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Math Help - the difference with T-distribution and normal

  1. #1
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    the difference with T-distribution and normal

    Hi, I'm confused, I don't know if I got the t- distribution right, and I can't fiqure out what value to use for the normal distribution table. Could someone please help me.

    Construct the indicated confidence interval for the population mean. Using (a) a t-distribution (b) if you had incorrectly used a normal distribution which interval would be wider?

    C = 0.95, x ( with the line on top) = 13.4, s= 0.85, n = 8

    This is what I got for the t-distribution interval
    3.13.4 + 2.364, is this even close?
    Thank you in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mkcar View Post
    Hi, I'm confused, I don't know if I got the t- distribution right, and I can't fiqure out what value to use for the normal distribution table. Could someone please help me.

    Construct the indicated confidence interval for the population mean. Using (a) a t-distribution (b) if you had incorrectly used a normal distribution which interval would be wider?

    C = 0.95, x ( with the line on top) = 13.4, s= 0.85, n = 8

    This is what I got for the t-distribution interval
    3.13.4 + 2.364, is this even close?
    Thank you in advance.
    I don't understand what you mean by your t-distribution interval, it should be like what follows:

    Such a confidence interval looks like:

    <br />
[\bar{x}-k~s/\sqrt{n},~ \bar{x}+k~s/\sqrt{n}]<br />

    where in one case k is determined from the table of the t-distribution with n-1 degrees of freedom, and in the other case from the normal table. These values are k\approx 2.36 for the t-distribution and k=1.96 for the normal distribution.

    The t-distribution confidence interval is always wider.

    RonL
    Last edited by CaptainBlack; August 6th 2007 at 12:56 PM. Reason: Misread number of degrees of freedom should be 7 not 84 which is what I used previously
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    I don't understand what you mean by your t-distribution interval, it should be like what follows:

    Such a confidence interval looks like:

    <br />
[\bar{x}-k~s/\sqrt{n},~ \bar{x}+k~s/\sqrt{n}]<br />

    where in one case k is determined from the table of the t-distribution with n-1 degrees of freedom, and in the other case from the normal table. These values are k\approx 2.36 for the t-distribution and k=1.96 for the normal distribution.

    The t-distribution confidence interval is always wider.

    RonL
    Putting in the numerical valuse we have:

    <br />
\left[13.4 -\frac{2.36 \times 0.85}{\sqrt{8}}, 13.4 +\frac{2.36 \times 0.85}{\sqrt{8}}\right] = [12.69, 14.11]<br />

    RonL
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