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  1. #1
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    inferential statistics




    In a recent national survey, the mean weekly allowance for a nine year old child from his parents was reported to be 3.65. A random sample of 45 nine year olds in norwestern Ohio revealed the mean allowance to be 3.69 with a standard deviation of .24. At the .05 level of significance, is there a difference in the mean allowances nationally and the mean allowances in northwestern Ohio for nine-year olds?

    I think you are supposed to do a t test but I am not sure how to go about it.

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    Hi

    You have to do a z-test given the sample size is above 30.

    You first state the null hypothesis which states that the mean allowance = 3.65. You then stae the alternative hypothesis which states that mean allowance is not equal to 3.65. Hence, you will do a two-tailed or sided z-test.

    The formula =

    z = (M - X)/sd/sq(n)

    M = Population mean = 3.65
    X = Sample mean = 3.69
    sd = Standard deviation = .24
    sq(n) = square root of N

    You will get a calculated z-value. You must see whether it falls within -1.96 to +1.96. If it does, you say that the means do not differ from each other. But if the value is outside this range, you say that means differs.

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    Dr. D

    so does this turn out to be 3.65-3.69/.24/.49

    then I get -.04/.24/.49 = .04/.49 = 0.08

    I still feel like I've done something wrong

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    Quote Originally Posted by dbardy View Post
    Dr. D

    so does this turn out to be 3.65-3.69/.24/.49

    then I get -.04/.24/.49 = .04/.49 = 0.08

    I still feel like I've done something wrong
    -.04/.24/.49 = .04/.49 = 0.08

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    I am sorry it should be


    X - M (and not M -X). So it should be

    3.69 - 3.65 = .04

    sd/sq(n) = .24/6.71 = .04

    Sq(n) = Sq(45) = 6.71 (I am not sure how you got .49...I think you made a mistake there).


    .04/.04 = 1

    Since 1 lies within -1.96 and +1.96, you must accept the null hypothesis, and say that there is no difference between the means

    That is the answer I am 110% sure...



    M = Population mean = 3.65
    X = Sample mean = 3.69
    sd = Standard deviation = .24
    sq(n) = square root of N

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    Just wanted to make a couple points here.

    1) We don't know for sure that the data is normally distributed. though I'm sure you were told that if the sample size is >30 you can assume the sample mean is normally distributed (this isn't always true). I can easily make a distribution that if you sampled 1,000,000 times and calculated the mean it wouldn't be normally distributed.

    2) the s.d. given is for the sample not the population. So if we assume normality we should use the t-test not the z-test

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    I am sorry but based on large sample theory, if the n > 30 (and even if population standard deviation is unknown), you use z-test.

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    Quote Originally Posted by Dr.D View Post
    I am sorry but based on large sample theory, if the n > 30 (and even if population standard deviation is unknown), you use z-test.
    That is not true.

    You can do that if you want and you may be "correct" for some intro stat class, but in reality you are wrong!

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    Well we all know that even though a sample size is larger than 30, it doesnt necessarily means that the data are normally distributed. But the theory of central limit theorem is what the dbardy is following in this assignment. It is has its flaws but if he wants to get his marks then he uses the z-test.

    It is not always fair in academics and stats,

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    Quote Originally Posted by Dr.D View Post
    I am sorry but based on large sample theory, if the n > 30 (and even if population standard deviation is unknown), you use z-test.
    See "Normal Approximation to the Binomial Distribution"

    http://www.ds.unifi.it/VL/VL_EN/sample/sample5.html

    "The rule of thumb is that n should be large enough for np >= 5 and n(1 - p) >= 5"



    Highly skewed distributions are a problem with the CLT

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    Quote Originally Posted by Dr.D View Post
    Well we all know that even though a sample size is larger than 30, it doesnt necessarily means that the data are normally distributed. But the theory of central limit theorem is what the dbardy is following in this assignment. It is has its flaws but if he wants to get his marks then he uses the z-test.

    It is not always fair in academics and stats,

    We don't know sigma...we only know the sample s.d. and the sample size it not large so we should at least use a t-test. A t table in one of my books goes up to a sample size of 120.

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    Based on sample size being larger than 30, we assum central limit theorem

    Hence we use the z-test

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    Read about central limit theorem

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    Quote Originally Posted by Dr.D View Post
    Based on sample size being larger than 30, we assum central limit theorem

    Hence we use the z-test


    Quote Originally Posted by Dr.D View Post
    Read about central limit theorem
    Let S_n=X_1+...+X_n

    The simplest CLT states that if X_1,X_2,... are iid with E(X_i)=mu and Var(X_i)=sigma^2 then


    (S_n-n*mu)/(sigma*sqrt(n))

    converges in distribution to a standard normal random variable.


    here mu and sigma are given!

    in the problem sigma is not given. therefore, even if we assume normality from the beginning we can't use a z-test because we are using an estimate for sigma. The sample size is small enough that we should use a t-test


    looking at the tables...

    at .05 the value of x such that P(X<x)=.95

    x= 1.645 for normal (known sigma)

    x= 1.671 for normal (unknown sigma sample size 60)

    they are different by quite a bit

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    Listen carefully,

    Based on central limit theorem, 30 or more in sample size allows the sample estimate of sigma to represent the population sigma. The answer is the z-test. Please do not misinform dbardy.

    See
    http://www.sixsigmafirst.com/sampling.htm for a review of this

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