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Math Help - [SOLVED] Large samples confidence interval for difference in means

  1. #1
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    [SOLVED] Large samples confidence interval for difference in means

    The following distinguishes TWO cases for large samples confidence interval for difference in means:


    where Sp^2 is the pooled estimate of the common variance, n1 is the sample size from the first population, n2 is the sample size from the second population, and z_alpha/2 is 100(1-alpha/2) th percentile of the standard normal.
    ==========================

    It seems to me that case 1 is a special case of case 2 with the population variances being equal. If this is the case, the formula for case 2 should reduce to the formula for case 1 when the population variances are equal. However, I have no way of seeing it being the case.
    [aside: I am trying to cut down on the number of formulas that I have to memorize. Instead of two different formulas, if case 2 contains case 1, then I only have to memorize the general case 2 formula which is nice.]

    Could somebody please show me how I can reduce case 2 to case 1?
    Any help would be appreciated!
    Last edited by kingwinner; February 20th 2009 at 10:40 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kingwinner View Post
    The following distinguishes TWO cases for large samples confidence interval for difference in means:


    where Sp^2 is the pooled estimate of the common variance, n1 is the sample size from the first population, n2 is the sample size from the second population, and z_alpha/2 is 100(1-alpha/2) th percentile of the standard normal.
    ==========================

    It seems to me that case 1 is a special case of case 2 with the population variances being equal. If this is the case, the formula for case 2 should reduce to the formula for case 1 when the population variances are equal. However, I have no way of seeing it being the case.
    [aside: I am trying to cut down on the number of formulas that I have to memorize. Instead of two different formulas, if case 2 contains case 1, then I only have to memorize the general case 2 formula which is nice.]

    Could somebody please show me how I can reduce case 2 to case 1.
    Any help would be appreciated!
    The two cases are not equivalent. In one case you have additional information, you know the variances are equal, so the pooled variance is the best estimate of the common variance. In the second case they just happen to be the same but you don't know that so you have to estimate both separately.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The two cases are not equivalent. In one case you have additional information, you know the variances are equal, so the pooled variance is the best estimate of the common variance. In the second case they just happen to be the same but you don't know that so you have to estimate both separately.

    CB
    I don't get your point...so for the second case, is it only for population variance unknown and unequal? (so that case 1 and case 2 are mutually exclusive?)
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  4. #4
    MHF Contributor matheagle's Avatar
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    Honestly, as both samples sizes go to infinity, the Central Limit Theorem
    makes both approximations normal.
    The point of the pooled estimator is the Chi-Square distribution you have in the denominatorr of the t distribution.
    (But here you must assume normality of the two independent samples, besides equality of the population variances.)
    That's where you really see the point of using S_p^2, when the n's are small.
    This pooled variance is a weighted average of the two sample variances.
    S_p^2= [(n_1-1)/(n_1+n_2-2)]S_1^2 + [(n_2-1)/(n_1+n_2-2)]S_2^2.
    So if the two sample sizes are equal
    S_p^2=(S_1^2 + S_2^2)/2.
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  5. #5
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    Quote Originally Posted by kingwinner View Post
    I don't get your point...so for the second case, is it only for population variance unknown and unequal? (so that case 1 and case 2 are mutually exclusive?)
    Case 1:
    The variance are unknown and not-known to be equal.

    Case 2:
    The variances are unknown but known to be equal.

    There is no case of unknown variances known to be unequal (unless you want to develop that theory).

    Also see what matheagle posted, they are asymtotically equivalent (when the variances are or just happen to be equal), which is all you need for large sample statistics.

    CB
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  6. #6
    MHF Contributor matheagle's Avatar
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    Satterthwaite has an approximation when
    unequal sample sizes, unequal variance
    which can be found in...
    Student's t-test - Wikipedia, the free encyclopedia
    this is really for small sample situations.
    When the n or n's are large you can do anything
    The CLT saves your butt, that and Slutsky's theorem.
    What's interesting is that if Sigma_1=c times Sigma_2,
    then we can obtain a similar result of the pooling situation.
    That can be found in a lot of textbooks, usually as a homework problem.
    It can be proved by the fact that adding two indep Chi-Squares we get another Chi-Square rv.
    Then using that in the denominator of our T stat.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    Satterthwaite has an approximation when
    unequal sample sizes, unequal variance
    which can be found in...
    Student's t-test - Wikipedia, the free encyclopedia
    this is really for small sample situations.
    When the n or n's are large you can do anything
    The CLT saves your butt, that and Slutsky's theorem.
    What's interesting is that if Sigma_1=c times Sigma_2,
    then we can obtain a similar result of the pooling situation.
    That can be found in a lot of textbooks, usually as a homework problem.
    It can be proved by the fact that adding two indep Chi-Squares we get another Chi-Square rv.
    Then using that in the denominator of our T stat.
    Since what we are calculating is an estimate of the variance of the difference of the two sample means almost any known relationship between the variances of the two processes will allow us (if we wish to make the effort) to develop a slightly better estimate of this variance that that used in the unknown variances case.

    CB
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  8. #8
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    Problem resolved! Thank you!
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