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Math Help - Need help calculating margin of sampling error and confidence interval!

  1. #1
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    Need help calculating margin of sampling error and confidence interval!

    Problem:
    Assume that the following random samples are taken from populations with normal distributions. In each case, find the margin of sampling error and a confidence interval for u for the indicated level.

    (a). n=12, X bar = 122, standard deviation = 25, 90% level of confidence.

    What I know:
    the confidence interval (CI) = point estimate +- margin of sampling error

    margin of sampling error = (quantile value) x (standard error of point estimator)

    standard error of point estimator = standard deviation divided by square root of n
    -or-
    CI = Point estimate +- margin of sampling error

    What I need help with:
    I have the answer, but cannot seem to arrive to it. I have tried to work this over and over again with no luck.

    According to what i know, the quantile value should .95 because (1-(.1/2)) and the standard error of point estimator should be 25 divided by square root of 12. But that is not the answer....Help!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Since n=12 is small you can't use the central limit theorem.
    But, the problem here is in your standard deviation.
    I don't know what you mean by that.
    Is that the population \sigma st deviation or the sample
    st deviation, S?

    If it's \sigma ,then the interval is

    (\bar X-z_{\alpha /2}\sigma/\sqrt{n}, \bar X+z_{\alpha /2}\sigma/\sqrt{n})

    and if you mean S, then the interval is

    (\bar X-t_{n-1,\alpha /2}S/\sqrt{n},\bar X+t_{n-1,\alpha /2}S/\sqrt{n}).
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  3. #3
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    here is what i am supposed to work with ... left side of the page
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  4. #4
    MHF Contributor matheagle's Avatar
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    that clearly says, when \sigma is known
    so use the first interval in my last post
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