Statistics with Probability

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  • #12964

    Ezra Halleck
    Participant

    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*.

    #16174

    Anonymous
    Inactive

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

    #16218

    rsanchez
    Member

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

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

    (2.95-3)/.03 = -1.5

    z(normsdist(-1.5))

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

    #16223

    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

    #16227

    Afzal
    Participant

    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.

    #16228

    Ezra Halleck
    Participant

    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.

    #16253

    Sha-Kenya
    Member

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

    #16229

    Ezra Halleck
    Participant

    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).

    #16254

    yuxiaolin
    Member

    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?

    #16261

    wantong
    Member

    “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.

    #16264

    Ezra Halleck
    Participant

    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”.

    #16284

    yuxiaolin
    Member

    thanks!

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