities per year are normal with mean 3 and standard deviation 1. A study of 900 children using the new version found an average of 2.95 cavities.
    *Can we, at the 5 percent level of significance, establish that the new version is better? (H0:μ≥3, we assume only as good or perhaps even worse than present formulation)

    (a) Calculate the standard deviation for the sample mean.
    (b) Draw the normal curve and label sample mean, mean plus/minus SD, mean plus/minus 2SD and the average value. Sketch in appropriate tail.
    (c) Normalize and add Z axis to the curve from b).
    (d) Write down but do not evaluate the appropriate Excel command.
    (e) The answer to (d) turns out to be 6.7%. Please use to answer the question*.



    a) I calculated 1/√900 = 0.03333
    c) Normalized TS = (2.95-3)/0.03333 = -1.5
    d) P = NORMSDIST(-1.5) = 6.6%



    u=3, x-bar=2.95, n=900, SD=1

    SD[x-bar]= 1/Rad(900)= .03

    (2.95-3)/.03 = -1.5


    This value can be resolved at randomness due to we don’t have enough evidence to reject it.


    given :

    u<3 u >3 a=1 n = 900

    SD(xbar) = 1/sqrt (900) = 1/30

    xbar 2.95

    z= (2.95-3)/(1/30) = -0.05/(1/30) = -1.5



    H0 less than μ = 3 S=1 n = 900 [ it is a 1 tail problem]

    SD(xbar) = s/sqrtn = 1/sqrt(900)=.03

    [x bar ]=2.95

    z= (2.95-3)/(1/30) = 0.05/(1/30) = 1.5
    [ i think we have to take absolute value of it other way i am getting different answer]

    so exle commend will be : =1-normsdist(1.5)=0.0668=6.7%.

    which is more than 5% so we can not establish the advertisement claim.


    Ezra Halleck

    Very good work everyone. The Null hypothesis is given as an inequality so it is a one-tailed problem. Therefore (rsanchez), there should not be a factor of 2 in front of normsdist.

    By the way, the null hypothesis was originally stated incorrectly. It should be (and now is) H0:mu>=3.



    For this problem, do we use 1-normsdist(1.5) or just normsdist(1.5) or does it not matter which one we use?


    Ezra Halleck

    In this problem, we are working with the left tail. Since normsdist gives you the area up until a certain point, you can use normsdist directly: normsdist(-1.5)

    If you decide to use the corresponding right tail instead, then it is the complement: 1-normsdist(1.5).



    how do you know the tail is on left side or right side?
    if the Ho is equality, it would be a 2 tails problem, right?



    “how do you know the tail is on left side or right side?”

    I think you know which side the tail is/are at based on the question’s H0 or the Z value, which is -1.5 for this question, so the tail must be at the left side.

    “if the Ho is equality, it would be a 2 tails problem, right?”

    ya it will be a 2 tails problem when it’s an equality, and when H0 is greater or less than, then it would be a 1 tail problem.


    Ezra Halleck

    Basically, the null hypothesis is what you want to overturn, so unless it is a trick question (which I won’t put on the final exam), the evidence will always be in the other direction. The question will always be “We have evidence, but do we have ENOUGH to overcome the null hypothesis”.




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