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Math Help - Type I error

  1. #1
    Lok
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    Type 1 and 2 errors

    In the problem I am given :
    <br />
H_0 : u=5 ; H_1:u\neq5
    n = 8 (sample size)

    The random variable follows a normal law with a standard deviation of 0.25
    The problem also states that we need to reject [tex]H_0[/Math] if \overline{x}<4.85 or \overline{x}>5.15.

    I need to find the probability of a type I error.

    What I've tried so far: (alpha represents my Type I error)
    \alpha=P(reject H_0|H_0 true)<br />
=P(\overline{x}<4.85 or \overline{x}>5.15 | u=5)

    So I tried to start and find the probability of \overline{x} not in that interval.
    \overline{x} \not \in [u-0.15;u+0.15]
    [u-\frac{s*Z_{\alpha/2}}{\sqrt{n}},u+\frac{s*Z_{\alpha/2}}{\sqrt{n}}]
    which gives me \frac{s*Z_{\alpha/2}}{\sqrt{n}} = 0.15
    when I plug my numbers and isolate Z_{\alpha/2}, i get Z_{\alpha/2}=0.5*\sqrt{8}/0.25^2=22.62<br />
    It seems to me that this value of Z can't be right since my table only goes to a maximum of 3.09.
    As for the rest of the problem, I could use a tip or two about how to find P(u=5) and how to calculate the conditionnal probability.

    Thanks
    Last edited by Lok; April 13th 2009 at 11:35 PM. Reason: changed the title by adding type 2 error
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  2. #2
    MHF Contributor matheagle's Avatar
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    alpha is the PROBABILTY of the type 1 error, it's not the error.

     P(\bar X>5.15)=P(Z>{5.15-5\over .25/\sqrt{8}}) = P(Z>{3\sqrt{8}\over 5})

    Likewise
     P(\bar X<4.85)= P(Z<-{3\sqrt{8}\over 5}) , which is the same as above.

    I am assuming that .25 is the POPULATION st deviation. If it's the sample st dev., we will have a t distribution with 7 degrees of freedom.
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  3. #3
    Lok
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    Ok wow I had it wrong, thanks!

    (you are correct, it's the standard deviation for the population)
    So I end up with P(Z< -1,697) or P(Z>1.697)
    which is a probability of 0.0448. Now I guess I add those 2 probabilities and I get alpha? Which is 0.0896 or 8,96%, right?
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  4. #4
    MHF Contributor matheagle's Avatar
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    yes, via a table
    you can always get a better approximation of these distributions via an online calculator...
    Free Two Tailed Area Under the Standard Normal Curve Calculator
    They give me... .0896966
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  5. #5
    Lok
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    Ok that's great.

    For the same problem I also need to find the probability of a type 2 error if \mu is 5.1 (I'll use B to represent the probability of this error)

    So I know that B(5.1)=P{accept H_0|H_0 is false}

    P(Z>\frac{4.85-5.1}{0.25/\sqrt{8}}) and P(Z<\frac{5.15-5.1}{0.25/\sqrt{8}})

    so I end up with P(Z>-2.828) and P(Z<0.567)
    that's

    (the regions -2.828 to 0 and 0 to 0.567 in the graph of the std normal law)
    so P=0.9977-0.5 and 0.7146-0.5
    gives me : P=0.4977+0.2146 = 0.7123

    This probability seems very high (71.23%) so I'm wondering if I did everything correctly?
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  6. #6
    MHF Contributor matheagle's Avatar
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    tex errors

    Quote Originally Posted by Lok View Post
    Ok that's great.

    For the same problem I also need to find the probability of a type 2 error if mu is 5.1 (I'll use B to represent the probability of this error)

    So I know that B(5.1)=P{accept H_0|H_0 is false}

    P(Z>{4.85-5.1\over 0.25/\sqrt{8}}) + P(Z<{5.15-5.1\over 0.25/\sqrt{8}})

    so I end up with P(Z>-2.828) and P(Z<0.567)
    that's

    (-2.828 to 0 and 0 to 0.567)
    so 0.9977-0.5 and 0.7146-0.5
    gives me : 0.4977+0.2146 = 0.7123

    This probability seems very high (71.23%) so I'm wondering if I did everything correctly?
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  7. #7
    MHF Contributor matheagle's Avatar
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    Use P(4.85<\bar X<5.15)=P\biggl({4.85-5.1\over 0.25/\sqrt{8}}<Z<{5.15-5.1\over 0.25/\sqrt{8}}\biggr) instead.
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  8. #8
    Lok
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    Yea, that's what I have (I think).
    Is it normal that the probability to make an error of type 2 be that high? (71.23%, I think my numbers are correct)
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  9. #9
    MHF Contributor matheagle's Avatar
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    it's high because 5.1 is near 5
    the closer you get to 5 the closer beta will be to 1-alpha
    so the beta at 5.00001 will be near .91
    GOOD NIGHT!!! it's 3am
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  10. #10
    Lok
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    Ok! Thanks a lot (:

    Good night.
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