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Math Help - df..stats

  1. #1
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    df..stats

    n1=28
    mean1=72.75
    SD1=5.37225

    n1=24
    mean1=72.625
    SD1=7.69987

    Im finding a 90% confidence interval of the difference.

    What is the df?
    What is the t*?
    I dont get it.
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  2. #2
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    Quote Originally Posted by Morgan82 View Post
    n1=28
    mean1=72.75
    SD1=5.37225

    n1=24
    mean1=72.625
    SD1=7.69987

    Im finding a 90% confidence interval of the difference.

    What is the df?
    What is the t*?
    I dont get it.
    Are the samples independent? I'll assume that they are.

    The samples are small size so the t-distribution should be used. Since s_1^2 and s_2^2 do NOT differ by an order of magnitude, it can be assumed that df = n_1 + n_2 - 2 = 50.

    \hat{\delta} = \mu_1 - \mu_2 = 0.125.

    se(\hat{\delta}) = \sqrt{\frac{5.37225^2}{28} + \frac{7.69987^2}{24}} = ......

    Two tailed therefore require t_{0.05} = 1.676 (refer table given at Student's t-distribution - Wikipedia, the free encyclopedia or use the calculator found at http://www.anu.edu.au/nceph/surfstat...e/tables/t.php).

    The confidence interval is \hat{\delta} \pm t_{\alpha} \, se(\hat{\delta}) = .....

    Since the degree of freedom is so large you could use the z-distribution without too much error.
    Last edited by mr fantastic; April 14th 2008 at 06:29 PM. Reason: Added the critical t-value calculator hyperlink
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Are the samples independent? I'll assume that they are.

    The samples are small size so the t-distribution should be used. Since s_1^2 and s_2^2 do NOT differ by an order of magnitude, it can be assumed that df = n_1 + n_2 - 2 = 50.

    \hat{\delta} = \mu_1 - \mu_2 = 0.125.

    se(\hat{\delta}) = \sqrt{\frac{5.37225^2}{28} + \frac{7.69987^2}{24}} = ......

    Two tailed therefore require t_{0.05} = 1.676 (refer table given at Student's t-distribution - Wikipedia, the free encyclopedia).

    The confidence interval is \hat{\delta} \pm t_{\alpha} \, se(\hat{\delta}) = .....

    Since the degree of freedom is so large you could use the z-distribution without too much error.

    Thanks alot! I really appreciate your help! I think i got it!
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