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Math Help - point estimates and endpoints of confidence interval

  1. #1
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    Red face point estimates and endpoints of confidence interval

    distribution of x is N(mu,4) n=10:
    55.95
    56.54
    57.58
    55.13
    57.48
    56.06
    59.93
    58.30
    52.57
    58.46

    point estimate?
    endpoints for 95% confidence interval for mu?
    probability when one is selected at random that it is less than 52?

    i'm lost and would loooove some help..... thankyou
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  2. #2
    Super Member Random Variable's Avatar
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     \bar{x} \pm z_{0.025} \ \Big( \frac {\sigma}{\sqrt{n}}\Big)

    where \bar{x} is the sample mean (which you have to calculate), \sigma is the standard deviation, and  n is the sample size
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  3. #3
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    i got a mean of 56.83, a variance of 3.814 and a standard deviation of 1.95295 but don't know what to do with them. where do you get the z0.025 from? thankyou for helping me!
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  4. #4
    Super Member Random Variable's Avatar
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     \sigma is the standard deviation of the population, not the sample. So in this problem  \sigma = \sqrt{4} = 2 .

    You get z_{0.025} from a standard normal distribution table. There is one in just about every statistics textbook. It's the value of z such that P(Z>z) = 0.025. The value is 1.96.

    Have you gone over this stuff in class?
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  5. #5
    MHF Contributor matheagle's Avatar
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    Chrissy, plug in 1.96 into Free One Tailed Area Under the Standard Normal Curve Calculator
    and see that the area to the right of 1.96 is .0249978.
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  6. #6
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    All I have is a textbook. No class. Would the average of the 10 values be considered a point estimate of \mu? Also, if the area to the right of 1.96 is 0.249978, would the endpoints of the 95% confidence interval be 1.9849978 on the right and 1.96 or 1.9350022 on the left?

    Thanks. I really don't know what I am doing and my text is no help. I need an actual example to learn from but it is very hard to find. I don't have enough calculus OR statistic background to solve these problems without a better book OR an actual teacher. Any tutors out there? =)
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  7. #7
    Super Member Random Variable's Avatar
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    From the data you calculated the sample mean ( \bar{x}) to be 56.83.

    So a 95% confidence inteveral for the population mean (  \mu ) is

     \Big[56.83 - 1.96* \frac{2}{\sqrt{10}}, \ 56.83 + 1.96*\frac{2}{\sqrt{10}}\Big]

    [55.59, 58.07]

    Now either this interval contains  \mu or it doesn't. But if you constructed many such intervals by taking different samples of the same size, 95% of them would contain  \mu .
    Last edited by Random Variable; June 13th 2009 at 02:58 PM.
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  8. #8
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    you are wonderful! Thankyou. So would that in fact be considered a point estimate (56.83)? I'm not clear on exactly what that is. My book only mentions that it is a point rather than an interval, but I don't know which point. Also, would the probability of randomly choosing a variable less than 52 be simply 0? Thanks once again! you have made this so much easier to understand.
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  9. #9
    Super Member Random Variable's Avatar
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    EDIT:

    If X is N(\mu, \sigma^{2}), then  \frac {X-\mu}{\sigma} is  N(0,1) (or just Z) .

    So  P(X<52) = P\Big(\frac{X -\mu}{\sigma} < \frac{52 - \mu}{2} \Big) = P(Z<?)

    But that can't be answered unless you know  \mu
    Last edited by Random Variable; June 13th 2009 at 04:14 PM.
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  10. #10
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    I feel so silly, I think I can just do the old (52-56.83)/2=-2.415, but 2.415 is not on my table so 2.42 becomes 0.0078. I'm not sure how to write that out. Is it \Phi2.42=0.0078?
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  11. #11
    MHF Contributor matheagle's Avatar
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    you can plug 2.415 into....
    Free One Tailed Area Under the Standard Normal Curve Calculator
    and get .0078676
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  12. #12
    Super Member Random Variable's Avatar
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    But I'm fairly certain (which probably means I'm wrong) that  \frac {X-\bar{x}}{\sigma} is not N(0,1).
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  13. #13
    MHF Contributor matheagle's Avatar
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    It has mean zero and it is a normal.
    I doubt the variance is 1.
    Is this X part of the sample, where \bar X was derived from or a different observaton?

    If it's part of the sample, then ....

    V\biggl({X_1-\bar X\over \sigma}\biggr)={V\biggl(X_1-\bar X\biggr) \over \sigma^2}

    ={V\biggl(X_1\bigl(1-{1\over n}\bigr)-{\sum_{k=2}^n X_k\over n}\biggr) \over \sigma^2}

    ={\bigl(1-{1\over n}\bigr)^2\sigma^2 +{n-1\over n^2}\sigma^2 \over \sigma^2}

     =1-{1\over n}

    Otherwise ...

    V\biggl({X_{n+1}-\bar X\over \sigma}\biggr)={\sigma^2+{\sigma^2\over n} \over \sigma^2}=1+{1\over n}

    which is exactly what you get in a prediction interval, inside the square root of course.
    Last edited by matheagle; June 14th 2009 at 08:30 PM.
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