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Math Help - T-score vs. Z-score

  1. #1
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    T-score vs. Z-score

    Random variables (RVs) X and Y represent annual incomes of well established unmarried young professional men and women. These two RVs are normally distributed and are statistically independent. The mean values of X and Y are $100K and $80K with standard deviations of $20K and $15K respectively. Here K represents units of thousands.


    a.A random sample of 25 professional men is taken to find their average income. What is the probability that this mean income of the sample is less than $90K. For professional men, the population standard deviation given is considered reliable. If this event does occur, what conclusion can you draw?


    b.A random sample of 20 professional women is taken to find their average income. However, the population standard deviation is considered unreliable and is estimated for this sample to be $22K. What is the probability that this mean income of the sample is more than $91.3932K. If this event does occur, what conclusion can you draw?


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    In the answers, a) uses the Z-score and b) uses the t-score. I'm confused because I thought the rule of thumb was for samples under 30 that the t-score should be used. Can someone further explain this to me please?
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  2. #2
    MHF Contributor matheagle's Avatar
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    People get this confused all the time.

    You use Z, when you have \sigma

    If the underlying population is normal, i.e.,
    X_1,...,X_n i.i.d\sim N(\mu ,\sigma^2), then

    {\bar X-\mu\over \sigma/\sqrt{n}}\sim N(0,1)

    BUT in real life, how would we know the st deviation when we usually don't know the \mu.

    So if we estimate \sigma with S and ....
    the underlying population is normal, i.e.,
    X_1,...,X_n i.i.d\sim N(\mu ,\sigma^2), then

    {\bar X-\mu\over s/\sqrt{n}}\sim t_{n-1}

    ---------------------------------------------------------------------
    NOW for the part of having a large sample.
    This n larger than 30 I assume is from the Berry-Esseen Theorem, which
    tells you how close you are to a normal distribution.
    http://en.wikipedia.org/wiki/Berry%E...Esseen_theorem

    If you have i.i.d random variables, then both

    {\bar X-\mu\over \sigma/\sqrt{n}} and {\bar X-\mu\over s/\sqrt{n}}

    approach a standard normal random variable as n goes to INFINITY (CLT).
    Now, we never sample an infinite number of items, but the larger the better.

    Note that a t density approaches a normal as n goes to infinity,
    so it doesn't make much difference which table you use for very large n.
    Last edited by matheagle; December 7th 2009 at 12:07 AM.
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  3. #3
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    but for b) we know the standard deviation to be $22k but it still uses the t-score in the answer.

    Ps. thanks for answering so fast!
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  4. #4
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    Quote Originally Posted by hello_math_ View Post
    but for b) we know the standard deviation to be $22k but it still uses the t-score in the answer.

    Ps. thanks for answering so fast!
    *Ahem* .....
    However, the population standard deviation is considered unreliable and is estimated for this sample to be $22K.
    How was it estimated I wonder ....?
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